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Evaluating $int_0^{pi/2}operatorname{arcsinh}(2tan x)dx$
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How to prove $$int_0^{pi/2}operatorname{arcsinh}(2tan x)dx=frac43G+frac13pilnleft(2+sqrt3right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $Imoperatorname{Li}_2left(2pmsqrt3right)$.
Attempt
$$int_0^{pi/2}operatorname{arcsinh}(2tan x)dx\=int_0^inftyfrac{operatorname{arcsinh}(2x)}{1+x^2}dx\
=2int_0^inftyfrac{xcosh x}{4+sinh^2x}dx\
=2int_0^inftyfrac{xcosh x}{3+cosh^2x}dx\
=2int_0^inftysum_{n=0}^infty x(-3)^ncosh^{-2n-1}(x)dx$$
I failed to integrate $xcosh^{-2n-1}(x)$. Mathematica returns a hypergeometric term while integrating it.
calculus integration definite-integrals polylogarithm
asked Jan 22 at 8:15
Kemono ChenKemono Chen
3,1941844
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How to prove $$int_0^{pi/2}operatorname{arcsinh}(2tan x)dx=frac43G+frac13pilnleft(2+sqrt3right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $Imoperatorname{Li}_2left(2pmsqrt3right)$.
Attempt
$$int_0^{pi/2}operatorname{arcsinh}(2tan x)dx\=int_0^inftyfrac{operatorname{arcsinh}(2x)}{1+x^2}dx\
=2int_0^inftyfrac{xcosh x}{4+sinh^2x}dx\
=2int_0^inftyfrac{xcosh x}{3+cosh^2x}dx\
=2int_0^inftysum_{n=0}^infty x(-3)^ncosh^{-2n-1}(x)dx$$
I failed to integrate $xcosh^{-2n-1}(x)$. Mathematica returns a hypergeometric term while integrating it.
calculus integration definite-integrals polylogarithm
asked Jan 22 at 8:15
Kemono ChenKemono Chen
3,1941844
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add a comment |
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How to prove $$int_0^{pi/2}operatorname{arcsinh}(2tan x)dx=frac43G+frac13pilnleft(2+sqrt3right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $Imoperatorname{Li}_2left(2pmsqrt3right)$.
Attempt
$$int_0^{pi/2}operatorname{arcsinh}(2tan x)dx\=int_0^inftyfrac{operatorname{arcsinh}(2x)}{1+x^2}dx\
=2int_0^inftyfrac{xcosh x}{4+sinh^2x}dx\
=2int_0^inftyfrac{xcosh x}{3+cosh^2x}dx\
=2int_0^inftysum_{n=0}^infty x(-3)^ncosh^{-2n-1}(x)dx$$
I failed to integrate $xcosh^{-2n-1}(x)$. Mathematica returns a hypergeometric term while integrating it.
calculus integration definite-integrals polylogarithm
asked Jan 22 at 8:15
Kemono ChenKemono Chen
3,1941844
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How to prove $$int_0^{pi/2}operatorname{arcsinh}(2tan x)dx=frac43G+frac13pilnleft(2+sqrt3right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $Imoperatorname{Li}_2left(2pmsqrt3right)$.
Attempt
$$int_0^{pi/2}operatorname{arcsinh}(2tan x)dx\=int_0^inftyfrac{operatorname{arcsinh}(2x)}{1+x^2}dx\
=2int_0^inftyfrac{xcosh x}{4+sinh^2x}dx\
=2int_0^inftyfrac{xcosh x}{3+cosh^2x}dx\
=2int_0^inftysum_{n=0}^infty x(-3)^ncosh^{-2n-1}(x)dx$$
I failed to integrate $xcosh^{-2n-1}(x)$. Mathematica returns a hypergeometric term while integrating it.
calculus integration definite-integrals polylogarithm
calculus integration definite-integrals polylogarithm
asked Jan 22 at 8:15
Kemono ChenKemono Chen
3,1941844
asked Jan 22 at 8:15
Kemono ChenKemono Chen
3,1941844
asked Jan 22 at 8:15
Kemono ChenKemono Chen
3,1941844
asked Jan 22 at 8:15
Kemono ChenKemono Chen
3,1941844
asked Jan 22 at 8:15
Kemono ChenKemono Chen
3,1941844
3,1941844
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6 Answers
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I will begin with the same approach as in @DavidG's answer, but will settle in a different argument. Let
$$ J(t) = int_{0}^{infty} frac{operatorname{arsinh}(tx)}{1+x^2} , mathrm{d}x. $$
Our goal is to compute $J(2)$ using Feynman's trick. Differentiating $J(t)$ and substituting $x=sqrt{u^{-2}-1}$, we obtain
begin{align*}
J'(t)
= int_{0}^{infty} frac{x}{(1+x^2)sqrt{1+t^2x^2}} , mathrm{d}x
= int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}u.
end{align*}
So it follows that
begin{align*}
J(2)
&= int_{0}^{2} int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}umathrm{d}t \
&= int_{0}^{1} int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}umathrm{d}t
+ int_{1}^{2} int_{0}^{1} frac{1}{sqrt{1 + (t^2-1)(1-u^2)}} , mathrm{d}umathrm{d}t.
end{align*}
The inner integral is easily computed, yielding
begin{align*}
J(2)
&= int_{0}^{1} frac{operatorname{artanh}left( sqrt{1 - t^2} right)}{sqrt{1 - t^2}} , mathrm{d}t
+ int_{1}^{2} frac{arctanleft(sqrt{t^2-1}right)}{sqrt{t^2 - 1}} , mathrm{d}t.
end{align*}
Now we substitute $t = operatorname{sech} varphi$ for the first integral and $t = sec theta$ for the second integral. This yields
begin{align*}
J(2)
&= int_{0}^{infty} frac{varphi}{coshvarphi} , mathrm{d}varphi
+ int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta.
end{align*}
These integrals can be computed as follows:
Using $ operatorname{sech}varphi = frac{2e^{-varphi}}{1 + e^{-2varphi}} = 2 sum_{n=0}^{infty} (-1)^n e^{-(2n+1)varphi} $, we obtain
$$ int_{0}^{infty} frac{varphi}{coshvarphi} , mathrm{d}varphi
= 2 sum_{n=0}^{infty} (-1)^n int_{0}^{infty} varphi e^{-(2n+1)varphi} , mathrm{d}varphi
= 2 sum_{n=0}^{infty} frac{(-1)^n}{(2n+1)^2}
= 2G. $$
Taking integration by parts,
begin{align*}
int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta
&= left[ - theta log left( tan left( frac{pi}{4} - frac{theta}{2} right) right) right]_{0}^{frac{pi}{3}} + int_{0}^{frac{pi}{3}} log left( tan left( frac{pi}{4} - frac{theta}{2} right) right) , mathrm{d}theta \
&= frac{pi}{3} log left( 2 + sqrt{3} right) + 2 int_{frac{pi}{12}}^{frac{pi}{4}} log left( tan theta right) , mathrm{d}theta.
end{align*}
This can be computed by using the Fourier series $log left( tan theta right) = - 2 sum_{n=0}^{infty} frac{cos(4n+2)theta}{2n+1} $ to yield
begin{align*}
int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta
&= frac{pi}{3} log left( 2 + sqrt{3} right) - 2 sum_{n=0}^{infty} frac{sinleft( frac{pi}{2} (2n+1) right) - sinleft( frac{pi}{6} (2n+1) right)}{(2n+1)^2} \
&= frac{pi}{3} log left( 2 + sqrt{3} right) - frac{2}{3}G.
end{align*}
Combining two result, we obtain the desired answer.
answered Jan 22 at 11:24
Sangchul LeeSangchul Lee
95.5k12171279
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This is beauty !
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– Claude Leibovici
Jan 22 at 16:10
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On the path of Kemono Chen...
begin{align}J&=int_0^{pi/2}operatorname{arcsinh}(2tan x)dxend{align}
Perform the change of variable $y=operatorname{arcsinh}(2tan x)$,
begin{align}J&=int_0^{+infty}frac{2xcosh x}{4+sinh^2 x},dx\
&=int_0^{+infty}frac{4xleft(text{e}^{x}+text{e}^{-x}right)}{14+text{e}^{2x}+text{e}^{-2x}},dx\
&=int_0^{+infty}frac{4xtext{e}^{-x}left(text{e}^{2x}+1right)}{14+text{e}^{2x}+text{e}^{-2x}},dx\
end{align}
Perform the change of variable $y=text{e}^{-x}$,
begin{align}J&=-int_0^1 frac{4ln xleft(1+frac{1}{x^2}right)}{14+x^2+frac{1}{x^2}}\
&=-int_0^1 frac{4ln xleft(1+x^2right)}{x^4+14x^2+1}\
&=left[-arctanleft(frac{4x}{1-x^2}right)ln xright]_0^1+int_0^1 frac{arctanleft(frac{4x}{1-x^2}right)}{x},dx\
&=int_0^1 frac{arctanleft(frac{4x}{1-x^2}right)}{x},dx\
&=int_0^1 frac{arctanleft(left(2+sqrt{3}right)xright)}{x},dx+int_0^1 frac{arctanleft(left(2-sqrt{3}right)xright)}{x},dx\
end{align}
In the first integral perform the change of variable $y=left(2+sqrt{3}right)x$,
In the second integral perform the change of variable $y=left(2-sqrt{3}right)x$,
begin{align}J&=int_0^{2+sqrt{3}}frac{arctan x}{x},dx+int_0^{2-sqrt{3}}frac{arctan x}{x},dx\
&=Big[arctan xln xBig]_0^{2+sqrt{3}}-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx+Big[arctan xln xBig]_0^{2-sqrt{3}}-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{5pi}{12}lnleft(2+sqrt{3}right)-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx+frac{pi}{12}lnleft(2-sqrt{3}right)-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx
end{align}
In the first integral perform the change of variable $y=dfrac{1}{x}$,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)+int_{2-sqrt{3}}^{+infty}frac{ln x}{1+x^2},dx-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)+int_0^{+infty}frac{ln x}{1+x^2},dx-2int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)-2int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
end{align}
Perform the change of variable $y=tan x$,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)-2int_0^{frac{pi}{12}}lnleft(tan xright),dx\
end{align}
It is well known that,
begin{align}
int_0^{frac{pi}{12}}lnleft(tan xright),dx=-frac{2}{3}text{G}
end{align}
(see: Integral: $int_0^{pi/12} ln(tan x),dx$ )
Thus,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)-2times -frac{2}{3}text{G}\
&=boxed{frac{pi}{3}lnleft(2+sqrt{3}right)+frac{4}{3}text{G}}
end{align}
NB:
Observe that,
begin{align}2-sqrt{3}&=frac{1}{2+sqrt{3}}\
lnleft(2-sqrt{3}right)&=-lnleft(2+sqrt{3}right)\
int_0^infty frac{ln x}{1+x^2},dx&=0
end{align}
(perform the change of variable $y=dfrac{1}{x}$ )
answered Jan 22 at 13:35
FDPFDP
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This is beauty !
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– Claude Leibovici
Jan 22 at 16:10
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How did you do $$intarctanfrac{4x}{1-x^2}frac{mathrm dx}{x}=intarctan(2+sqrt3)xfrac{mathrm dx}{x}+intarctan(2-sqrt3)xfrac{mathrm dx}{x}$$
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– clathratus
Jan 22 at 16:10
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@Clathratus: if $ab<1$then begin{align}arctan a+arctan b=arctanleft(frac{a+b}{1-ab}right)end{align}
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– FDP
Jan 22 at 16:23
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Here is another variation along a theme which like @DavidG approach uses Feynman's trick of differentiating under the integral sign.
Let
$$I(a) = int_0^{frac{pi}{2}} operatorname{arcsinh} (a tan x) , dx, qquad a > 1.$$
We are required to find $I(2)$. We start by finding $I(1)$ first, which will be needed later on.
For $a = 1$ we have
begin{align}
I(1) &= int_0^{frac{pi}{2}} operatorname{arcsinh} (tan x) , dx\
&= int_0^{frac{pi}{2}} ln left (frac{1 + sin x}{cos x} right ) , dx\
&= int_0^{frac{pi}{2}} ln (1 + sin x) , dx - int_0^{frac{pi}{2}} ln (cos x) , dx.
end{align}
Now the first of these integrals can be found, for example, by rewriting it as
begin{align}
int_0^{frac{pi}{2}} ln (1 + sin x) , dx &= int_0^{frac{pi}{2}} ln (1 + cos x) , dx\
&= int_0^{frac{pi}{2}} ln left (frac{1}{2} cos^2 frac{x}{2} right ) , dx\
&= frac{pi}{2} ln 2 + 4 int_0^{frac{pi}{4}} ln (cos x) , dx
end{align}
and then employing the Fourier series representation for $ln (cos x)$ found here. The final result is
$$int_0^{frac{pi}{2}} ln (1 + sin x) , dx = 2 mathbf{G} - frac{pi}{2} ln 2.$$
Here $mathbf{G}$ is Catalan's constant.
The second of the integrals is very well known. Here
$$int_0^{frac{pi}{2}} ln (cos x) , dx = - frac{pi}{2} ln 2.$$
Thus $I(1) = 2 mathbf{G}$.
