Mixing
Integral $int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx$
$begingroup$
Last year I wondered about this integral:$$int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx$$
That is because it looks very similar to this integral
and this one. Surprisingly the result is quite nice and an approach can be found here.
$$boxed{int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx=frac{sqrt{2}pi(5pi^2+12piln 2 - 12ln^22)}{96}}$$
Although the approach there is quite skillful, I believed that an elementary approach can be found for this integral.
Here is my idea. First we will consider the following two integrals: $$I=int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx ,;quad J=int_0^frac{pi}{2} x^2sqrt{cot x},mathrm dx$$
$$Rightarrow I=frac12 left((I-J)+(I+J)right)$$
Thust we need to evaluate the sum and the difference of those two from above.
I also saw from here that the "sister" integral differs only by a minus sign: $$boxed{int_0^frac{pi}{2} x^2sqrt{cot x},mathrm dx=frac{sqrt{2}pi(5pi^2-12piln 2 - 12ln^22)}{96}}$$
Thus using those two boxed answer we expect to find: $$I-J=frac{pi^2 ln 2}{2sqrt 2};quad I+J=frac{5pi^3}{24sqrt 2}-frac{pi ln^2 2}{2sqrt 2}tag1$$
$$I-J=int_0^frac{pi}{2} x^2left(sqrt{tan x}-sqrt{cot x}right),mathrm dx=sqrt 2int_0^frac{pi}{2} x^2 cdot frac{sin x-cos x}{sqrt{sin (2x)}}dx$$
$$=-sqrt 2int_0^frac{pi}{2} x^2 left(operatorname{arccosh}(sin x+cos x) right)'dx=2sqrt 2 int_0^frac{pi}{2} xoperatorname{arccosh} (sin x+cos x)dx$$
Let us also denote the last integral with $I_1$ and do a $frac{pi}{2}-x=x$ substitution:
$$I_1=int_0^frac{pi}{2} xoperatorname{arccosh} (sin x+cos x)dx=int_0^frac{pi}{2} left(frac{pi}{2}-xright)operatorname{arccosh} (sin x+cos x)dx$$
$$2I_1=frac{pi}{2} int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dxRightarrow I-J=frac{pi}{sqrt 2}int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dx$$
By using $(1)$ we can easily deduce that: $$bbox[10pt,#000, border:2px solid green ]{color{orange}{int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dx=frac{pi}{2}ln 2}}$$
Doing something similar for $I+J$ we get:
$$I+J=int_0^frac{pi}{2} x^2left(sqrt{tan x}+sqrt{cot x}right),mathrm dx=sqrt 2int_0^frac{pi}{2} x^2 cdot frac{sin x+cos x}{sqrt{sin (2x)}}dx$$
$$=sqrt 2 int_0^frac{pi}{2} x^2 left( arcsin left(sin x-cos xright)right)'dx=frac{pi^3 sqrt 2}{8}-2sqrt 2 int_0^frac{pi}{2} x arcsin left(sin x-cos xright)dx$$
Unfortunately, we're not lucky this time and the substitution used for $I-J$ doesn't help in this case.
Of course using $(1)$ we can again deduce that:
$$bbox[10pt,#000, border:2px solid green ]{color{red}{int_0^frac{pi}{2} x arcsin left(sin x-cos xright)dx=frac{pi^3}{96}+frac{pi}{8}ln^2 2}}$$
In the meantime I found a way for the first one, mainly using: $$frac{arctan x}{x}=int_0^1 frac{dy}{1+x^2y^2}$$ Let us denote: $$I_1=int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dxoverset{IBP}= int_0^frac{pi}{2} x cdot frac{sin x-cos x}{sqrt{sin(2x)}}dx$$
$$overset{tan xrightarrow x}=frac{1}{sqrt 2}int_0^infty frac{arctan x}{1+x^2}frac{x-1}{sqrt x}dx=frac1{sqrt 2}int_0^infty int_0^1 frac{dy}{1+x^2y^2} frac{sqrt x(x-1)}{1+x^2}dx$$
$$=frac1{sqrt 2}int_0^1 int_0^infty frac{1}{1+y^2x^2} frac{sqrt x(x-1)}{1+x^2} dxdy$$
$$=frac{1}{sqrt 2}int_0^1 frac{{pi}}{sqrt 2}left(frac{2}{y^2-1}-frac{1}{sqrt y (y^2-1)}-frac{sqrt y}{y^2-1}right)dy=frac{pi}{2}ln 2$$
Although the integral in the third row looks quite unpleasant, it can be done quite elementary.