Moving to the main event, by differentiating under the integral sign with respect to $a$ we have
begin{align}
I'(a) &= int_0^{frac{pi}{2}} frac{tan x}{sqrt{a^2 tan^2 x + 1}} , dx\
&= int_0^{frac{pi}{2}} frac{sin x}{sqrt{a^2 - (a^2 - 1) cos^2 x}} , dx.
end{align}
Observe the term $(a^2 - 1)$ is positive since $a > 1$. Letting $u = cos x$ one has
$$I'(a) = int_0^1 frac{du}{sqrt{a^2 - (a^2 - 1) u^2}} = frac{1}{sqrt{a^2 - 1}} sin^{-1} left (frac{sqrt{a^2 - 1}}{a} right ).$$
As we require $I(2)$ observe that
$$I(2) - I(1) = int_1^2 I'(a) , da.$$
Thus
$$I(2) = 2 mathbf{G} + int_1^2 frac{1}{sqrt{a^2 - 1}} sin^{-1} left (frac{sqrt{a^2 - 1}}{a} right ) , da.$$
Integrating by parts leads to
$$I(2) = 2 mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - int_1^2 frac{cosh^{-1} a}{a sqrt{a^2 - 1}} , da.$$
Now let $a = cosh t$. This gives
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - int_0^{ln(2 + sqrt{3})} frac{t}{cosh t} , dt.$$
Then let $t = ln y$. This gives
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2int_1^{2 + sqrt{3}} frac{ln y}{1 + y^2} , dy.$$
Now let $y = tan theta$. Then we have
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2int_{frac{pi}{4}}^{frac{5pi}{12}} ln (tan theta) , dtheta.$$
Finally, enforcing a substitution of $theta mapsto dfrac{pi}{2} - theta$ leads to
begin{align}
I(2) &= 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) + 2 int_{frac{pi}{12}}^{frac{pi}{4}} ln (tan theta) , dtheta\
&= 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) + 2int_0^{frac{pi}{4}} ln (tan theta) , dtheta - 2 int_0^{frac{pi}{12}} ln (tan theta) , dtheta.
end{align}
For the first of the integrals we have
$$int_0^{frac{pi}{4}} ln (tan x) , dx = -mathbf{G}.$$
While for the second of the integrals we have
$$int_0^{frac{pi}{12}} ln (tan x) , dx = -frac{2}{3} mathbf{G}.$$
So finally
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2mathbf{G} + frac{4}{3} mathbf{G},$$
or
$$int_0^{frac{pi}{2}} operatorname{arcsinh} (2 tan x) , dx = frac{pi}{3} ln (2 + sqrt{3}) + frac{4}{3} mathbf{G},$$
as announced.
answered Jan 23 at 2:51
omegadotomegadot
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begin{align}A&=int_0^{frac{pi}{4}}ln(cos x),dx\ B&=int_0^{frac{pi}{4}}ln(sin x),dx\ B-A&=int_0^{frac{pi}{4}}ln(tan x),dx\ &=-text{G}\ A+B&=int_0^{frac{pi}{4}}ln(sin xcos x),dx\ &=int_0^{frac{pi}{4}}lnleft(frac{sin(2x)}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(frac{sin x}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(sin xright),dx-frac{pi}{4}ln 2\ &=-frac{1}{2}piln 2\ A&=frac{1}{2}left(A+B-left(B-Aright)right)\ &=frac{1}{2}text{G}-frac{1}{4}piln 2 end{align}
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– FDP
Jan 23 at 13:44
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Yes, very nice.
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– omegadot
Jan 24 at 1:29
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Sorry in a rush, so only a PARTIAL SOLUTION:
Here I will employ Feynman's Trick:
begin{equation}
I = int_0^{infty}frac{operatorname{arcsinh(2x)}}{1 + x^2}:dx
end{equation}
Let
begin{equation}
J(t) = int_0^{infty}frac{operatorname{arcsinh(tx)}}{1 + x^2}:dx
end{equation}
We observe that $J(2) = I$ and $J(0) = 0$. Using Leibniz's Integral rule we differentiate with respect to '$t$':
begin{equation}
J'(t) = int_0^{infty}frac{x}{sqrt{1 + t^2x^2}}frac{1}{1 + x^2}:dx = left[frac{1}{sqrt{t^2 - 1}} cdot arctanleft(sqrt{frac{1 + t^2x^2}{t^2 - 1}} right) right]_0^{infty} = frac{1}{sqrt{t^2 - 1}}left[frac{pi}{2} - arctanleft(frac{1}{sqrt{t^2 -1}} right) right]
end{equation}
Thus,
begin{align}
J(t) &= int frac{1}{sqrt{t^2 - 1}}left[frac{pi}{2} - arctanleft(frac{x}{sqrt{t^2 -1}} right) right]:dt \
&= int frac{pi}{2}cdot frac{1}{sqrt{t^2 - 1}}:dt - int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt = I_1 - I_2
end{align}
For $I_1$:
begin{equation}
I_1 = int frac{pi}{2}cdot frac{1}{sqrt{t^2 - 1}}:dt = frac{pi}{2}lnleft| sqrt{t^2 - 1} + tright| + C_1
end{equation}
Where $C_1$ is the constant of integration. Note that $I_1(2) = lnleft|2 + sqrt{3} right|$
For $I_2$:
begin{equation}
I_2 = int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt
end{equation}
Unfortunately this is not so easy to evaluate. I will first make the substitution $u = frac{1}{sqrt{t^2 - 1}}$:
begin{align}
I_2 &= int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt = int u cdot arctan(u) cdot frac{-u^4}{sqrt{1 +u^2}}:du\
& = - int frac{-u^5}{sqrt{1 +u^2}} cdot arctan(u) :du
end{align}
answered Jan 22 at 10:07
DavidGDavidG
2,5061726
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Will post up full solution later.
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– DavidG
Jan 22 at 10:08
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$$
begin{align}newcommand{arcsinh}{operatorname{arcsinh}}
int_0^inftyfrac{arcsinh(x)}{1+x^2},mathrm{d}x
&=int_0^inftyfrac{x,mathrm{d}x}{cosh(x)}tag{1a}\
&=int_0^inftyfrac{2x,mathrm{d}x}{e^x+e^{-x}}tag{1b}\
&=2sum_{k=0}^infty(-1)^kint_0^infty x,e^{-(2k+1)x},mathrm{d}xtag{1c}\ &=2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}tag{1d}\[6pt]
&=2mathrm{G}tag{1e}
end{align}
$$
Explanation:
$text{(1a)}$: substitute $xmapstosinh(x)$
$text{(1b)}$: write $cosh(x)$ as exponentials
$text{(1c)}$: expand $left(e^x+e^{-x}right)^{-1}$ as a power series
$text{(1d)}$: evaluate the integral
$text{(1e)}$: use the definition of Catalan's Constant
$$
begin{align}
frac{mathrm{d}}{mathrm{d}a}int_0^inftyfrac{arcsinh(ax)}{1+x^2},mathrm{d}x
&=int_0^inftyfrac{x}{1+x^2}frac{mathrm{d}x}{sqrt{1+a^2x^2}}tag{2a}\
&=int_0^inftyfrac{x}{a^2+x^2}frac{mathrm{d}x}{sqrt{1+x^2}}tag{2b}\
&=int_0^{pi/2}frac{mathrm{d}sec(x)}{a^2+tan^2(x)}tag{2c}\
&=int_1^inftyfrac{mathrm{d}x}{x^2+left(a^2-1right)}tag{2d}\
&=frac{arctanleft(sqrt{a^2-1}right)}{sqrt{a^2-1}}tag{2e}
end{align}
$$
Explanation:
$text{(2a)}$: $frac{mathrm{d}}{mathrm{d}a}arcsinh(ax)=frac{x}{sqrt{1+a^2x^2}}$
$text{(2b)}$: substitute $xmapsto x/a$
$text{(2c)}$: substitute $xmapstotan(x)$
$text{(2d)}$: substitute $sec(x)mapsto x$
$text{(2e)}$: arctan integral
Therefore,
$$
begin{align}newcommand{Li}{operatorname{Li}}
int_0^inftyfrac{arcsinh(2x)}{1+x^2},mathrm{d}x
&=2mathrm{G}+int_1^2frac{arctanleft(sqrt{a^2-1}right)}{sqrt{a^2-1}},mathrm{d}atag{3a}\
&=2mathrm{G}+int_0^{pi/3}frac{a}{cos(a)},mathrm{d}atag{3b}\
&=2mathrm{G}+2int_0^{pi/3}frac{a}{e^{ia}+e^{-ia}},mathrm{d}atag{3c}\
&=2mathrm{G}+2sum_{k=0}^infty(-1)^kint_0^{pi/3}ae^{-i(2k+1)a},mathrm{d}atag{3d}\
&=2mathrm{G}+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}int_0^{(2k+1)pi/3}ae^{-ia},mathrm{d}atag{3e}\
&=2mathrm{G}+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}left[(1+i(2k+1)pi/3)e^{-i(2k+1)pi/3}-1right]tag{3f}\
&=2mathrm{G}+fracpi3logleft(frac{2+sqrt3+i}{2-sqrt3-i}right)+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}left[e^{-i(2k+1)pi/3}-1right]tag{3g}\
&=2mathrm{G}+fracpi3logleft(left(2+sqrt3right)iright)
+sum_{k=0}^infty(-1)^kscriptsizeleft[frac{-1-isqrt3}{(6k+1)^2}-frac{-4}{(6k+3)^2}+frac{-1+isqrt3}{(6k+5)^2}right]tag{3h}\
&=frac43mathrm{G}+fracpi3logleft(2+sqrt3right)+ileft[frac{pi^2}6-sqrt3sum_{k=0}^infty(-1)^kscriptsizeleft(frac1{(6k+1)^2}-frac1{(6k+5)^2}right)right]tag{3i}\
&=frac43mathrm{G}+fracpi3logleft(2+sqrt3right)+ileft[frac{pi^2}6-frac{sqrt3}{36}sum_{kinmathbb{Z}}frac{(-1)^k}{left(k+frac16right)^2}right]tag{3j}\
&=bbox[5px,border:2px solid #C0A000]{frac43mathrm{G}+fracpi3logleft(2+sqrt3right)}tag{3k}
end{align}
$$
Explanation:
$text{(3a)}$: apply $(1)$ and $(2)$
$text{(3b)}$: substitute $amapstosec(a)$
$text{(3c)}$: write $cos(x)$ as exponentials
$text{(3d)}$: expand $left(e^{ix}+e^{-ix}right)^{-1}$ as a power series
$text{(3e)}$: substitute $amapsto a/(2k+1)$
$text{(3f)}$: integrate
$text{(3g)}$: recognize the series for $arctan(x)=frac1{2i}logleft(frac{1+ix}{1-ix}right)$
$text{(3h)}$: evaluate the exponentials
$text{(3i)}$: separate the real and imaginary parts
$text{(3j)}$: rewrite the sum
$text{(3k)}$: $sum_{kinmathbb{Z}}frac{(-1)^k}{(k+x)^2}=pi^2cot(pi x)csc(pi x)$
answered Jan 24 at 8:44
robjohn♦robjohn
269k27309635
$endgroup$
$begingroup$
Seems like there should be a^2in the denominator of the sum of the explanation (3k)
$endgroup$
– Kemono Chen
Jan 24 at 9:03
$begingroup$
@KemonoChen: Indeed. I put the parentheses there because I meant to square the denominator. Thanks. While it's true that we can just take the real part of the right side since the quantity is real, I wanted to make sure I hadn't made a mistake and verify that the imaginary part was $0$.
$endgroup$
– robjohn♦
Jan 24 at 9:25
add a comment |
$begingroup$
This is not an answer.
Your premonition seems to be good. Using another CAS,
$$2int_0^inftyfrac{xcosh( x)}{3+cosh^2(x)},dx=-frac{pi}{4} log left(7-4 sqrt{3}right)-i left(text{Li}_2left(-i left(-2+sqrt{3}right)right)-text{Li}_2left(i
left(-2+sqrt{3}right)right)right)$$ Now, ???
answered Jan 22 at 9:12
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
add a comment |
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$begingroup$
I will begin with the same approach as in @DavidG's answer, but will settle in a different argument. Let
$$ J(t) = int_{0}^{infty} frac{operatorname{arsinh}(tx)}{1+x^2} , mathrm{d}x. $$
Our goal is to compute $J(2)$ using Feynman's trick. Differentiating $J(t)$ and substituting $x=sqrt{u^{-2}-1}$, we obtain
begin{align*}
J'(t)
= int_{0}^{infty} frac{x}{(1+x^2)sqrt{1+t^2x^2}} , mathrm{d}x
= int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}u.
end{align*}
So it follows that
begin{align*}
J(2)
&= int_{0}^{2} int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}umathrm{d}t \
&= int_{0}^{1} int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}umathrm{d}t
+ int_{1}^{2} int_{0}^{1} frac{1}{sqrt{1 + (t^2-1)(1-u^2)}} , mathrm{d}umathrm{d}t.
end{align*}
The inner integral is easily computed, yielding
begin{align*}
J(2)
&= int_{0}^{1} frac{operatorname{artanh}left( sqrt{1 - t^2} right)}{sqrt{1 - t^2}} , mathrm{d}t
+ int_{1}^{2} frac{arctanleft(sqrt{t^2-1}right)}{sqrt{t^2 - 1}} , mathrm{d}t.
end{align*}
Now we substitute $t = operatorname{sech} varphi$ for the first integral and $t = sec theta$ for the second integral. This yields
begin{align*}
J(2)
&= int_{0}^{infty} frac{varphi}{coshvarphi} , mathrm{d}varphi
+ int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta.
end{align*}
These integrals can be computed as follows:
Using $ operatorname{sech}varphi = frac{2e^{-varphi}}{1 + e^{-2varphi}} = 2 sum_{n=0}^{infty} (-1)^n e^{-(2n+1)varphi} $, we obtain
$$ int_{0}^{infty} frac{varphi}{coshvarphi} , mathrm{d}varphi
= 2 sum_{n=0}^{infty} (-1)^n int_{0}^{infty} varphi e^{-(2n+1)varphi} , mathrm{d}varphi
= 2 sum_{n=0}^{infty} frac{(-1)^n}{(2n+1)^2}
= 2G. $$
Taking integration by parts,
begin{align*}
int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta
&= left[ - theta log left( tan left( frac{pi}{4} - frac{theta}{2} right) right) right]_{0}^{frac{pi}{3}} + int_{0}^{frac{pi}{3}} log left( tan left( frac{pi}{4} - frac{theta}{2} right) right) , mathrm{d}theta \
&= frac{pi}{3} log left( 2 + sqrt{3} right) + 2 int_{frac{pi}{12}}^{frac{pi}{4}} log left( tan theta right) , mathrm{d}theta.
end{align*}
This can be computed by using the Fourier series $log left( tan theta right) = - 2 sum_{n=0}^{infty} frac{cos(4n+2)theta}{2n+1} $ to yield
begin{align*}
int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta
&= frac{pi}{3} log left( 2 + sqrt{3} right) - 2 sum_{n=0}^{infty} frac{sinleft( frac{pi}{2} (2n+1) right) - sinleft( frac{pi}{6} (2n+1) right)}{(2n+1)^2} \
&= frac{pi}{3} log left( 2 + sqrt{3} right) - frac{2}{3}G.
end{align*}
Combining two result, we obtain the desired answer.
answered Jan 22 at 11:24
Sangchul LeeSangchul Lee
95.5k12171279
$endgroup$
1
$begingroup$
This is beauty !
$endgroup$
– Claude Leibovici
Jan 22 at 16:10
add a comment |
$begingroup$
I will begin with the same approach as in @DavidG's answer, but will settle in a different argument. Let
$$ J(t) = int_{0}^{infty} frac{operatorname{arsinh}(tx)}{1+x^2} , mathrm{d}x. $$
Our goal is to compute $J(2)$ using Feynman's trick. Differentiating $J(t)$ and substituting $x=sqrt{u^{-2}-1}$, we obtain
begin{align*}
J'(t)
= int_{0}^{infty} frac{x}{(1+x^2)sqrt{1+t^2x^2}} , mathrm{d}x
= int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}u.
end{align*}
So it follows that
begin{align*}
J(2)
&= int_{0}^{2} int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}umathrm{d}t \
&= int_{0}^{1} int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}umathrm{d}t
+ int_{1}^{2} int_{0}^{1} frac{1}{sqrt{1 + (t^2-1)(1-u^2)}} , mathrm{d}umathrm{d}t.
end{align*}
The inner integral is easily computed, yielding
begin{align*}
J(2)
&= int_{0}^{1} frac{operatorname{artanh}left( sqrt{1 - t^2} right)}{sqrt{1 - t^2}} , mathrm{d}t
+ int_{1}^{2} frac{arctanleft(sqrt{t^2-1}right)}{sqrt{t^2 - 1}} , mathrm{d}t.
end{align*}
Now we substitute $t = operatorname{sech} varphi$ for the first integral and $t = sec theta$ for the second integral. This yields
begin{align*}
J(2)
&= int_{0}^{infty} frac{varphi}{coshvarphi} , mathrm{d}varphi
+ int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta.
end{align*}
These integrals can be computed as follows:
Using $ operatorname{sech}varphi = frac{2e^{-varphi}}{1 + e^{-2varphi}} = 2 sum_{n=0}^{infty} (-1)^n e^{-(2n+1)varphi} $, we obtain
$$ int_{0}^{infty} frac{varphi}{coshvarphi} , mathrm{d}varphi
= 2 sum_{n=0}^{infty} (-1)^n int_{0}^{infty} varphi e^{-(2n+1)varphi} , mathrm{d}varphi
= 2 sum_{n=0}^{infty} frac{(-1)^n}{(2n+1)^2}
= 2G. $$
Taking integration by parts,
begin{align*}
int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta
&= left[ - theta log left( tan left( frac{pi}{4} - frac{theta}{2} right) right) right]_{0}^{frac{pi}{3}} + int_{0}^{frac{pi}{3}} log left( tan left( frac{pi}{4} - frac{theta}{2} right) right) , mathrm{d}theta \
&= frac{pi}{3} log left( 2 + sqrt{3} right) + 2 int_{frac{pi}{12}}^{frac{pi}{4}} log left( tan theta right) , mathrm{d}theta.
end{align*}
This can be computed by using the Fourier series $log left( tan theta right) = - 2 sum_{n=0}^{infty} frac{cos(4n+2)theta}{2n+1} $ to yield
begin{align*}
int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta
&= frac{pi}{3} log left( 2 + sqrt{3} right) - 2 sum_{n=0}^{infty} frac{sinleft( frac{pi}{2} (2n+1) right) - sinleft( frac{pi}{6} (2n+1) right)}{(2n+1)^2} \
&= frac{pi}{3} log left( 2 + sqrt{3} right) - frac{2}{3}G.
end{align*}
Combining two result, we obtain the desired answer.
answered Jan 22 at 11:24
Sangchul LeeSangchul Lee
95.5k12171279
$endgroup$
1
$begingroup$
This is beauty !
$endgroup$
– Claude Leibovici
Jan 22 at 16:10
add a comment |
$begingroup$
I will begin with the same approach as in @DavidG's answer, but will settle in a different argument. Let
$$ J(t) = int_{0}^{infty} frac{operatorname{arsinh}(tx)}{1+x^2} , mathrm{d}x. $$
Our goal is to compute $J(2)$ using Feynman's trick. Differentiating $J(t)$ and substituting $x=sqrt{u^{-2}-1}$, we obtain
begin{align*}
J'(t)
= int_{0}^{infty} frac{x}{(1+x^2)sqrt{1+t^2x^2}} , mathrm{d}x
= int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}u.
end{align*}
So it follows that
begin{align*}
J(2)
&= int_{0}^{2} int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}umathrm{d}t \
&= int_{0}^{1} int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}umathrm{d}t
+ int_{1}^{2} int_{0}^{1} frac{1}{sqrt{1 + (t^2-1)(1-u^2)}} , mathrm{d}umathrm{d}t.
end{align*}
The inner integral is easily computed, yielding
begin{align*}
J(2)
&= int_{0}^{1} frac{operatorname{artanh}left( sqrt{1 - t^2} right)}{sqrt{1 - t^2}} , mathrm{d}t
+ int_{1}^{2} frac{arctanleft(sqrt{t^2-1}right)}{sqrt{t^2 - 1}} , mathrm{d}t.
end{align*}
Now we substitute $t = operatorname{sech} varphi$ for the first integral and $t = sec theta$ for the second integral. This yields
begin{align*}
J(2)
&= int_{0}^{infty} frac{varphi}{coshvarphi} , mathrm{d}varphi
+ int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta.
end{align*}
These integrals can be computed as follows:
Using $ operatorname{sech}varphi = frac{2e^{-varphi}}{1 + e^{-2varphi}} = 2 sum_{n=0}^{infty} (-1)^n e^{-(2n+1)varphi} $, we obtain
$$ int_{0}^{infty} frac{varphi}{coshvarphi} , mathrm{d}varphi
= 2 sum_{n=0}^{infty} (-1)^n int_{0}^{infty} varphi e^{-(2n+1)varphi} , mathrm{d}varphi
= 2 sum_{n=0}^{infty} frac{(-1)^n}{(2n+1)^2}
= 2G. $$
Taking integration by parts,
begin{align*}
int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta
&= left[ - theta log left( tan left( frac{pi}{4} - frac{theta}{2} right) right) right]_{0}^{frac{pi}{3}} + int_{0}^{frac{pi}{3}} log left( tan left( frac{pi}{4} - frac{theta}{2} right) right) , mathrm{d}theta \
&= frac{pi}{3} log left( 2 + sqrt{3} right) + 2 int_{frac{pi}{12}}^{frac{pi}{4}} log left( tan theta right) , mathrm{d}theta.
end{align*}
This can be computed by using the Fourier series $log left( tan theta right) = - 2 sum_{n=0}^{infty} frac{cos(4n+2)theta}{2n+1} $ to yield
begin{align*}
int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta
&= frac{pi}{3} log left( 2 + sqrt{3} right) - 2 sum_{n=0}^{infty} frac{sinleft( frac{pi}{2} (2n+1) right) - sinleft( frac{pi}{6} (2n+1) right)}{(2n+1)^2} \
&= frac{pi}{3} log left( 2 + sqrt{3} right) - frac{2}{3}G.
end{align*}
Combining two result, we obtain the desired answer.
answered Jan 22 at 11:24
Sangchul LeeSangchul Lee
95.5k12171279
$endgroup$
I will begin with the same approach as in @DavidG's answer, but will settle in a different argument. Let
$$ J(t) = int_{0}^{infty} frac{operatorname{arsinh}(tx)}{1+x^2} , mathrm{d}x. $$
Our goal is to compute $J(2)$ using Feynman's trick. Differentiating $J(t)$ and substituting $x=sqrt{u^{-2}-1}$, we obtain
begin{align*}
J'(t)
= int_{0}^{infty} frac{x}{(1+x^2)sqrt{1+t^2x^2}} , mathrm{d}x
= int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}u.
end{align*}
So it follows that
begin{align*}
J(2)
&= int_{0}^{2} int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}umathrm{d}t \
&= int_{0}^{1} int_{0}^{1} frac{1}{sqrt{1 - (1-t^2)(1-u^2)}} , mathrm{d}umathrm{d}t
+ int_{1}^{2} int_{0}^{1} frac{1}{sqrt{1 + (t^2-1)(1-u^2)}} , mathrm{d}umathrm{d}t.
end{align*}
The inner integral is easily computed, yielding
begin{align*}
J(2)
&= int_{0}^{1} frac{operatorname{artanh}left( sqrt{1 - t^2} right)}{sqrt{1 - t^2}} , mathrm{d}t
+ int_{1}^{2} frac{arctanleft(sqrt{t^2-1}right)}{sqrt{t^2 - 1}} , mathrm{d}t.
end{align*}
Now we substitute $t = operatorname{sech} varphi$ for the first integral and $t = sec theta$ for the second integral. This yields
begin{align*}
J(2)
&= int_{0}^{infty} frac{varphi}{coshvarphi} , mathrm{d}varphi
+ int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta.
end{align*}
These integrals can be computed as follows:
Using $ operatorname{sech}varphi = frac{2e^{-varphi}}{1 + e^{-2varphi}} = 2 sum_{n=0}^{infty} (-1)^n e^{-(2n+1)varphi} $, we obtain
$$ int_{0}^{infty} frac{varphi}{coshvarphi} , mathrm{d}varphi
= 2 sum_{n=0}^{infty} (-1)^n int_{0}^{infty} varphi e^{-(2n+1)varphi} , mathrm{d}varphi
= 2 sum_{n=0}^{infty} frac{(-1)^n}{(2n+1)^2}
= 2G. $$
Taking integration by parts,
begin{align*}
int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta
&= left[ - theta log left( tan left( frac{pi}{4} - frac{theta}{2} right) right) right]_{0}^{frac{pi}{3}} + int_{0}^{frac{pi}{3}} log left( tan left( frac{pi}{4} - frac{theta}{2} right) right) , mathrm{d}theta \
&= frac{pi}{3} log left( 2 + sqrt{3} right) + 2 int_{frac{pi}{12}}^{frac{pi}{4}} log left( tan theta right) , mathrm{d}theta.
end{align*}
This can be computed by using the Fourier series $log left( tan theta right) = - 2 sum_{n=0}^{infty} frac{cos(4n+2)theta}{2n+1} $ to yield
begin{align*}
int_{0}^{frac{pi}{3}} frac{theta}{costheta} , mathrm{d}theta
&= frac{pi}{3} log left( 2 + sqrt{3} right) - 2 sum_{n=0}^{infty} frac{sinleft( frac{pi}{2} (2n+1) right) - sinleft( frac{pi}{6} (2n+1) right)}{(2n+1)^2} \
&= frac{pi}{3} log left( 2 + sqrt{3} right) - frac{2}{3}G.
end{align*}
Combining two result, we obtain the desired answer.
answered Jan 22 at 11:24
Sangchul LeeSangchul Lee
95.5k12171279
answered Jan 22 at 11:24
Sangchul LeeSangchul Lee
95.5k12171279
answered Jan 22 at 11:24
Sangchul LeeSangchul Lee
95.5k12171279
answered Jan 22 at 11:24
Sangchul LeeSangchul Lee
95.5k12171279
95.5k12171279
1
$begingroup$
This is beauty !
$endgroup$
– Claude Leibovici
Jan 22 at 16:10
add a comment |
1
$begingroup$
This is beauty !
$endgroup$
– Claude Leibovici
Jan 22 at 16:10
1
1
$begingroup$
This is beauty !
$endgroup$
– Claude Leibovici
Jan 22 at 16:10
$begingroup$
This is beauty !
$endgroup$
– Claude Leibovici
Jan 22 at 16:10
add a comment |
$begingroup$
On the path of Kemono Chen...
begin{align}J&=int_0^{pi/2}operatorname{arcsinh}(2tan x)dxend{align}
Perform the change of variable $y=operatorname{arcsinh}(2tan x)$,
begin{align}J&=int_0^{+infty}frac{2xcosh x}{4+sinh^2 x},dx\
&=int_0^{+infty}frac{4xleft(text{e}^{x}+text{e}^{-x}right)}{14+text{e}^{2x}+text{e}^{-2x}},dx\
&=int_0^{+infty}frac{4xtext{e}^{-x}left(text{e}^{2x}+1right)}{14+text{e}^{2x}+text{e}^{-2x}},dx\
end{align}
Perform the change of variable $y=text{e}^{-x}$,
begin{align}J&=-int_0^1 frac{4ln xleft(1+frac{1}{x^2}right)}{14+x^2+frac{1}{x^2}}\
&=-int_0^1 frac{4ln xleft(1+x^2right)}{x^4+14x^2+1}\
&=left[-arctanleft(frac{4x}{1-x^2}right)ln xright]_0^1+int_0^1 frac{arctanleft(frac{4x}{1-x^2}right)}{x},dx\
&=int_0^1 frac{arctanleft(frac{4x}{1-x^2}right)}{x},dx\
&=int_0^1 frac{arctanleft(left(2+sqrt{3}right)xright)}{x},dx+int_0^1 frac{arctanleft(left(2-sqrt{3}right)xright)}{x},dx\
end{align}
In the first integral perform the change of variable $y=left(2+sqrt{3}right)x$,
In the second integral perform the change of variable $y=left(2-sqrt{3}right)x$,
begin{align}J&=int_0^{2+sqrt{3}}frac{arctan x}{x},dx+int_0^{2-sqrt{3}}frac{arctan x}{x},dx\
&=Big[arctan xln xBig]_0^{2+sqrt{3}}-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx+Big[arctan xln xBig]_0^{2-sqrt{3}}-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{5pi}{12}lnleft(2+sqrt{3}right)-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx+frac{pi}{12}lnleft(2-sqrt{3}right)-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx
end{align}
In the first integral perform the change of variable $y=dfrac{1}{x}$,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)+int_{2-sqrt{3}}^{+infty}frac{ln x}{1+x^2},dx-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)+int_0^{+infty}frac{ln x}{1+x^2},dx-2int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)-2int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
end{align}
Perform the change of variable $y=tan x$,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)-2int_0^{frac{pi}{12}}lnleft(tan xright),dx\
end{align}
It is well known that,
begin{align}
int_0^{frac{pi}{12}}lnleft(tan xright),dx=-frac{2}{3}text{G}
end{align}
(see: Integral: $int_0^{pi/12} ln(tan x),dx$ )
Thus,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)-2times -frac{2}{3}text{G}\
&=boxed{frac{pi}{3}lnleft(2+sqrt{3}right)+frac{4}{3}text{G}}
end{align}
NB:
Observe that,
begin{align}2-sqrt{3}&=frac{1}{2+sqrt{3}}\
lnleft(2-sqrt{3}right)&=-lnleft(2+sqrt{3}right)\
int_0^infty frac{ln x}{1+x^2},dx&=0
end{align}
(perform the change of variable $y=dfrac{1}{x}$ )
answered Jan 22 at 13:35
FDPFDP
6,02211829
$endgroup$
$begingroup$
This is beauty !