Sadly a similar approach for the second one is madness, because we would have:
$$I_2=int_0^1 int_0^1 int_0^infty frac{sqrt x (x+1)}{1+x^2}frac{1}{1+y^2x^2}frac{1}{1+z^2x^2} dxdydz$$
But atleast it gives hope that an elementary approach exists.
For this question I would like to see an elementary approach (without relying on special functions) for the second integral (red one).
If possible please avoid contour integration, although this might be
included in elementary.
integration definite-integrals closed-form alternative-proof
asked Jan 12 at 15:41
ZackyZacky
6,5351858
$endgroup$
add a comment |
$begingroup$
Last year I wondered about this integral:$$int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx$$
That is because it looks very similar to this integral
and this one. Surprisingly the result is quite nice and an approach can be found here.
$$boxed{int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx=frac{sqrt{2}pi(5pi^2+12piln 2 - 12ln^22)}{96}}$$
Although the approach there is quite skillful, I believed that an elementary approach can be found for this integral.
Here is my idea. First we will consider the following two integrals: $$I=int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx ,;quad J=int_0^frac{pi}{2} x^2sqrt{cot x},mathrm dx$$
$$Rightarrow I=frac12 left((I-J)+(I+J)right)$$
Thust we need to evaluate the sum and the difference of those two from above.
I also saw from here that the "sister" integral differs only by a minus sign: $$boxed{int_0^frac{pi}{2} x^2sqrt{cot x},mathrm dx=frac{sqrt{2}pi(5pi^2-12piln 2 - 12ln^22)}{96}}$$
Thus using those two boxed answer we expect to find: $$I-J=frac{pi^2 ln 2}{2sqrt 2};quad I+J=frac{5pi^3}{24sqrt 2}-frac{pi ln^2 2}{2sqrt 2}tag1$$
$$I-J=int_0^frac{pi}{2} x^2left(sqrt{tan x}-sqrt{cot x}right),mathrm dx=sqrt 2int_0^frac{pi}{2} x^2 cdot frac{sin x-cos x}{sqrt{sin (2x)}}dx$$
$$=-sqrt 2int_0^frac{pi}{2} x^2 left(operatorname{arccosh}(sin x+cos x) right)'dx=2sqrt 2 int_0^frac{pi}{2} xoperatorname{arccosh} (sin x+cos x)dx$$
Let us also denote the last integral with $I_1$ and do a $frac{pi}{2}-x=x$ substitution:
$$I_1=int_0^frac{pi}{2} xoperatorname{arccosh} (sin x+cos x)dx=int_0^frac{pi}{2} left(frac{pi}{2}-xright)operatorname{arccosh} (sin x+cos x)dx$$
$$2I_1=frac{pi}{2} int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dxRightarrow I-J=frac{pi}{sqrt 2}int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dx$$
By using $(1)$ we can easily deduce that: $$bbox[10pt,#000, border:2px solid green ]{color{orange}{int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dx=frac{pi}{2}ln 2}}$$
Doing something similar for $I+J$ we get:
$$I+J=int_0^frac{pi}{2} x^2left(sqrt{tan x}+sqrt{cot x}right),mathrm dx=sqrt 2int_0^frac{pi}{2} x^2 cdot frac{sin x+cos x}{sqrt{sin (2x)}}dx$$
$$=sqrt 2 int_0^frac{pi}{2} x^2 left( arcsin left(sin x-cos xright)right)'dx=frac{pi^3 sqrt 2}{8}-2sqrt 2 int_0^frac{pi}{2} x arcsin left(sin x-cos xright)dx$$
Unfortunately, we're not lucky this time and the substitution used for $I-J$ doesn't help in this case.