$endgroup$
– Claude Leibovici
Jan 22 at 16:10
$begingroup$
How did you do $$intarctanfrac{4x}{1-x^2}frac{mathrm dx}{x}=intarctan(2+sqrt3)xfrac{mathrm dx}{x}+intarctan(2-sqrt3)xfrac{mathrm dx}{x}$$
$endgroup$
– clathratus
Jan 22 at 16:10
1
$begingroup$
@Clathratus: if $ab<1$then begin{align}arctan a+arctan b=arctanleft(frac{a+b}{1-ab}right)end{align}
$endgroup$
– FDP
Jan 22 at 16:23
add a comment |
$begingroup$
On the path of Kemono Chen...
begin{align}J&=int_0^{pi/2}operatorname{arcsinh}(2tan x)dxend{align}
Perform the change of variable $y=operatorname{arcsinh}(2tan x)$,
begin{align}J&=int_0^{+infty}frac{2xcosh x}{4+sinh^2 x},dx\
&=int_0^{+infty}frac{4xleft(text{e}^{x}+text{e}^{-x}right)}{14+text{e}^{2x}+text{e}^{-2x}},dx\
&=int_0^{+infty}frac{4xtext{e}^{-x}left(text{e}^{2x}+1right)}{14+text{e}^{2x}+text{e}^{-2x}},dx\
end{align}
Perform the change of variable $y=text{e}^{-x}$,
begin{align}J&=-int_0^1 frac{4ln xleft(1+frac{1}{x^2}right)}{14+x^2+frac{1}{x^2}}\
&=-int_0^1 frac{4ln xleft(1+x^2right)}{x^4+14x^2+1}\
&=left[-arctanleft(frac{4x}{1-x^2}right)ln xright]_0^1+int_0^1 frac{arctanleft(frac{4x}{1-x^2}right)}{x},dx\
&=int_0^1 frac{arctanleft(frac{4x}{1-x^2}right)}{x},dx\
&=int_0^1 frac{arctanleft(left(2+sqrt{3}right)xright)}{x},dx+int_0^1 frac{arctanleft(left(2-sqrt{3}right)xright)}{x},dx\
end{align}
In the first integral perform the change of variable $y=left(2+sqrt{3}right)x$,
In the second integral perform the change of variable $y=left(2-sqrt{3}right)x$,
begin{align}J&=int_0^{2+sqrt{3}}frac{arctan x}{x},dx+int_0^{2-sqrt{3}}frac{arctan x}{x},dx\
&=Big[arctan xln xBig]_0^{2+sqrt{3}}-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx+Big[arctan xln xBig]_0^{2-sqrt{3}}-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{5pi}{12}lnleft(2+sqrt{3}right)-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx+frac{pi}{12}lnleft(2-sqrt{3}right)-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx
end{align}
In the first integral perform the change of variable $y=dfrac{1}{x}$,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)+int_{2-sqrt{3}}^{+infty}frac{ln x}{1+x^2},dx-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)+int_0^{+infty}frac{ln x}{1+x^2},dx-2int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)-2int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
end{align}
Perform the change of variable $y=tan x$,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)-2int_0^{frac{pi}{12}}lnleft(tan xright),dx\
end{align}
It is well known that,
begin{align}
int_0^{frac{pi}{12}}lnleft(tan xright),dx=-frac{2}{3}text{G}
end{align}
(see: Integral: $int_0^{pi/12} ln(tan x),dx$ )
Thus,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)-2times -frac{2}{3}text{G}\
&=boxed{frac{pi}{3}lnleft(2+sqrt{3}right)+frac{4}{3}text{G}}
end{align}
NB:
Observe that,
begin{align}2-sqrt{3}&=frac{1}{2+sqrt{3}}\
lnleft(2-sqrt{3}right)&=-lnleft(2+sqrt{3}right)\
int_0^infty frac{ln x}{1+x^2},dx&=0
end{align}
(perform the change of variable $y=dfrac{1}{x}$ )
answered Jan 22 at 13:35
FDPFDP
6,02211829
$endgroup$
$begingroup$
This is beauty !
$endgroup$
– Claude Leibovici
Jan 22 at 16:10
$begingroup$
How did you do $$intarctanfrac{4x}{1-x^2}frac{mathrm dx}{x}=intarctan(2+sqrt3)xfrac{mathrm dx}{x}+intarctan(2-sqrt3)xfrac{mathrm dx}{x}$$
$endgroup$
– clathratus
Jan 22 at 16:10
1
$begingroup$
@Clathratus: if $ab<1$then begin{align}arctan a+arctan b=arctanleft(frac{a+b}{1-ab}right)end{align}
$endgroup$
– FDP
Jan 22 at 16:23
add a comment |
$begingroup$
On the path of Kemono Chen...
begin{align}J&=int_0^{pi/2}operatorname{arcsinh}(2tan x)dxend{align}
Perform the change of variable $y=operatorname{arcsinh}(2tan x)$,
begin{align}J&=int_0^{+infty}frac{2xcosh x}{4+sinh^2 x},dx\
&=int_0^{+infty}frac{4xleft(text{e}^{x}+text{e}^{-x}right)}{14+text{e}^{2x}+text{e}^{-2x}},dx\
&=int_0^{+infty}frac{4xtext{e}^{-x}left(text{e}^{2x}+1right)}{14+text{e}^{2x}+text{e}^{-2x}},dx\
end{align}
Perform the change of variable $y=text{e}^{-x}$,
begin{align}J&=-int_0^1 frac{4ln xleft(1+frac{1}{x^2}right)}{14+x^2+frac{1}{x^2}}\
&=-int_0^1 frac{4ln xleft(1+x^2right)}{x^4+14x^2+1}\
&=left[-arctanleft(frac{4x}{1-x^2}right)ln xright]_0^1+int_0^1 frac{arctanleft(frac{4x}{1-x^2}right)}{x},dx\
&=int_0^1 frac{arctanleft(frac{4x}{1-x^2}right)}{x},dx\
&=int_0^1 frac{arctanleft(left(2+sqrt{3}right)xright)}{x},dx+int_0^1 frac{arctanleft(left(2-sqrt{3}right)xright)}{x},dx\
end{align}
In the first integral perform the change of variable $y=left(2+sqrt{3}right)x$,
In the second integral perform the change of variable $y=left(2-sqrt{3}right)x$,
begin{align}J&=int_0^{2+sqrt{3}}frac{arctan x}{x},dx+int_0^{2-sqrt{3}}frac{arctan x}{x},dx\
&=Big[arctan xln xBig]_0^{2+sqrt{3}}-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx+Big[arctan xln xBig]_0^{2-sqrt{3}}-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{5pi}{12}lnleft(2+sqrt{3}right)-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx+frac{pi}{12}lnleft(2-sqrt{3}right)-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx
end{align}
In the first integral perform the change of variable $y=dfrac{1}{x}$,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)+int_{2-sqrt{3}}^{+infty}frac{ln x}{1+x^2},dx-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)+int_0^{+infty}frac{ln x}{1+x^2},dx-2int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)-2int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
end{align}
Perform the change of variable $y=tan x$,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)-2int_0^{frac{pi}{12}}lnleft(tan xright),dx\
end{align}
It is well known that,
begin{align}
int_0^{frac{pi}{12}}lnleft(tan xright),dx=-frac{2}{3}text{G}
end{align}
(see: Integral: $int_0^{pi/12} ln(tan x),dx$ )
Thus,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)-2times -frac{2}{3}text{G}\
&=boxed{frac{pi}{3}lnleft(2+sqrt{3}right)+frac{4}{3}text{G}}
end{align}
NB:
Observe that,
begin{align}2-sqrt{3}&=frac{1}{2+sqrt{3}}\
lnleft(2-sqrt{3}right)&=-lnleft(2+sqrt{3}right)\
int_0^infty frac{ln x}{1+x^2},dx&=0
end{align}
(perform the change of variable $y=dfrac{1}{x}$ )
answered Jan 22 at 13:35
FDPFDP
6,02211829
$endgroup$
On the path of Kemono Chen...
begin{align}J&=int_0^{pi/2}operatorname{arcsinh}(2tan x)dxend{align}
Perform the change of variable $y=operatorname{arcsinh}(2tan x)$,
begin{align}J&=int_0^{+infty}frac{2xcosh x}{4+sinh^2 x},dx\
&=int_0^{+infty}frac{4xleft(text{e}^{x}+text{e}^{-x}right)}{14+text{e}^{2x}+text{e}^{-2x}},dx\
&=int_0^{+infty}frac{4xtext{e}^{-x}left(text{e}^{2x}+1right)}{14+text{e}^{2x}+text{e}^{-2x}},dx\
end{align}
Perform the change of variable $y=text{e}^{-x}$,
begin{align}J&=-int_0^1 frac{4ln xleft(1+frac{1}{x^2}right)}{14+x^2+frac{1}{x^2}}\
&=-int_0^1 frac{4ln xleft(1+x^2right)}{x^4+14x^2+1}\
&=left[-arctanleft(frac{4x}{1-x^2}right)ln xright]_0^1+int_0^1 frac{arctanleft(frac{4x}{1-x^2}right)}{x},dx\
&=int_0^1 frac{arctanleft(frac{4x}{1-x^2}right)}{x},dx\
&=int_0^1 frac{arctanleft(left(2+sqrt{3}right)xright)}{x},dx+int_0^1 frac{arctanleft(left(2-sqrt{3}right)xright)}{x},dx\
end{align}
In the first integral perform the change of variable $y=left(2+sqrt{3}right)x$,
In the second integral perform the change of variable $y=left(2-sqrt{3}right)x$,
begin{align}J&=int_0^{2+sqrt{3}}frac{arctan x}{x},dx+int_0^{2-sqrt{3}}frac{arctan x}{x},dx\
&=Big[arctan xln xBig]_0^{2+sqrt{3}}-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx+Big[arctan xln xBig]_0^{2-sqrt{3}}-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{5pi}{12}lnleft(2+sqrt{3}right)-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx+frac{pi}{12}lnleft(2-sqrt{3}right)-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)-int_0^{2+sqrt{3}}frac{ln x}{1+x^2},dx-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx
end{align}
In the first integral perform the change of variable $y=dfrac{1}{x}$,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)+int_{2-sqrt{3}}^{+infty}frac{ln x}{1+x^2},dx-int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)+int_0^{+infty}frac{ln x}{1+x^2},dx-2int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
&=frac{pi}{3}lnleft(2+sqrt{3}right)-2int_0^{2-sqrt{3}}frac{ln x}{1+x^2},dx\
end{align}
Perform the change of variable $y=tan x$,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)-2int_0^{frac{pi}{12}}lnleft(tan xright),dx\
end{align}
It is well known that,
begin{align}
int_0^{frac{pi}{12}}lnleft(tan xright),dx=-frac{2}{3}text{G}
end{align}
(see: Integral: $int_0^{pi/12} ln(tan x),dx$ )
Thus,
begin{align}J&=frac{pi}{3}lnleft(2+sqrt{3}right)-2times -frac{2}{3}text{G}\
&=boxed{frac{pi}{3}lnleft(2+sqrt{3}right)+frac{4}{3}text{G}}
end{align}
NB:
Observe that,
begin{align}2-sqrt{3}&=frac{1}{2+sqrt{3}}\
lnleft(2-sqrt{3}right)&=-lnleft(2+sqrt{3}right)\
int_0^infty frac{ln x}{1+x^2},dx&=0
end{align}
(perform the change of variable $y=dfrac{1}{x}$ )
answered Jan 22 at 13:35
FDPFDP
6,02211829
edited Jan 22 at 13:40
answered Jan 22 at 13:35
FDPFDP
6,02211829
answered Jan 22 at 13:35
FDPFDP
6,02211829
answered Jan 22 at 13:35
FDPFDP
6,02211829
6,02211829
$begingroup$
This is beauty !
$endgroup$
– Claude Leibovici
Jan 22 at 16:10
$begingroup$
How did you do $$intarctanfrac{4x}{1-x^2}frac{mathrm dx}{x}=intarctan(2+sqrt3)xfrac{mathrm dx}{x}+intarctan(2-sqrt3)xfrac{mathrm dx}{x}$$
$endgroup$
– clathratus
Jan 22 at 16:10
1
$begingroup$
@Clathratus: if $ab<1$then begin{align}arctan a+arctan b=arctanleft(frac{a+b}{1-ab}right)end{align}
$endgroup$
– FDP
Jan 22 at 16:23
add a comment |
$begingroup$
This is beauty !
$endgroup$
– Claude Leibovici
Jan 22 at 16:10
$begingroup$
How did you do $$intarctanfrac{4x}{1-x^2}frac{mathrm dx}{x}=intarctan(2+sqrt3)xfrac{mathrm dx}{x}+intarctan(2-sqrt3)xfrac{mathrm dx}{x}$$
$endgroup$
– clathratus
Jan 22 at 16:10
1
$begingroup$
@Clathratus: if $ab<1$then begin{align}arctan a+arctan b=arctanleft(frac{a+b}{1-ab}right)end{align}
$endgroup$
– FDP
Jan 22 at 16:23
$begingroup$
This is beauty !
$endgroup$
– Claude Leibovici
Jan 22 at 16:10
$begingroup$
This is beauty !
$endgroup$
– Claude Leibovici
Jan 22 at 16:10
$begingroup$
How did you do $$intarctanfrac{4x}{1-x^2}frac{mathrm dx}{x}=intarctan(2+sqrt3)xfrac{mathrm dx}{x}+intarctan(2-sqrt3)xfrac{mathrm dx}{x}$$
$endgroup$
– clathratus
Jan 22 at 16:10
$begingroup$
How did you do $$intarctanfrac{4x}{1-x^2}frac{mathrm dx}{x}=intarctan(2+sqrt3)xfrac{mathrm dx}{x}+intarctan(2-sqrt3)xfrac{mathrm dx}{x}$$
$endgroup$
– clathratus
Jan 22 at 16:10
1
1
$begingroup$
@Clathratus: if $ab<1$then begin{align}arctan a+arctan b=arctanleft(frac{a+b}{1-ab}right)end{align}
$endgroup$
– FDP
Jan 22 at 16:23
$begingroup$
@Clathratus: if $ab<1$then begin{align}arctan a+arctan b=arctanleft(frac{a+b}{1-ab}right)end{align}
$endgroup$
– FDP
Jan 22 at 16:23
add a comment |
$begingroup$
Here is another variation along a theme which like @DavidG approach uses Feynman's trick of differentiating under the integral sign.
Let
$$I(a) = int_0^{frac{pi}{2}} operatorname{arcsinh} (a tan x) , dx, qquad a > 1.$$
We are required to find $I(2)$. We start by finding $I(1)$ first, which will be needed later on.
For $a = 1$ we have
begin{align}
I(1) &= int_0^{frac{pi}{2}} operatorname{arcsinh} (tan x) , dx\
&= int_0^{frac{pi}{2}} ln left (frac{1 + sin x}{cos x} right ) , dx\
&= int_0^{frac{pi}{2}} ln (1 + sin x) , dx - int_0^{frac{pi}{2}} ln (cos x) , dx.
end{align}
Now the first of these integrals can be found, for example, by rewriting it as
begin{align}
int_0^{frac{pi}{2}} ln (1 + sin x) , dx &= int_0^{frac{pi}{2}} ln (1 + cos x) , dx\
&= int_0^{frac{pi}{2}} ln left (frac{1}{2} cos^2 frac{x}{2} right ) , dx\
&= frac{pi}{2} ln 2 + 4 int_0^{frac{pi}{4}} ln (cos x) , dx
end{align}
and then employing the Fourier series representation for $ln (cos x)$ found here. The final result is
$$int_0^{frac{pi}{2}} ln (1 + sin x) , dx = 2 mathbf{G} - frac{pi}{2} ln 2.$$
Here $mathbf{G}$ is Catalan's constant.