Of course using $(1)$ we can again deduce that:
$$bbox[10pt,#000, border:2px solid green ]{color{red}{int_0^frac{pi}{2} x arcsin left(sin x-cos xright)dx=frac{pi^3}{96}+frac{pi}{8}ln^2 2}}$$
In the meantime I found a way for the first one, mainly using: $$frac{arctan x}{x}=int_0^1 frac{dy}{1+x^2y^2}$$ Let us denote: $$I_1=int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dxoverset{IBP}= int_0^frac{pi}{2} x cdot frac{sin x-cos x}{sqrt{sin(2x)}}dx$$
$$overset{tan xrightarrow x}=frac{1}{sqrt 2}int_0^infty frac{arctan x}{1+x^2}frac{x-1}{sqrt x}dx=frac1{sqrt 2}int_0^infty int_0^1 frac{dy}{1+x^2y^2} frac{sqrt x(x-1)}{1+x^2}dx$$
$$=frac1{sqrt 2}int_0^1 int_0^infty frac{1}{1+y^2x^2} frac{sqrt x(x-1)}{1+x^2} dxdy$$
$$=frac{1}{sqrt 2}int_0^1 frac{{pi}}{sqrt 2}left(frac{2}{y^2-1}-frac{1}{sqrt y (y^2-1)}-frac{sqrt y}{y^2-1}right)dy=frac{pi}{2}ln 2$$
Although the integral in the third row looks quite unpleasant, it can be done quite elementary.
Sadly a similar approach for the second one is madness, because we would have:
$$I_2=int_0^1 int_0^1 int_0^infty frac{sqrt x (x+1)}{1+x^2}frac{1}{1+y^2x^2}frac{1}{1+z^2x^2} dxdydz$$
But atleast it gives hope that an elementary approach exists.
For this question I would like to see an elementary approach (without relying on special functions) for the second integral (red one).
If possible please avoid contour integration, although this might be
included in elementary.
integration definite-integrals closed-form alternative-proof
asked Jan 12 at 15:41
ZackyZacky
6,5351858
$endgroup$
2
$begingroup$
I look forward to being about to integrate as you do. Very nice work here.
$endgroup$
– DavidG
Jan 13 at 6:42
add a comment |
$begingroup$
Last year I wondered about this integral:$$int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx$$
That is because it looks very similar to this integral
and this one. Surprisingly the result is quite nice and an approach can be found here.
$$boxed{int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx=frac{sqrt{2}pi(5pi^2+12piln 2 - 12ln^22)}{96}}$$
Although the approach there is quite skillful, I believed that an elementary approach can be found for this integral.
Here is my idea. First we will consider the following two integrals: $$I=int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx ,;quad J=int_0^frac{pi}{2} x^2sqrt{cot x},mathrm dx$$
$$Rightarrow I=frac12 left((I-J)+(I+J)right)$$
Thust we need to evaluate the sum and the difference of those two from above.