The second of the integrals is very well known. Here
$$int_0^{frac{pi}{2}} ln (cos x) , dx = - frac{pi}{2} ln 2.$$
Thus $I(1) = 2 mathbf{G}$.
Moving to the main event, by differentiating under the integral sign with respect to $a$ we have
begin{align}
I'(a) &= int_0^{frac{pi}{2}} frac{tan x}{sqrt{a^2 tan^2 x + 1}} , dx\
&= int_0^{frac{pi}{2}} frac{sin x}{sqrt{a^2 - (a^2 - 1) cos^2 x}} , dx.
end{align}
Observe the term $(a^2 - 1)$ is positive since $a > 1$. Letting $u = cos x$ one has
$$I'(a) = int_0^1 frac{du}{sqrt{a^2 - (a^2 - 1) u^2}} = frac{1}{sqrt{a^2 - 1}} sin^{-1} left (frac{sqrt{a^2 - 1}}{a} right ).$$
As we require $I(2)$ observe that
$$I(2) - I(1) = int_1^2 I'(a) , da.$$
Thus
$$I(2) = 2 mathbf{G} + int_1^2 frac{1}{sqrt{a^2 - 1}} sin^{-1} left (frac{sqrt{a^2 - 1}}{a} right ) , da.$$
Integrating by parts leads to
$$I(2) = 2 mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - int_1^2 frac{cosh^{-1} a}{a sqrt{a^2 - 1}} , da.$$
Now let $a = cosh t$. This gives
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - int_0^{ln(2 + sqrt{3})} frac{t}{cosh t} , dt.$$
Then let $t = ln y$. This gives
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2int_1^{2 + sqrt{3}} frac{ln y}{1 + y^2} , dy.$$
Now let $y = tan theta$. Then we have
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2int_{frac{pi}{4}}^{frac{5pi}{12}} ln (tan theta) , dtheta.$$
Finally, enforcing a substitution of $theta mapsto dfrac{pi}{2} - theta$ leads to
begin{align}
I(2) &= 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) + 2 int_{frac{pi}{12}}^{frac{pi}{4}} ln (tan theta) , dtheta\
&= 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) + 2int_0^{frac{pi}{4}} ln (tan theta) , dtheta - 2 int_0^{frac{pi}{12}} ln (tan theta) , dtheta.
end{align}
For the first of the integrals we have
$$int_0^{frac{pi}{4}} ln (tan x) , dx = -mathbf{G}.$$
While for the second of the integrals we have
$$int_0^{frac{pi}{12}} ln (tan x) , dx = -frac{2}{3} mathbf{G}.$$
So finally
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2mathbf{G} + frac{4}{3} mathbf{G},$$
or
$$int_0^{frac{pi}{2}} operatorname{arcsinh} (2 tan x) , dx = frac{pi}{3} ln (2 + sqrt{3}) + frac{4}{3} mathbf{G},$$
as announced.
answered Jan 23 at 2:51
omegadotomegadot
6,3972829
$endgroup$
1
$begingroup$
begin{align}A&=int_0^{frac{pi}{4}}ln(cos x),dx\ B&=int_0^{frac{pi}{4}}ln(sin x),dx\ B-A&=int_0^{frac{pi}{4}}ln(tan x),dx\ &=-text{G}\ A+B&=int_0^{frac{pi}{4}}ln(sin xcos x),dx\ &=int_0^{frac{pi}{4}}lnleft(frac{sin(2x)}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(frac{sin x}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(sin xright),dx-frac{pi}{4}ln 2\ &=-frac{1}{2}piln 2\ A&=frac{1}{2}left(A+B-left(B-Aright)right)\ &=frac{1}{2}text{G}-frac{1}{4}piln 2 end{align}
$endgroup$
– FDP
Jan 23 at 13:44
$begingroup$
Yes, very nice.
$endgroup$
– omegadot
Jan 24 at 1:29
add a comment |
$begingroup$
Here is another variation along a theme which like @DavidG approach uses Feynman's trick of differentiating under the integral sign.
Let
$$I(a) = int_0^{frac{pi}{2}} operatorname{arcsinh} (a tan x) , dx, qquad a > 1.$$
We are required to find $I(2)$. We start by finding $I(1)$ first, which will be needed later on.
For $a = 1$ we have
begin{align}
I(1) &= int_0^{frac{pi}{2}} operatorname{arcsinh} (tan x) , dx\
&= int_0^{frac{pi}{2}} ln left (frac{1 + sin x}{cos x} right ) , dx\
&= int_0^{frac{pi}{2}} ln (1 + sin x) , dx - int_0^{frac{pi}{2}} ln (cos x) , dx.
end{align}
Now the first of these integrals can be found, for example, by rewriting it as
begin{align}
int_0^{frac{pi}{2}} ln (1 + sin x) , dx &= int_0^{frac{pi}{2}} ln (1 + cos x) , dx\
&= int_0^{frac{pi}{2}} ln left (frac{1}{2} cos^2 frac{x}{2} right ) , dx\
&= frac{pi}{2} ln 2 + 4 int_0^{frac{pi}{4}} ln (cos x) , dx
end{align}
and then employing the Fourier series representation for $ln (cos x)$ found here. The final result is
$$int_0^{frac{pi}{2}} ln (1 + sin x) , dx = 2 mathbf{G} - frac{pi}{2} ln 2.$$
Here $mathbf{G}$ is Catalan's constant.
The second of the integrals is very well known. Here
$$int_0^{frac{pi}{2}} ln (cos x) , dx = - frac{pi}{2} ln 2.$$
Thus $I(1) = 2 mathbf{G}$.
Moving to the main event, by differentiating under the integral sign with respect to $a$ we have
begin{align}
I'(a) &= int_0^{frac{pi}{2}} frac{tan x}{sqrt{a^2 tan^2 x + 1}} , dx\
&= int_0^{frac{pi}{2}} frac{sin x}{sqrt{a^2 - (a^2 - 1) cos^2 x}} , dx.
end{align}
Observe the term $(a^2 - 1)$ is positive since $a > 1$. Letting $u = cos x$ one has
$$I'(a) = int_0^1 frac{du}{sqrt{a^2 - (a^2 - 1) u^2}} = frac{1}{sqrt{a^2 - 1}} sin^{-1} left (frac{sqrt{a^2 - 1}}{a} right ).$$
As we require $I(2)$ observe that
$$I(2) - I(1) = int_1^2 I'(a) , da.$$
Thus
$$I(2) = 2 mathbf{G} + int_1^2 frac{1}{sqrt{a^2 - 1}} sin^{-1} left (frac{sqrt{a^2 - 1}}{a} right ) , da.$$
Integrating by parts leads to
$$I(2) = 2 mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - int_1^2 frac{cosh^{-1} a}{a sqrt{a^2 - 1}} , da.$$
Now let $a = cosh t$. This gives
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - int_0^{ln(2 + sqrt{3})} frac{t}{cosh t} , dt.$$
Then let $t = ln y$. This gives
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2int_1^{2 + sqrt{3}} frac{ln y}{1 + y^2} , dy.$$
Now let $y = tan theta$. Then we have
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2int_{frac{pi}{4}}^{frac{5pi}{12}} ln (tan theta) , dtheta.$$
Finally, enforcing a substitution of $theta mapsto dfrac{pi}{2} - theta$ leads to
begin{align}
I(2) &= 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) + 2 int_{frac{pi}{12}}^{frac{pi}{4}} ln (tan theta) , dtheta\
&= 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) + 2int_0^{frac{pi}{4}} ln (tan theta) , dtheta - 2 int_0^{frac{pi}{12}} ln (tan theta) , dtheta.
end{align}
For the first of the integrals we have
$$int_0^{frac{pi}{4}} ln (tan x) , dx = -mathbf{G}.$$
While for the second of the integrals we have
$$int_0^{frac{pi}{12}} ln (tan x) , dx = -frac{2}{3} mathbf{G}.$$
So finally
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2mathbf{G} + frac{4}{3} mathbf{G},$$
or
$$int_0^{frac{pi}{2}} operatorname{arcsinh} (2 tan x) , dx = frac{pi}{3} ln (2 + sqrt{3}) + frac{4}{3} mathbf{G},$$
as announced.
answered Jan 23 at 2:51
omegadotomegadot
6,3972829
$endgroup$
1
$begingroup$
begin{align}A&=int_0^{frac{pi}{4}}ln(cos x),dx\ B&=int_0^{frac{pi}{4}}ln(sin x),dx\ B-A&=int_0^{frac{pi}{4}}ln(tan x),dx\ &=-text{G}\ A+B&=int_0^{frac{pi}{4}}ln(sin xcos x),dx\ &=int_0^{frac{pi}{4}}lnleft(frac{sin(2x)}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(frac{sin x}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(sin xright),dx-frac{pi}{4}ln 2\ &=-frac{1}{2}piln 2\ A&=frac{1}{2}left(A+B-left(B-Aright)right)\ &=frac{1}{2}text{G}-frac{1}{4}piln 2 end{align}
$endgroup$
– FDP
Jan 23 at 13:44
$begingroup$
Yes, very nice.
$endgroup$
– omegadot
Jan 24 at 1:29
add a comment |
$begingroup$
Here is another variation along a theme which like @DavidG approach uses Feynman's trick of differentiating under the integral sign.
Let
$$I(a) = int_0^{frac{pi}{2}} operatorname{arcsinh} (a tan x) , dx, qquad a > 1.$$
We are required to find $I(2)$. We start by finding $I(1)$ first, which will be needed later on.
For $a = 1$ we have
begin{align}
I(1) &= int_0^{frac{pi}{2}} operatorname{arcsinh} (tan x) , dx\
&= int_0^{frac{pi}{2}} ln left (frac{1 + sin x}{cos x} right ) , dx\
&= int_0^{frac{pi}{2}} ln (1 + sin x) , dx - int_0^{frac{pi}{2}} ln (cos x) , dx.
end{align}
Now the first of these integrals can be found, for example, by rewriting it as
begin{align}
int_0^{frac{pi}{2}} ln (1 + sin x) , dx &= int_0^{frac{pi}{2}} ln (1 + cos x) , dx\
&= int_0^{frac{pi}{2}} ln left (frac{1}{2} cos^2 frac{x}{2} right ) , dx\
&= frac{pi}{2} ln 2 + 4 int_0^{frac{pi}{4}} ln (cos x) , dx
end{align}
and then employing the Fourier series representation for $ln (cos x)$ found here. The final result is
$$int_0^{frac{pi}{2}} ln (1 + sin x) , dx = 2 mathbf{G} - frac{pi}{2} ln 2.$$
Here $mathbf{G}$ is Catalan's constant.
The second of the integrals is very well known. Here
$$int_0^{frac{pi}{2}} ln (cos x) , dx = - frac{pi}{2} ln 2.$$
Thus $I(1) = 2 mathbf{G}$.
Moving to the main event, by differentiating under the integral sign with respect to $a$ we have
begin{align}
I'(a) &= int_0^{frac{pi}{2}} frac{tan x}{sqrt{a^2 tan^2 x + 1}} , dx\
&= int_0^{frac{pi}{2}} frac{sin x}{sqrt{a^2 - (a^2 - 1) cos^2 x}} , dx.
end{align}
Observe the term $(a^2 - 1)$ is positive since $a > 1$. Letting $u = cos x$ one has
$$I'(a) = int_0^1 frac{du}{sqrt{a^2 - (a^2 - 1) u^2}} = frac{1}{sqrt{a^2 - 1}} sin^{-1} left (frac{sqrt{a^2 - 1}}{a} right ).$$
As we require $I(2)$ observe that
$$I(2) - I(1) = int_1^2 I'(a) , da.$$
Thus
$$I(2) = 2 mathbf{G} + int_1^2 frac{1}{sqrt{a^2 - 1}} sin^{-1} left (frac{sqrt{a^2 - 1}}{a} right ) , da.$$
Integrating by parts leads to
$$I(2) = 2 mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - int_1^2 frac{cosh^{-1} a}{a sqrt{a^2 - 1}} , da.$$
Now let $a = cosh t$. This gives
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - int_0^{ln(2 + sqrt{3})} frac{t}{cosh t} , dt.$$
Then let $t = ln y$. This gives
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2int_1^{2 + sqrt{3}} frac{ln y}{1 + y^2} , dy.$$
Now let $y = tan theta$. Then we have
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2int_{frac{pi}{4}}^{frac{5pi}{12}} ln (tan theta) , dtheta.$$
Finally, enforcing a substitution of $theta mapsto dfrac{pi}{2} - theta$ leads to
begin{align}
I(2) &= 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) + 2 int_{frac{pi}{12}}^{frac{pi}{4}} ln (tan theta) , dtheta\
&= 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) + 2int_0^{frac{pi}{4}} ln (tan theta) , dtheta - 2 int_0^{frac{pi}{12}} ln (tan theta) , dtheta.
end{align}
For the first of the integrals we have
$$int_0^{frac{pi}{4}} ln (tan x) , dx = -mathbf{G}.$$
While for the second of the integrals we have
$$int_0^{frac{pi}{12}} ln (tan x) , dx = -frac{2}{3} mathbf{G}.$$
So finally
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2mathbf{G} + frac{4}{3} mathbf{G},$$
or
$$int_0^{frac{pi}{2}} operatorname{arcsinh} (2 tan x) , dx = frac{pi}{3} ln (2 + sqrt{3}) + frac{4}{3} mathbf{G},$$
as announced.
answered Jan 23 at 2:51
omegadotomegadot
6,3972829
$endgroup$
Here is another variation along a theme which like @DavidG approach uses Feynman's trick of differentiating under the integral sign.
Let
$$I(a) = int_0^{frac{pi}{2}} operatorname{arcsinh} (a tan x) , dx, qquad a > 1.$$
We are required to find $I(2)$. We start by finding $I(1)$ first, which will be needed later on.
For $a = 1$ we have
begin{align}
I(1) &= int_0^{frac{pi}{2}} operatorname{arcsinh} (tan x) , dx\
&= int_0^{frac{pi}{2}} ln left (frac{1 + sin x}{cos x} right ) , dx\
&= int_0^{frac{pi}{2}} ln (1 + sin x) , dx - int_0^{frac{pi}{2}} ln (cos x) , dx.
end{align}
Now the first of these integrals can be found, for example, by rewriting it as
begin{align}
int_0^{frac{pi}{2}} ln (1 + sin x) , dx &= int_0^{frac{pi}{2}} ln (1 + cos x) , dx\
&= int_0^{frac{pi}{2}} ln left (frac{1}{2} cos^2 frac{x}{2} right ) , dx\
&= frac{pi}{2} ln 2 + 4 int_0^{frac{pi}{4}} ln (cos x) , dx
end{align}
and then employing the Fourier series representation for $ln (cos x)$ found here. The final result is
$$int_0^{frac{pi}{2}} ln (1 + sin x) , dx = 2 mathbf{G} - frac{pi}{2} ln 2.$$
Here $mathbf{G}$ is Catalan's constant.
The second of the integrals is very well known. Here
$$int_0^{frac{pi}{2}} ln (cos x) , dx = - frac{pi}{2} ln 2.$$
Thus $I(1) = 2 mathbf{G}$.
Moving to the main event, by differentiating under the integral sign with respect to $a$ we have
begin{align}
I'(a) &= int_0^{frac{pi}{2}} frac{tan x}{sqrt{a^2 tan^2 x + 1}} , dx\
&= int_0^{frac{pi}{2}} frac{sin x}{sqrt{a^2 - (a^2 - 1) cos^2 x}} , dx.
end{align}
Observe the term $(a^2 - 1)$ is positive since $a > 1$. Letting $u = cos x$ one has
$$I'(a) = int_0^1 frac{du}{sqrt{a^2 - (a^2 - 1) u^2}} = frac{1}{sqrt{a^2 - 1}} sin^{-1} left (frac{sqrt{a^2 - 1}}{a} right ).$$
As we require $I(2)$ observe that
$$I(2) - I(1) = int_1^2 I'(a) , da.$$
Thus
$$I(2) = 2 mathbf{G} + int_1^2 frac{1}{sqrt{a^2 - 1}} sin^{-1} left (frac{sqrt{a^2 - 1}}{a} right ) , da.$$
Integrating by parts leads to
$$I(2) = 2 mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - int_1^2 frac{cosh^{-1} a}{a sqrt{a^2 - 1}} , da.$$
Now let $a = cosh t$. This gives
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - int_0^{ln(2 + sqrt{3})} frac{t}{cosh t} , dt.$$
Then let $t = ln y$. This gives
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2int_1^{2 + sqrt{3}} frac{ln y}{1 + y^2} , dy.$$
Now let $y = tan theta$. Then we have
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2int_{frac{pi}{4}}^{frac{5pi}{12}} ln (tan theta) , dtheta.$$
Finally, enforcing a substitution of $theta mapsto dfrac{pi}{2} - theta$ leads to
begin{align}
I(2) &= 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) + 2 int_{frac{pi}{12}}^{frac{pi}{4}} ln (tan theta) , dtheta\
&= 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) + 2int_0^{frac{pi}{4}} ln (tan theta) , dtheta - 2 int_0^{frac{pi}{12}} ln (tan theta) , dtheta.
end{align}
For the first of the integrals we have
$$int_0^{frac{pi}{4}} ln (tan x) , dx = -mathbf{G}.$$
While for the second of the integrals we have
$$int_0^{frac{pi}{12}} ln (tan x) , dx = -frac{2}{3} mathbf{G}.$$
So finally
$$I(2) = 2mathbf{G} + frac{pi}{3} ln (2 + sqrt{3}) - 2mathbf{G} + frac{4}{3} mathbf{G},$$
or
$$int_0^{frac{pi}{2}} operatorname{arcsinh} (2 tan x) , dx = frac{pi}{3} ln (2 + sqrt{3}) + frac{4}{3} mathbf{G},$$
as announced.
answered Jan 23 at 2:51
omegadotomegadot
6,3972829
answered Jan 23 at 2:51
omegadotomegadot
6,3972829
answered Jan 23 at 2:51
omegadotomegadot
6,3972829
answered Jan 23 at 2:51
omegadotomegadot
6,3972829
6,3972829
1
$begingroup$
begin{align}A&=int_0^{frac{pi}{4}}ln(cos x),dx\ B&=int_0^{frac{pi}{4}}ln(sin x),dx\ B-A&=int_0^{frac{pi}{4}}ln(tan x),dx\ &=-text{G}\ A+B&=int_0^{frac{pi}{4}}ln(sin xcos x),dx\ &=int_0^{frac{pi}{4}}lnleft(frac{sin(2x)}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(frac{sin x}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(sin xright),dx-frac{pi}{4}ln 2\ &=-frac{1}{2}piln 2\ A&=frac{1}{2}left(A+B-left(B-Aright)right)\ &=frac{1}{2}text{G}-frac{1}{4}piln 2 end{align}
$endgroup$
– FDP
Jan 23 at 13:44
$begingroup$
Yes, very nice.
$endgroup$
– omegadot
Jan 24 at 1:29
add a comment |
1
$begingroup$
begin{align}A&=int_0^{frac{pi}{4}}ln(cos x),dx\ B&=int_0^{frac{pi}{4}}ln(sin x),dx\ B-A&=int_0^{frac{pi}{4}}ln(tan x),dx\ &=-text{G}\ A+B&=int_0^{frac{pi}{4}}ln(sin xcos x),dx\ &=int_0^{frac{pi}{4}}lnleft(frac{sin(2x)}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(frac{sin x}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(sin xright),dx-frac{pi}{4}ln 2\ &=-frac{1}{2}piln 2\ A&=frac{1}{2}left(A+B-left(B-Aright)right)\ &=frac{1}{2}text{G}-frac{1}{4}piln 2 end{align}
$endgroup$
– FDP
Jan 23 at 13:44
$begingroup$
Yes, very nice.
$endgroup$
– omegadot
Jan 24 at 1:29
1
1
$begingroup$
begin{align}A&=int_0^{frac{pi}{4}}ln(cos x),dx\ B&=int_0^{frac{pi}{4}}ln(sin x),dx\ B-A&=int_0^{frac{pi}{4}}ln(tan x),dx\ &=-text{G}\ A+B&=int_0^{frac{pi}{4}}ln(sin xcos x),dx\ &=int_0^{frac{pi}{4}}lnleft(frac{sin(2x)}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(frac{sin x}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(sin xright),dx-frac{pi}{4}ln 2\ &=-frac{1}{2}piln 2\ A&=frac{1}{2}left(A+B-left(B-Aright)right)\ &=frac{1}{2}text{G}-frac{1}{4}piln 2 end{align}
$endgroup$
– FDP
Jan 23 at 13:44
$begingroup$
begin{align}A&=int_0^{frac{pi}{4}}ln(cos x),dx\ B&=int_0^{frac{pi}{4}}ln(sin x),dx\ B-A&=int_0^{frac{pi}{4}}ln(tan x),dx\ &=-text{G}\ A+B&=int_0^{frac{pi}{4}}ln(sin xcos x),dx\ &=int_0^{frac{pi}{4}}lnleft(frac{sin(2x)}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(frac{sin x}{2}right),dx\ &=frac{1}{2}int_0^{frac{pi}{2}}lnleft(sin xright),dx-frac{pi}{4}ln 2\ &=-frac{1}{2}piln 2\ A&=frac{1}{2}left(A+B-left(B-Aright)right)\ &=frac{1}{2}text{G}-frac{1}{4}piln 2 end{align}
$endgroup$
– FDP
Jan 23 at 13:44
$begingroup$
Yes, very nice.
$endgroup$
– omegadot
Jan 24 at 1:29
$begingroup$
Yes, very nice.
$endgroup$
– omegadot
Jan 24 at 1:29
add a comment |
$begingroup$
Sorry in a rush, so only a PARTIAL SOLUTION:
Here I will employ Feynman's Trick:
begin{equation}
I = int_0^{infty}frac{operatorname{arcsinh(2x)}}{1 + x^2}:dx
end{equation}
Let
begin{equation}
J(t) = int_0^{infty}frac{operatorname{arcsinh(tx)}}{1 + x^2}:dx
end{equation}
We observe that $J(2) = I$ and $J(0) = 0$. Using Leibniz's Integral rule we differentiate with respect to '$t$':
begin{equation}
J'(t) = int_0^{infty}frac{x}{sqrt{1 + t^2x^2}}frac{1}{1 + x^2}:dx = left[frac{1}{sqrt{t^2 - 1}} cdot arctanleft(sqrt{frac{1 + t^2x^2}{t^2 - 1}} right) right]_0^{infty} = frac{1}{sqrt{t^2 - 1}}left[frac{pi}{2} - arctanleft(frac{1}{sqrt{t^2 -1}} right) right]
end{equation}
Thus,
begin{align}
J(t) &= int frac{1}{sqrt{t^2 - 1}}left[frac{pi}{2} - arctanleft(frac{x}{sqrt{t^2 -1}} right) right]:dt \
&= int frac{pi}{2}cdot frac{1}{sqrt{t^2 - 1}}:dt - int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt = I_1 - I_2
end{align}
For $I_1$:
begin{equation}
I_1 = int frac{pi}{2}cdot frac{1}{sqrt{t^2 - 1}}:dt = frac{pi}{2}lnleft| sqrt{t^2 - 1} + tright| + C_1
end{equation}
Where $C_1$ is the constant of integration. Note that $I_1(2) = lnleft|2 + sqrt{3} right|$
For $I_2$:
begin{equation}
I_2 = int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt
end{equation}
Unfortunately this is not so easy to evaluate. I will first make the substitution $u = frac{1}{sqrt{t^2 - 1}}$:
begin{align}
I_2 &= int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt = int u cdot arctan(u) cdot frac{-u^4}{sqrt{1 +u^2}}:du\
& = - int frac{-u^5}{sqrt{1 +u^2}} cdot arctan(u) :du
end{align}
answered Jan 22 at 10:07
DavidGDavidG
2,5061726
$endgroup$
$begingroup$
Will post up full solution later.
$endgroup$
– DavidG
Jan 22 at 10:08
add a comment |
$begingroup$
Sorry in a rush, so only a PARTIAL SOLUTION:
Here I will employ Feynman's Trick:
begin{equation}
I = int_0^{infty}frac{operatorname{arcsinh(2x)}}{1 + x^2}:dx
end{equation}
Let
begin{equation}
J(t) = int_0^{infty}frac{operatorname{arcsinh(tx)}}{1 + x^2}:dx
end{equation}
We observe that $J(2) = I$ and $J(0) = 0$. Using Leibniz's Integral rule we differentiate with respect to '$t$':
begin{equation}
J'(t) = int_0^{infty}frac{x}{sqrt{1 + t^2x^2}}frac{1}{1 + x^2}:dx = left[frac{1}{sqrt{t^2 - 1}} cdot arctanleft(sqrt{frac{1 + t^2x^2}{t^2 - 1}} right) right]_0^{infty} = frac{1}{sqrt{t^2 - 1}}left[frac{pi}{2} - arctanleft(frac{1}{sqrt{t^2 -1}} right) right]
end{equation}
Thus,
begin{align}
J(t) &= int frac{1}{sqrt{t^2 - 1}}left[frac{pi}{2} - arctanleft(frac{x}{sqrt{t^2 -1}} right) right]:dt \
&= int frac{pi}{2}cdot frac{1}{sqrt{t^2 - 1}}:dt - int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt = I_1 - I_2
end{align}
For $I_1$:
begin{equation}
I_1 = int frac{pi}{2}cdot frac{1}{sqrt{t^2 - 1}}:dt = frac{pi}{2}lnleft| sqrt{t^2 - 1} + tright| + C_1
end{equation}
Where $C_1$ is the constant of integration. Note that $I_1(2) = lnleft|2 + sqrt{3} right|$
For $I_2$:
begin{equation}
I_2 = int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt
end{equation}
Unfortunately this is not so easy to evaluate. I will first make the substitution $u = frac{1}{sqrt{t^2 - 1}}$:
begin{align}
I_2 &= int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt = int u cdot arctan(u) cdot frac{-u^4}{sqrt{1 +u^2}}:du\
& = - int frac{-u^5}{sqrt{1 +u^2}} cdot arctan(u) :du
end{align}
answered Jan 22 at 10:07
DavidGDavidG
2,5061726
$endgroup$
$begingroup$
Will post up full solution later.
$endgroup$
– DavidG
Jan 22 at 10:08
add a comment |
$begingroup$
Sorry in a rush, so only a PARTIAL SOLUTION:
Here I will employ Feynman's Trick:
begin{equation}
I = int_0^{infty}frac{operatorname{arcsinh(2x)}}{1 + x^2}:dx
end{equation}
Let
begin{equation}
J(t) = int_0^{infty}frac{operatorname{arcsinh(tx)}}{1 + x^2}:dx
end{equation}
We observe that $J(2) = I$ and $J(0) = 0$. Using Leibniz's Integral rule we differentiate with respect to '$t$':
begin{equation}
J'(t) = int_0^{infty}frac{x}{sqrt{1 + t^2x^2}}frac{1}{1 + x^2}:dx = left[frac{1}{sqrt{t^2 - 1}} cdot arctanleft(sqrt{frac{1 + t^2x^2}{t^2 - 1}} right) right]_0^{infty} = frac{1}{sqrt{t^2 - 1}}left[frac{pi}{2} - arctanleft(frac{1}{sqrt{t^2 -1}} right) right]
end{equation}
Thus,
begin{align}
J(t) &= int frac{1}{sqrt{t^2 - 1}}left[frac{pi}{2} - arctanleft(frac{x}{sqrt{t^2 -1}} right) right]:dt \
&= int frac{pi}{2}cdot frac{1}{sqrt{t^2 - 1}}:dt - int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt = I_1 - I_2
end{align}
For $I_1$:
begin{equation}
I_1 = int frac{pi}{2}cdot frac{1}{sqrt{t^2 - 1}}:dt = frac{pi}{2}lnleft| sqrt{t^2 - 1} + tright| + C_1
end{equation}
Where $C_1$ is the constant of integration. Note that $I_1(2) = lnleft|2 + sqrt{3} right|$
For $I_2$:
begin{equation}
I_2 = int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt
end{equation}
Unfortunately this is not so easy to evaluate. I will first make the substitution $u = frac{1}{sqrt{t^2 - 1}}$:
begin{align}
I_2 &= int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt = int u cdot arctan(u) cdot frac{-u^4}{sqrt{1 +u^2}}:du\
& = - int frac{-u^5}{sqrt{1 +u^2}} cdot arctan(u) :du
end{align}
answered Jan 22 at 10:07
DavidGDavidG
2,5061726
$endgroup$
Sorry in a rush, so only a PARTIAL SOLUTION:
Here I will employ Feynman's Trick:
begin{equation}
I = int_0^{infty}frac{operatorname{arcsinh(2x)}}{1 + x^2}:dx
end{equation}
Let
begin{equation}
J(t) = int_0^{infty}frac{operatorname{arcsinh(tx)}}{1 + x^2}:dx
end{equation}
We observe that $J(2) = I$ and $J(0) = 0$. Using Leibniz's Integral rule we differentiate with respect to '$t$':
begin{equation}
J'(t) = int_0^{infty}frac{x}{sqrt{1 + t^2x^2}}frac{1}{1 + x^2}:dx = left[frac{1}{sqrt{t^2 - 1}} cdot arctanleft(sqrt{frac{1 + t^2x^2}{t^2 - 1}} right) right]_0^{infty} = frac{1}{sqrt{t^2 - 1}}left[frac{pi}{2} - arctanleft(frac{1}{sqrt{t^2 -1}} right) right]
end{equation}
Thus,
begin{align}
J(t) &= int frac{1}{sqrt{t^2 - 1}}left[frac{pi}{2} - arctanleft(frac{x}{sqrt{t^2 -1}} right) right]:dt \
&= int frac{pi}{2}cdot frac{1}{sqrt{t^2 - 1}}:dt - int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt = I_1 - I_2
end{align}
For $I_1$:
begin{equation}
I_1 = int frac{pi}{2}cdot frac{1}{sqrt{t^2 - 1}}:dt = frac{pi}{2}lnleft| sqrt{t^2 - 1} + tright| + C_1
end{equation}
Where $C_1$ is the constant of integration. Note that $I_1(2) = lnleft|2 + sqrt{3} right|$
For $I_2$:
begin{equation}
I_2 = int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt
end{equation}
Unfortunately this is not so easy to evaluate. I will first make the substitution $u = frac{1}{sqrt{t^2 - 1}}$:
begin{align}
I_2 &= int frac{1}{sqrt{t^2 - 1}}arctanleft(frac{1}{sqrt{t^2 -1}} right):dt = int u cdot arctan(u) cdot frac{-u^4}{sqrt{1 +u^2}}:du\
& = - int frac{-u^5}{sqrt{1 +u^2}} cdot arctan(u) :du
end{align}
answered Jan 22 at 10:07
DavidGDavidG
2,5061726
edited Jan 23 at 6:07
answered Jan 22 at 10:07
DavidGDavidG
2,5061726
answered Jan 22 at 10:07
DavidGDavidG
2,5061726
answered Jan 22 at 10:07
DavidGDavidG
2,5061726
2,5061726
$begingroup$
Will post up full solution later.