I also saw from here that the "sister" integral differs only by a minus sign: $$boxed{int_0^frac{pi}{2} x^2sqrt{cot x},mathrm dx=frac{sqrt{2}pi(5pi^2-12piln 2 - 12ln^22)}{96}}$$
Thus using those two boxed answer we expect to find: $$I-J=frac{pi^2 ln 2}{2sqrt 2};quad I+J=frac{5pi^3}{24sqrt 2}-frac{pi ln^2 2}{2sqrt 2}tag1$$
$$I-J=int_0^frac{pi}{2} x^2left(sqrt{tan x}-sqrt{cot x}right),mathrm dx=sqrt 2int_0^frac{pi}{2} x^2 cdot frac{sin x-cos x}{sqrt{sin (2x)}}dx$$
$$=-sqrt 2int_0^frac{pi}{2} x^2 left(operatorname{arccosh}(sin x+cos x) right)'dx=2sqrt 2 int_0^frac{pi}{2} xoperatorname{arccosh} (sin x+cos x)dx$$
Let us also denote the last integral with $I_1$ and do a $frac{pi}{2}-x=x$ substitution:
$$I_1=int_0^frac{pi}{2} xoperatorname{arccosh} (sin x+cos x)dx=int_0^frac{pi}{2} left(frac{pi}{2}-xright)operatorname{arccosh} (sin x+cos x)dx$$
$$2I_1=frac{pi}{2} int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dxRightarrow I-J=frac{pi}{sqrt 2}int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dx$$
By using $(1)$ we can easily deduce that: $$bbox[10pt,#000, border:2px solid green ]{color{orange}{int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dx=frac{pi}{2}ln 2}}$$
Doing something similar for $I+J$ we get:
$$I+J=int_0^frac{pi}{2} x^2left(sqrt{tan x}+sqrt{cot x}right),mathrm dx=sqrt 2int_0^frac{pi}{2} x^2 cdot frac{sin x+cos x}{sqrt{sin (2x)}}dx$$
$$=sqrt 2 int_0^frac{pi}{2} x^2 left( arcsin left(sin x-cos xright)right)'dx=frac{pi^3 sqrt 2}{8}-2sqrt 2 int_0^frac{pi}{2} x arcsin left(sin x-cos xright)dx$$
Unfortunately, we're not lucky this time and the substitution used for $I-J$ doesn't help in this case.
Of course using $(1)$ we can again deduce that:
$$bbox[10pt,#000, border:2px solid green ]{color{red}{int_0^frac{pi}{2} x arcsin left(sin x-cos xright)dx=frac{pi^3}{96}+frac{pi}{8}ln^2 2}}$$
In the meantime I found a way for the first one, mainly using: $$frac{arctan x}{x}=int_0^1 frac{dy}{1+x^2y^2}$$ Let us denote: $$I_1=int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dxoverset{IBP}= int_0^frac{pi}{2} x cdot frac{sin x-cos x}{sqrt{sin(2x)}}dx$$
$$overset{tan xrightarrow x}=frac{1}{sqrt 2}int_0^infty frac{arctan x}{1+x^2}frac{x-1}{sqrt x}dx=frac1{sqrt 2}int_0^infty int_0^1 frac{dy}{1+x^2y^2} frac{sqrt x(x-1)}{1+x^2}dx$$
$$=frac1{sqrt 2}int_0^1 int_0^infty frac{1}{1+y^2x^2} frac{sqrt x(x-1)}{1+x^2} dxdy$$
$$=frac{1}{sqrt 2}int_0^1 frac{{pi}}{sqrt 2}left(frac{2}{y^2-1}-frac{1}{sqrt y (y^2-1)}-frac{sqrt y}{y^2-1}right)dy=frac{pi}{2}ln 2$$
Although the integral in the third row looks quite unpleasant, it can be done quite elementary.
Sadly a similar approach for the second one is madness, because we would have:
$$I_2=int_0^1 int_0^1 int_0^infty frac{sqrt x (x+1)}{1+x^2}frac{1}{1+y^2x^2}frac{1}{1+z^2x^2} dxdydz$$
But atleast it gives hope that an elementary approach exists.
For this question I would like to see an elementary approach (without relying on special functions) for the second integral (red one).
If possible please avoid contour integration, although this might be
included in elementary.
integration definite-integrals closed-form alternative-proof
asked Jan 12 at 15:41
ZackyZacky
6,5351858
$endgroup$
Last year I wondered about this integral:$$int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx$$
That is because it looks very similar to this integral
and this one. Surprisingly the result is quite nice and an approach can be found here.
$$boxed{int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx=frac{sqrt{2}pi(5pi^2+12piln 2 - 12ln^22)}{96}}$$
Although the approach there is quite skillful, I believed that an elementary approach can be found for this integral.
Here is my idea. First we will consider the following two integrals: $$I=int_0^frac{pi}{2} x^2sqrt{tan x},mathrm dx ,;quad J=int_0^frac{pi}{2} x^2sqrt{cot x},mathrm dx$$
$$Rightarrow I=frac12 left((I-J)+(I+J)right)$$
Thust we need to evaluate the sum and the difference of those two from above.
I also saw from here that the "sister" integral differs only by a minus sign: $$boxed{int_0^frac{pi}{2} x^2sqrt{cot x},mathrm dx=frac{sqrt{2}pi(5pi^2-12piln 2 - 12ln^22)}{96}}$$
Thus using those two boxed answer we expect to find: $$I-J=frac{pi^2 ln 2}{2sqrt 2};quad I+J=frac{5pi^3}{24sqrt 2}-frac{pi ln^2 2}{2sqrt 2}tag1$$
$$I-J=int_0^frac{pi}{2} x^2left(sqrt{tan x}-sqrt{cot x}right),mathrm dx=sqrt 2int_0^frac{pi}{2} x^2 cdot frac{sin x-cos x}{sqrt{sin (2x)}}dx$$
$$=-sqrt 2int_0^frac{pi}{2} x^2 left(operatorname{arccosh}(sin x+cos x) right)'dx=2sqrt 2 int_0^frac{pi}{2} xoperatorname{arccosh} (sin x+cos x)dx$$
Let us also denote the last integral with $I_1$ and do a $frac{pi}{2}-x=x$ substitution:
$$I_1=int_0^frac{pi}{2} xoperatorname{arccosh} (sin x+cos x)dx=int_0^frac{pi}{2} left(frac{pi}{2}-xright)operatorname{arccosh} (sin x+cos x)dx$$
$$2I_1=frac{pi}{2} int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dxRightarrow I-J=frac{pi}{sqrt 2}int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dx$$
By using $(1)$ we can easily deduce that: $$bbox[10pt,#000, border:2px solid green ]{color{orange}{int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dx=frac{pi}{2}ln 2}}$$
Doing something similar for $I+J$ we get:
$$I+J=int_0^frac{pi}{2} x^2left(sqrt{tan x}+sqrt{cot x}right),mathrm dx=sqrt 2int_0^frac{pi}{2} x^2 cdot frac{sin x+cos x}{sqrt{sin (2x)}}dx$$
$$=sqrt 2 int_0^frac{pi}{2} x^2 left( arcsin left(sin x-cos xright)right)'dx=frac{pi^3 sqrt 2}{8}-2sqrt 2 int_0^frac{pi}{2} x arcsin left(sin x-cos xright)dx$$
Unfortunately, we're not lucky this time and the substitution used for $I-J$ doesn't help in this case.
Of course using $(1)$ we can again deduce that:
$$bbox[10pt,#000, border:2px solid green ]{color{red}{int_0^frac{pi}{2} x arcsin left(sin x-cos xright)dx=frac{pi^3}{96}+frac{pi}{8}ln^2 2}}$$
In the meantime I found a way for the first one, mainly using: $$frac{arctan x}{x}=int_0^1 frac{dy}{1+x^2y^2}$$ Let us denote: $$I_1=int_0^frac{pi}{2} operatorname{arccosh} (sin x+cos x)dxoverset{IBP}= int_0^frac{pi}{2} x cdot frac{sin x-cos x}{sqrt{sin(2x)}}dx$$
$$overset{tan xrightarrow x}=frac{1}{sqrt 2}int_0^infty frac{arctan x}{1+x^2}frac{x-1}{sqrt x}dx=frac1{sqrt 2}int_0^infty int_0^1 frac{dy}{1+x^2y^2} frac{sqrt x(x-1)}{1+x^2}dx$$
$$=frac1{sqrt 2}int_0^1 int_0^infty frac{1}{1+y^2x^2} frac{sqrt x(x-1)}{1+x^2} dxdy$$
$$=frac{1}{sqrt 2}int_0^1 frac{{pi}}{sqrt 2}left(frac{2}{y^2-1}-frac{1}{sqrt y (y^2-1)}-frac{sqrt y}{y^2-1}right)dy=frac{pi}{2}ln 2$$
Although the integral in the third row looks quite unpleasant, it can be done quite elementary.