$endgroup$
– DavidG
Jan 22 at 10:08
add a comment |
$begingroup$
Will post up full solution later.
$endgroup$
– DavidG
Jan 22 at 10:08
$begingroup$
Will post up full solution later.
$endgroup$
– DavidG
Jan 22 at 10:08
$begingroup$
Will post up full solution later.
$endgroup$
– DavidG
Jan 22 at 10:08
add a comment |
$begingroup$
$$
begin{align}newcommand{arcsinh}{operatorname{arcsinh}}
int_0^inftyfrac{arcsinh(x)}{1+x^2},mathrm{d}x
&=int_0^inftyfrac{x,mathrm{d}x}{cosh(x)}tag{1a}\
&=int_0^inftyfrac{2x,mathrm{d}x}{e^x+e^{-x}}tag{1b}\
&=2sum_{k=0}^infty(-1)^kint_0^infty x,e^{-(2k+1)x},mathrm{d}xtag{1c}\ &=2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}tag{1d}\[6pt]
&=2mathrm{G}tag{1e}
end{align}
$$
Explanation:
$text{(1a)}$: substitute $xmapstosinh(x)$
$text{(1b)}$: write $cosh(x)$ as exponentials
$text{(1c)}$: expand $left(e^x+e^{-x}right)^{-1}$ as a power series
$text{(1d)}$: evaluate the integral
$text{(1e)}$: use the definition of Catalan's Constant
$$
begin{align}
frac{mathrm{d}}{mathrm{d}a}int_0^inftyfrac{arcsinh(ax)}{1+x^2},mathrm{d}x
&=int_0^inftyfrac{x}{1+x^2}frac{mathrm{d}x}{sqrt{1+a^2x^2}}tag{2a}\
&=int_0^inftyfrac{x}{a^2+x^2}frac{mathrm{d}x}{sqrt{1+x^2}}tag{2b}\
&=int_0^{pi/2}frac{mathrm{d}sec(x)}{a^2+tan^2(x)}tag{2c}\
&=int_1^inftyfrac{mathrm{d}x}{x^2+left(a^2-1right)}tag{2d}\
&=frac{arctanleft(sqrt{a^2-1}right)}{sqrt{a^2-1}}tag{2e}
end{align}
$$
Explanation:
$text{(2a)}$: $frac{mathrm{d}}{mathrm{d}a}arcsinh(ax)=frac{x}{sqrt{1+a^2x^2}}$
$text{(2b)}$: substitute $xmapsto x/a$
$text{(2c)}$: substitute $xmapstotan(x)$
$text{(2d)}$: substitute $sec(x)mapsto x$
$text{(2e)}$: arctan integral
Therefore,
$$
begin{align}newcommand{Li}{operatorname{Li}}
int_0^inftyfrac{arcsinh(2x)}{1+x^2},mathrm{d}x
&=2mathrm{G}+int_1^2frac{arctanleft(sqrt{a^2-1}right)}{sqrt{a^2-1}},mathrm{d}atag{3a}\
&=2mathrm{G}+int_0^{pi/3}frac{a}{cos(a)},mathrm{d}atag{3b}\
&=2mathrm{G}+2int_0^{pi/3}frac{a}{e^{ia}+e^{-ia}},mathrm{d}atag{3c}\
&=2mathrm{G}+2sum_{k=0}^infty(-1)^kint_0^{pi/3}ae^{-i(2k+1)a},mathrm{d}atag{3d}\
&=2mathrm{G}+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}int_0^{(2k+1)pi/3}ae^{-ia},mathrm{d}atag{3e}\
&=2mathrm{G}+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}left[(1+i(2k+1)pi/3)e^{-i(2k+1)pi/3}-1right]tag{3f}\
&=2mathrm{G}+fracpi3logleft(frac{2+sqrt3+i}{2-sqrt3-i}right)+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}left[e^{-i(2k+1)pi/3}-1right]tag{3g}\
&=2mathrm{G}+fracpi3logleft(left(2+sqrt3right)iright)
+sum_{k=0}^infty(-1)^kscriptsizeleft[frac{-1-isqrt3}{(6k+1)^2}-frac{-4}{(6k+3)^2}+frac{-1+isqrt3}{(6k+5)^2}right]tag{3h}\
&=frac43mathrm{G}+fracpi3logleft(2+sqrt3right)+ileft[frac{pi^2}6-sqrt3sum_{k=0}^infty(-1)^kscriptsizeleft(frac1{(6k+1)^2}-frac1{(6k+5)^2}right)right]tag{3i}\
&=frac43mathrm{G}+fracpi3logleft(2+sqrt3right)+ileft[frac{pi^2}6-frac{sqrt3}{36}sum_{kinmathbb{Z}}frac{(-1)^k}{left(k+frac16right)^2}right]tag{3j}\
&=bbox[5px,border:2px solid #C0A000]{frac43mathrm{G}+fracpi3logleft(2+sqrt3right)}tag{3k}
end{align}
$$
Explanation:
$text{(3a)}$: apply $(1)$ and $(2)$
$text{(3b)}$: substitute $amapstosec(a)$
$text{(3c)}$: write $cos(x)$ as exponentials
$text{(3d)}$: expand $left(e^{ix}+e^{-ix}right)^{-1}$ as a power series
$text{(3e)}$: substitute $amapsto a/(2k+1)$
$text{(3f)}$: integrate
$text{(3g)}$: recognize the series for $arctan(x)=frac1{2i}logleft(frac{1+ix}{1-ix}right)$
$text{(3h)}$: evaluate the exponentials
$text{(3i)}$: separate the real and imaginary parts
$text{(3j)}$: rewrite the sum
$text{(3k)}$: $sum_{kinmathbb{Z}}frac{(-1)^k}{(k+x)^2}=pi^2cot(pi x)csc(pi x)$
answered Jan 24 at 8:44
robjohn♦robjohn
269k27309635
$endgroup$
$begingroup$
Seems like there should be a^2in the denominator of the sum of the explanation (3k)
$endgroup$
– Kemono Chen
Jan 24 at 9:03
$begingroup$
@KemonoChen: Indeed. I put the parentheses there because I meant to square the denominator. Thanks. While it's true that we can just take the real part of the right side since the quantity is real, I wanted to make sure I hadn't made a mistake and verify that the imaginary part was $0$.
$endgroup$
– robjohn♦
Jan 24 at 9:25
add a comment |
$begingroup$
$$
begin{align}newcommand{arcsinh}{operatorname{arcsinh}}
int_0^inftyfrac{arcsinh(x)}{1+x^2},mathrm{d}x
&=int_0^inftyfrac{x,mathrm{d}x}{cosh(x)}tag{1a}\
&=int_0^inftyfrac{2x,mathrm{d}x}{e^x+e^{-x}}tag{1b}\
&=2sum_{k=0}^infty(-1)^kint_0^infty x,e^{-(2k+1)x},mathrm{d}xtag{1c}\ &=2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}tag{1d}\[6pt]
&=2mathrm{G}tag{1e}
end{align}
$$
Explanation:
$text{(1a)}$: substitute $xmapstosinh(x)$
$text{(1b)}$: write $cosh(x)$ as exponentials
$text{(1c)}$: expand $left(e^x+e^{-x}right)^{-1}$ as a power series
$text{(1d)}$: evaluate the integral
$text{(1e)}$: use the definition of Catalan's Constant
$$
begin{align}
frac{mathrm{d}}{mathrm{d}a}int_0^inftyfrac{arcsinh(ax)}{1+x^2},mathrm{d}x
&=int_0^inftyfrac{x}{1+x^2}frac{mathrm{d}x}{sqrt{1+a^2x^2}}tag{2a}\
&=int_0^inftyfrac{x}{a^2+x^2}frac{mathrm{d}x}{sqrt{1+x^2}}tag{2b}\
&=int_0^{pi/2}frac{mathrm{d}sec(x)}{a^2+tan^2(x)}tag{2c}\
&=int_1^inftyfrac{mathrm{d}x}{x^2+left(a^2-1right)}tag{2d}\
&=frac{arctanleft(sqrt{a^2-1}right)}{sqrt{a^2-1}}tag{2e}
end{align}
$$
Explanation:
$text{(2a)}$: $frac{mathrm{d}}{mathrm{d}a}arcsinh(ax)=frac{x}{sqrt{1+a^2x^2}}$
$text{(2b)}$: substitute $xmapsto x/a$
$text{(2c)}$: substitute $xmapstotan(x)$
$text{(2d)}$: substitute $sec(x)mapsto x$
$text{(2e)}$: arctan integral
Therefore,
$$
begin{align}newcommand{Li}{operatorname{Li}}
int_0^inftyfrac{arcsinh(2x)}{1+x^2},mathrm{d}x
&=2mathrm{G}+int_1^2frac{arctanleft(sqrt{a^2-1}right)}{sqrt{a^2-1}},mathrm{d}atag{3a}\
&=2mathrm{G}+int_0^{pi/3}frac{a}{cos(a)},mathrm{d}atag{3b}\
&=2mathrm{G}+2int_0^{pi/3}frac{a}{e^{ia}+e^{-ia}},mathrm{d}atag{3c}\
&=2mathrm{G}+2sum_{k=0}^infty(-1)^kint_0^{pi/3}ae^{-i(2k+1)a},mathrm{d}atag{3d}\
&=2mathrm{G}+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}int_0^{(2k+1)pi/3}ae^{-ia},mathrm{d}atag{3e}\
&=2mathrm{G}+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}left[(1+i(2k+1)pi/3)e^{-i(2k+1)pi/3}-1right]tag{3f}\
&=2mathrm{G}+fracpi3logleft(frac{2+sqrt3+i}{2-sqrt3-i}right)+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}left[e^{-i(2k+1)pi/3}-1right]tag{3g}\
&=2mathrm{G}+fracpi3logleft(left(2+sqrt3right)iright)
+sum_{k=0}^infty(-1)^kscriptsizeleft[frac{-1-isqrt3}{(6k+1)^2}-frac{-4}{(6k+3)^2}+frac{-1+isqrt3}{(6k+5)^2}right]tag{3h}\
&=frac43mathrm{G}+fracpi3logleft(2+sqrt3right)+ileft[frac{pi^2}6-sqrt3sum_{k=0}^infty(-1)^kscriptsizeleft(frac1{(6k+1)^2}-frac1{(6k+5)^2}right)right]tag{3i}\
&=frac43mathrm{G}+fracpi3logleft(2+sqrt3right)+ileft[frac{pi^2}6-frac{sqrt3}{36}sum_{kinmathbb{Z}}frac{(-1)^k}{left(k+frac16right)^2}right]tag{3j}\
&=bbox[5px,border:2px solid #C0A000]{frac43mathrm{G}+fracpi3logleft(2+sqrt3right)}tag{3k}
end{align}
$$
Explanation:
$text{(3a)}$: apply $(1)$ and $(2)$
$text{(3b)}$: substitute $amapstosec(a)$
$text{(3c)}$: write $cos(x)$ as exponentials
$text{(3d)}$: expand $left(e^{ix}+e^{-ix}right)^{-1}$ as a power series
$text{(3e)}$: substitute $amapsto a/(2k+1)$
$text{(3f)}$: integrate
$text{(3g)}$: recognize the series for $arctan(x)=frac1{2i}logleft(frac{1+ix}{1-ix}right)$
$text{(3h)}$: evaluate the exponentials
$text{(3i)}$: separate the real and imaginary parts
$text{(3j)}$: rewrite the sum
$text{(3k)}$: $sum_{kinmathbb{Z}}frac{(-1)^k}{(k+x)^2}=pi^2cot(pi x)csc(pi x)$
answered Jan 24 at 8:44
robjohn♦robjohn
269k27309635
$endgroup$
$begingroup$
Seems like there should be a^2in the denominator of the sum of the explanation (3k)
$endgroup$
– Kemono Chen
Jan 24 at 9:03
$begingroup$
@KemonoChen: Indeed. I put the parentheses there because I meant to square the denominator. Thanks. While it's true that we can just take the real part of the right side since the quantity is real, I wanted to make sure I hadn't made a mistake and verify that the imaginary part was $0$.