Sadly a similar approach for the second one is madness, because we would have:
$$I_2=int_0^1 int_0^1 int_0^infty frac{sqrt x (x+1)}{1+x^2}frac{1}{1+y^2x^2}frac{1}{1+z^2x^2} dxdydz$$
But atleast it gives hope that an elementary approach exists.
For this question I would like to see an elementary approach (without relying on special functions) for the second integral (red one).
If possible please avoid contour integration, although this might be
included in elementary.
integration definite-integrals closed-form alternative-proof
integration definite-integrals closed-form alternative-proof
asked Jan 12 at 15:41
ZackyZacky
6,5351858
asked Jan 12 at 15:41
ZackyZacky
6,5351858
edited Jan 15 at 21:00
Zacky
asked Jan 12 at 15:41
ZackyZacky
6,5351858
asked Jan 12 at 15:41
ZackyZacky
6,5351858
asked Jan 12 at 15:41
ZackyZacky
6,5351858
6,5351858
2
$begingroup$
I look forward to being about to integrate as you do. Very nice work here.
$endgroup$
– DavidG
Jan 13 at 6:42
add a comment |
2
$begingroup$
I look forward to being about to integrate as you do. Very nice work here.
$endgroup$
– DavidG
Jan 13 at 6:42
2
2
$begingroup$
I look forward to being about to integrate as you do. Very nice work here.
$endgroup$
– DavidG
Jan 13 at 6:42
$begingroup$
I look forward to being about to integrate as you do. Very nice work here.
$endgroup$
– DavidG
Jan 13 at 6:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
On the path of Zacky, the missing part...
Let,
begin{align}I&=int_0^{frac{pi}{2}}x^2sqrt{tan x},dx\
J&=int_0^{frac{pi}{2}}frac{x^2}{sqrt{tan x}},dx\
end{align}
Perform the change of variable $y=sqrt{tan x}$,
begin{align}I&=int_0^{infty}frac{2x^2arctan^2left(x^2right)}{1+x^4},dx\\
J&=int_0^{infty}frac{2x^2arctan^2left(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
begin{align}
text{I+J}&=int_0^{infty}frac{2x^2left(arctanleft(x^2right)+arctanleft(frac{1}{x^2}right)right)^2}{1+x^4},dx-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
&=frac{pi^2}{4}int_0^{infty}frac{2x^2}{1+x^4},dx-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
Perform the change of variable $y=dfrac{1}{x}$,
begin{align}
text{K}&=int_0^{infty}frac{2x^2}{1+x^4},dx\
&=int_0^{infty}frac{2}{1+x^4},dx\
end{align}
Therefore,
begin{align}
text{2K}=int_0^{infty}frac{2left(1+frac{1}{x^2}right)}{left(x-frac{1}{x}right)^2+2},dx
end{align}
Perform the change of variable $y=x-dfrac{1}{x}$,
begin{align}text{2K}&=2int_{-infty}^{+infty}frac{1}{2+x^2},dx\
&=2left[frac{1}{sqrt{2}}arctanleft(frac{x}{sqrt{2}}right)right]_{-infty}^{+infty}\
&=2times frac{pi}{sqrt{2}}
end{align}
therefore,
begin{align}
text{I+J}&=frac{pi^3}{4sqrt{2}}-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
Let $a>0$,
begin{align}
text{K}_1(a)&=int_0^{infty}frac{x^2}{a+x^4},dx\
&=frac{1}{a}int_0^{infty}frac{x^2}{1+left(a^{-frac{1}{4}}xright)^4},dx\
end{align}
Perform the change of variable $y=a^{-frac{1}{4}}x$,
begin{align}
text{K}_1(a)&=a^{-frac{1}{4}}int_0^{infty}frac{x^2}{1+x^4},dx\
&=frac{a^{-frac{1}{4}}pi}{2sqrt{2}}
end{align}
In the same manner,