$endgroup$
– robjohn♦
Jan 24 at 9:25
add a comment |
$begingroup$
$$
begin{align}newcommand{arcsinh}{operatorname{arcsinh}}
int_0^inftyfrac{arcsinh(x)}{1+x^2},mathrm{d}x
&=int_0^inftyfrac{x,mathrm{d}x}{cosh(x)}tag{1a}\
&=int_0^inftyfrac{2x,mathrm{d}x}{e^x+e^{-x}}tag{1b}\
&=2sum_{k=0}^infty(-1)^kint_0^infty x,e^{-(2k+1)x},mathrm{d}xtag{1c}\ &=2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}tag{1d}\[6pt]
&=2mathrm{G}tag{1e}
end{align}
$$
Explanation:
$text{(1a)}$: substitute $xmapstosinh(x)$
$text{(1b)}$: write $cosh(x)$ as exponentials
$text{(1c)}$: expand $left(e^x+e^{-x}right)^{-1}$ as a power series
$text{(1d)}$: evaluate the integral
$text{(1e)}$: use the definition of Catalan's Constant
$$
begin{align}
frac{mathrm{d}}{mathrm{d}a}int_0^inftyfrac{arcsinh(ax)}{1+x^2},mathrm{d}x
&=int_0^inftyfrac{x}{1+x^2}frac{mathrm{d}x}{sqrt{1+a^2x^2}}tag{2a}\
&=int_0^inftyfrac{x}{a^2+x^2}frac{mathrm{d}x}{sqrt{1+x^2}}tag{2b}\
&=int_0^{pi/2}frac{mathrm{d}sec(x)}{a^2+tan^2(x)}tag{2c}\
&=int_1^inftyfrac{mathrm{d}x}{x^2+left(a^2-1right)}tag{2d}\
&=frac{arctanleft(sqrt{a^2-1}right)}{sqrt{a^2-1}}tag{2e}
end{align}
$$
Explanation:
$text{(2a)}$: $frac{mathrm{d}}{mathrm{d}a}arcsinh(ax)=frac{x}{sqrt{1+a^2x^2}}$
$text{(2b)}$: substitute $xmapsto x/a$
$text{(2c)}$: substitute $xmapstotan(x)$
$text{(2d)}$: substitute $sec(x)mapsto x$
$text{(2e)}$: arctan integral
Therefore,
$$
begin{align}newcommand{Li}{operatorname{Li}}
int_0^inftyfrac{arcsinh(2x)}{1+x^2},mathrm{d}x
&=2mathrm{G}+int_1^2frac{arctanleft(sqrt{a^2-1}right)}{sqrt{a^2-1}},mathrm{d}atag{3a}\
&=2mathrm{G}+int_0^{pi/3}frac{a}{cos(a)},mathrm{d}atag{3b}\
&=2mathrm{G}+2int_0^{pi/3}frac{a}{e^{ia}+e^{-ia}},mathrm{d}atag{3c}\
&=2mathrm{G}+2sum_{k=0}^infty(-1)^kint_0^{pi/3}ae^{-i(2k+1)a},mathrm{d}atag{3d}\
&=2mathrm{G}+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}int_0^{(2k+1)pi/3}ae^{-ia},mathrm{d}atag{3e}\
&=2mathrm{G}+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}left[(1+i(2k+1)pi/3)e^{-i(2k+1)pi/3}-1right]tag{3f}\
&=2mathrm{G}+fracpi3logleft(frac{2+sqrt3+i}{2-sqrt3-i}right)+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}left[e^{-i(2k+1)pi/3}-1right]tag{3g}\
&=2mathrm{G}+fracpi3logleft(left(2+sqrt3right)iright)
+sum_{k=0}^infty(-1)^kscriptsizeleft[frac{-1-isqrt3}{(6k+1)^2}-frac{-4}{(6k+3)^2}+frac{-1+isqrt3}{(6k+5)^2}right]tag{3h}\
&=frac43mathrm{G}+fracpi3logleft(2+sqrt3right)+ileft[frac{pi^2}6-sqrt3sum_{k=0}^infty(-1)^kscriptsizeleft(frac1{(6k+1)^2}-frac1{(6k+5)^2}right)right]tag{3i}\
&=frac43mathrm{G}+fracpi3logleft(2+sqrt3right)+ileft[frac{pi^2}6-frac{sqrt3}{36}sum_{kinmathbb{Z}}frac{(-1)^k}{left(k+frac16right)^2}right]tag{3j}\
&=bbox[5px,border:2px solid #C0A000]{frac43mathrm{G}+fracpi3logleft(2+sqrt3right)}tag{3k}
end{align}
$$
Explanation:
$text{(3a)}$: apply $(1)$ and $(2)$
$text{(3b)}$: substitute $amapstosec(a)$
$text{(3c)}$: write $cos(x)$ as exponentials
$text{(3d)}$: expand $left(e^{ix}+e^{-ix}right)^{-1}$ as a power series
$text{(3e)}$: substitute $amapsto a/(2k+1)$
$text{(3f)}$: integrate
$text{(3g)}$: recognize the series for $arctan(x)=frac1{2i}logleft(frac{1+ix}{1-ix}right)$
$text{(3h)}$: evaluate the exponentials
$text{(3i)}$: separate the real and imaginary parts
$text{(3j)}$: rewrite the sum
$text{(3k)}$: $sum_{kinmathbb{Z}}frac{(-1)^k}{(k+x)^2}=pi^2cot(pi x)csc(pi x)$
answered Jan 24 at 8:44
robjohn♦robjohn
269k27309635
$endgroup$
$$
begin{align}newcommand{arcsinh}{operatorname{arcsinh}}
int_0^inftyfrac{arcsinh(x)}{1+x^2},mathrm{d}x
&=int_0^inftyfrac{x,mathrm{d}x}{cosh(x)}tag{1a}\
&=int_0^inftyfrac{2x,mathrm{d}x}{e^x+e^{-x}}tag{1b}\
&=2sum_{k=0}^infty(-1)^kint_0^infty x,e^{-(2k+1)x},mathrm{d}xtag{1c}\ &=2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}tag{1d}\[6pt]
&=2mathrm{G}tag{1e}
end{align}
$$
Explanation:
$text{(1a)}$: substitute $xmapstosinh(x)$
$text{(1b)}$: write $cosh(x)$ as exponentials
$text{(1c)}$: expand $left(e^x+e^{-x}right)^{-1}$ as a power series
$text{(1d)}$: evaluate the integral
$text{(1e)}$: use the definition of Catalan's Constant
$$
begin{align}
frac{mathrm{d}}{mathrm{d}a}int_0^inftyfrac{arcsinh(ax)}{1+x^2},mathrm{d}x
&=int_0^inftyfrac{x}{1+x^2}frac{mathrm{d}x}{sqrt{1+a^2x^2}}tag{2a}\
&=int_0^inftyfrac{x}{a^2+x^2}frac{mathrm{d}x}{sqrt{1+x^2}}tag{2b}\
&=int_0^{pi/2}frac{mathrm{d}sec(x)}{a^2+tan^2(x)}tag{2c}\
&=int_1^inftyfrac{mathrm{d}x}{x^2+left(a^2-1right)}tag{2d}\
&=frac{arctanleft(sqrt{a^2-1}right)}{sqrt{a^2-1}}tag{2e}
end{align}
$$
Explanation:
$text{(2a)}$: $frac{mathrm{d}}{mathrm{d}a}arcsinh(ax)=frac{x}{sqrt{1+a^2x^2}}$
$text{(2b)}$: substitute $xmapsto x/a$
$text{(2c)}$: substitute $xmapstotan(x)$
$text{(2d)}$: substitute $sec(x)mapsto x$
$text{(2e)}$: arctan integral
Therefore,
$$
begin{align}newcommand{Li}{operatorname{Li}}
int_0^inftyfrac{arcsinh(2x)}{1+x^2},mathrm{d}x
&=2mathrm{G}+int_1^2frac{arctanleft(sqrt{a^2-1}right)}{sqrt{a^2-1}},mathrm{d}atag{3a}\
&=2mathrm{G}+int_0^{pi/3}frac{a}{cos(a)},mathrm{d}atag{3b}\
&=2mathrm{G}+2int_0^{pi/3}frac{a}{e^{ia}+e^{-ia}},mathrm{d}atag{3c}\
&=2mathrm{G}+2sum_{k=0}^infty(-1)^kint_0^{pi/3}ae^{-i(2k+1)a},mathrm{d}atag{3d}\
&=2mathrm{G}+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}int_0^{(2k+1)pi/3}ae^{-ia},mathrm{d}atag{3e}\
&=2mathrm{G}+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}left[(1+i(2k+1)pi/3)e^{-i(2k+1)pi/3}-1right]tag{3f}\
&=2mathrm{G}+fracpi3logleft(frac{2+sqrt3+i}{2-sqrt3-i}right)+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}left[e^{-i(2k+1)pi/3}-1right]tag{3g}\
&=2mathrm{G}+fracpi3logleft(left(2+sqrt3right)iright)
+sum_{k=0}^infty(-1)^kscriptsizeleft[frac{-1-isqrt3}{(6k+1)^2}-frac{-4}{(6k+3)^2}+frac{-1+isqrt3}{(6k+5)^2}right]tag{3h}\
&=frac43mathrm{G}+fracpi3logleft(2+sqrt3right)+ileft[frac{pi^2}6-sqrt3sum_{k=0}^infty(-1)^kscriptsizeleft(frac1{(6k+1)^2}-frac1{(6k+5)^2}right)right]tag{3i}\
&=frac43mathrm{G}+fracpi3logleft(2+sqrt3right)+ileft[frac{pi^2}6-frac{sqrt3}{36}sum_{kinmathbb{Z}}frac{(-1)^k}{left(k+frac16right)^2}right]tag{3j}\
&=bbox[5px,border:2px solid #C0A000]{frac43mathrm{G}+fracpi3logleft(2+sqrt3right)}tag{3k}
end{align}
$$
Explanation:
$text{(3a)}$: apply $(1)$ and $(2)$
$text{(3b)}$: substitute $amapstosec(a)$
$text{(3c)}$: write $cos(x)$ as exponentials
$text{(3d)}$: expand $left(e^{ix}+e^{-ix}right)^{-1}$ as a power series
$text{(3e)}$: substitute $amapsto a/(2k+1)$
$text{(3f)}$: integrate
$text{(3g)}$: recognize the series for $arctan(x)=frac1{2i}logleft(frac{1+ix}{1-ix}right)$
$text{(3h)}$: evaluate the exponentials
$text{(3i)}$: separate the real and imaginary parts
$text{(3j)}$: rewrite the sum
$text{(3k)}$: $sum_{kinmathbb{Z}}frac{(-1)^k}{(k+x)^2}=pi^2cot(pi x)csc(pi x)$
answered Jan 24 at 8:44
robjohn♦robjohn
269k27309635
edited Jan 24 at 9:24
answered Jan 24 at 8:44
robjohn♦robjohn
269k27309635
answered Jan 24 at 8:44
robjohn♦robjohn
269k27309635
answered Jan 24 at 8:44
robjohn♦robjohn
269k27309635
269k27309635
$begingroup$
Seems like there should be a^2in the denominator of the sum of the explanation (3k)
$endgroup$
– Kemono Chen
Jan 24 at 9:03
$begingroup$
@KemonoChen: Indeed. I put the parentheses there because I meant to square the denominator. Thanks. While it's true that we can just take the real part of the right side since the quantity is real, I wanted to make sure I hadn't made a mistake and verify that the imaginary part was $0$.
$endgroup$
– robjohn♦
Jan 24 at 9:25
add a comment |
$begingroup$
Seems like there should be a^2in the denominator of the sum of the explanation (3k)
$endgroup$
– Kemono Chen
Jan 24 at 9:03
$begingroup$
@KemonoChen: Indeed. I put the parentheses there because I meant to square the denominator. Thanks. While it's true that we can just take the real part of the right side since the quantity is real, I wanted to make sure I hadn't made a mistake and verify that the imaginary part was $0$.
$endgroup$
– robjohn♦
Jan 24 at 9:25
$begingroup$
Seems like there should be a
^2 in the denominator of the sum of the explanation (3k)$endgroup$
– Kemono Chen
Jan 24 at 9:03
$begingroup$
Seems like there should be a
^2 in the denominator of the sum of the explanation (3k)$endgroup$
– Kemono Chen
Jan 24 at 9:03
$begingroup$
@KemonoChen: Indeed. I put the parentheses there because I meant to square the denominator. Thanks. While it's true that we can just take the real part of the right side since the quantity is real, I wanted to make sure I hadn't made a mistake and verify that the imaginary part was $0$.
$endgroup$
– robjohn♦
Jan 24 at 9:25
$begingroup$
@KemonoChen: Indeed. I put the parentheses there because I meant to square the denominator. Thanks. While it's true that we can just take the real part of the right side since the quantity is real, I wanted to make sure I hadn't made a mistake and verify that the imaginary part was $0$.
$endgroup$
– robjohn♦
Jan 24 at 9:25
add a comment |
$begingroup$
This is not an answer.
Your premonition seems to be good. Using another CAS,
$$2int_0^inftyfrac{xcosh( x)}{3+cosh^2(x)},dx=-frac{pi}{4} log left(7-4 sqrt{3}right)-i left(text{Li}_2left(-i left(-2+sqrt{3}right)right)-text{Li}_2left(i
left(-2+sqrt{3}right)right)right)$$ Now, ???
answered Jan 22 at 9:12
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
add a comment |
$begingroup$
This is not an answer.
Your premonition seems to be good. Using another CAS,
$$2int_0^inftyfrac{xcosh( x)}{3+cosh^2(x)},dx=-frac{pi}{4} log left(7-4 sqrt{3}right)-i left(text{Li}_2left(-i left(-2+sqrt{3}right)right)-text{Li}_2left(i
left(-2+sqrt{3}right)right)right)$$ Now, ???
answered Jan 22 at 9:12
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
add a comment |
$begingroup$
This is not an answer.
Your premonition seems to be good. Using another CAS,
$$2int_0^inftyfrac{xcosh( x)}{3+cosh^2(x)},dx=-frac{pi}{4} log left(7-4 sqrt{3}right)-i left(text{Li}_2left(-i left(-2+sqrt{3}right)right)-text{Li}_2left(i
left(-2+sqrt{3}right)right)right)$$ Now, ???
answered Jan 22 at 9:12
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
This is not an answer.
Your premonition seems to be good. Using another CAS,
$$2int_0^inftyfrac{xcosh( x)}{3+cosh^2(x)},dx=-frac{pi}{4} log left(7-4 sqrt{3}right)-i left(text{Li}_2left(-i left(-2+sqrt{3}right)right)-text{Li}_2left(i
left(-2+sqrt{3}right)right)right)$$ Now, ???
answered Jan 22 at 9:12
Claude LeiboviciClaude Leibovici
123k1157135
answered Jan 22 at 9:12
Claude LeiboviciClaude Leibovici
123k1157135
answered Jan 22 at 9:12
Claude LeiboviciClaude Leibovici
123k1157135
answered Jan 22 at 9:12
Claude LeiboviciClaude Leibovici
123k1157135
123k1157135
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