begin{align}
text{K}_2(a)&=int_0^{infty}frac{x^2}{1+ax^4},dx\
&=frac{a^{-frac{3}{4}}pi}{2sqrt{2}}
end{align}
Since, for $a$ real,
begin{align}arctan a=int_0^1 frac{a}{1+a^2t^2},dtend{align}
then,
begin{align}text{L}&=int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
&=int_0^{infty}left(int_0^1 int_0^1 frac{x^2}{(1+u^2x^4)left(1+frac{v^2}{x^4}right)(1+x^4)},du,dvright),dx\
&=\
&int_0^{infty}left(int_0^1int_0^1 left(frac{x^2}{(1-u^2)(1-v^2)(1+x^4)}-frac{x^2}{1-u^2v^2}left(frac{u^2}{(1-u^2)(1+u^2x^4)}+frac{v^2}{(1-v^2)(v^2+x^4)}right)
right)dudvright)dx\
&=int_0^1int_0^1 left(frac{pi}{2sqrt{2}(1-u^2)(1-v^2)}-frac{1}{1-u^2v^2}left(frac{u^2text{K}_2(u^2)}{1-u^2}+frac{v^2text{K}_1(v^2)}{1-v^2}right)right)dudv\
&=frac{pi}{2sqrt{2}}int_0^1int_0^1 left(frac{1}{(1-u^2)(1-v^2)}-frac{1}{(1-u^2v^2)}left(frac{u^{frac{1}{2}}}{1-u^2}+frac{v^{frac{3}{2}}}{1-v^2}right)right)dudv\
&=piint_0^1left[frac{sqrt{v}left(text{ arctanh}left(sqrt{uv}right)-text{ arctan}left(sqrt{uv}right)-text{ arctanh}left(uvright)right)+arctanleft(sqrt{u}right)+lnleft(frac{sqrt{1+u}}{1+sqrt{u}}right)}{2sqrt{2}(1-v^2)}right]_{u=0}^{u=1},dv\
&=frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}big(text{ arctanh}left(sqrt{v}right)-text{ arctan}left(sqrt{v}right)-text{ arctanh}left(vright)big)+frac{pi}{4}-frac{1}{2}ln 2}{1-v^2},dv\
&=frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}arctanleft(frac{1-sqrt{v}}{1+sqrt{v}}right)}{1-v^2},dv+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right)int_0^1 frac{1-sqrt{v}}{1-v^2},dv+\
&frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}lnleft(frac{1+sqrt{v}}{2}right)}{1-v^2},dv-frac{pi}{4sqrt{2}}int_0^1frac{sqrt{v}lnleft(frac{1+v}{2}right)}{1-v^2},dv
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}text{R}_1&=int_0^1frac{sqrt{v}arctanleft(frac{1-sqrt{v}}{1+sqrt{v}}right)}{1-v^2},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2arctan v}{v(1+v^2)},dv\
&=frac{1}{2}int_0^1 frac{arctan v}{v},dv-int_0^1 frac{arctan v}{1+v^2},dv\
&=frac{1}{2}text{G}-frac{1}{2}Big[arctan^2 vBig]_0^1\
&=frac{1}{2}text{G}-frac{pi^2}{32}\
text{R}_2&=int_0^1 frac{1-sqrt{v}}{1-v^2},dv\
&=left[lnleft(frac{sqrt{1+v}}{1+sqrt{v}}right)+arctanleft(sqrt{v}right)right]_0^1\
&=frac{pi}{4}-frac{1}{2}ln 2\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}text{R}_3&=int_0^1frac{sqrt{v}lnleft(frac{1+sqrt{v}}{2}right)}{1-v^2},dv\
&=-frac{1}{2}int_0^1frac{(1-v)^2ln(1+v)}{v(1+v^2)},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{1}{2}int_0^1 frac{ln(1+v
)}{v},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{1}{4}int_0^1 frac{2vln(1-v^2)}{v^2},dv+frac{1}{2}int_0^1 frac{ln(1-v)}{v},dv\
end{align}
In the second integral perform the change of variable $y=v^2$,
begin{align}text{R}_3&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}int_0^1 frac{ln(1-v)}{v},dv\
end{align}
In the second integral perform the change of variable $y=1-v$,
begin{align}text{R}_3&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}int_0^1 frac{ln v}{1-v},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}times -zeta(2)\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{pi^2}{24}\
end{align}
Perform the change of variable $y=dfrac{1-v}{1+v}$,
begin{align}
text{S}_1&=int_0^1frac{ln(1+v)}{1+v^2},dv\
&=int_0^1frac{ln(frac{2}{1+v})}{1+v^2},dv\
&=ln 2int_0^1 frac{1}{1+v^2},dv-text{S}_1\
&=frac{pi}{4}ln 2-text{S}_1
end{align}
Therefore,
begin{align}
text{S}_1&=frac{pi}{8}ln 2\
text{R}_3&=frac{pi}{8}ln 2-frac{pi^2}{24}\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}
text{R}_4&=int_0^1frac{sqrt{v}lnleft(frac{1+v}{2}right)}{1-v^2},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2lnleft(frac{1+v^2}{(1+v)^2}right)}{v(1+v^2)},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2lnleft(1+v^2right)}{v(1+v^2)},dv+2text{R}_3\
&=frac{1}{2}int_0^1frac{ln(1+v^2)}{v},dv-int_0^1frac{ln(1+v^2)}{1+v^2},dv+frac{pi}{4}ln 2-frac{pi^2}{12}\
&=frac{1}{2}times frac{1}{4}zeta(2)-int_0^1frac{ln(1+v^2)}{1+v^2},dv+frac{pi}{4}ln 2-frac{pi^2}{12}\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1frac{ln(1+v^2)}{1+v^2},dv\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1int_0^1frac{v^2}{(1+v^2)(1+v^2t)},dt,dv\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1 left[frac{arctanleft(vright)sqrt{t}-arctanleft(vsqrt{t}right)}{(t-1)sqrt{t}}right]_{v=0}^{v=1},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1 frac{frac{pisqrt{t}}{4}-arctanleft(sqrt{t}right)}{(t-1)sqrt{t}},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}+int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}int_0^1 frac{sqrt{t}-1}{(t-1)sqrt{t}},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}+int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}Big[2lnleft(1+sqrt{t}right)Big]_0^1\
&=int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}ln 2-frac{pi^2}{16}\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{t}}{1+sqrt{t}}$,
begin{align}
text{R}_4&=int_0^1 frac{arctan t}{t},dt-frac{pi}{4}ln 2-frac{pi^2}{16}\
&=text{G}-frac{pi}{4}ln 2-frac{pi^2}{16}\
end{align}
Therefore,
begin{align}L&=frac{pi}{2sqrt{2}}text{R}_1+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right) text{R}_2+frac{pi}{2sqrt{2}}text{R}_3-frac{pi}{4sqrt{2}}text{R}_4\
&=frac{pi}{2sqrt{2}}left(frac{text{G}}{2}-frac{pi^2}{32}right)+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right)^2+frac{pi}{2sqrt{2}}left(frac{pi}{8}ln 2-frac{pi^2}{24}right)-\
&frac{pi}{4sqrt{2}}left(text{G}-frac{pi}{4}ln 2-frac{pi^2}{16}right)\
&=frac{pi^3}{96sqrt{2}}+frac{piln^2 2}{8sqrt{2}}
end{align}
Thus,
begin{align}text{I+J}&=frac{pi^3}{4sqrt{2}}-4text{L}\
&=frac{pi^3}{4sqrt{2}}-4left(frac{pi^3}{96sqrt{2}}+frac{piln^2 2}{8sqrt{2}}right)\
&=boxed{frac{5pi^3}{24sqrt{2}}-frac{piln^2 2}{2sqrt{2}}}
end{align}
answered Jan 18 at 14:45
FDPFDP
5,35711524
$endgroup$
$begingroup$
I wish I could up-vote this multiple times. Excellent work
$endgroup$
– clathratus
yesterday
add a comment |
$begingroup$
Both integrals depend on
$$int_{0}^{pi} x^2 frac{sqrt{1+cos x}pm sqrt{1-cos x}}{sqrt{sin x}},dx=\int_{0}^{pi/2}frac{dx}{sqrt{sin x}}left[(x^2pm(pi-x)^2)sqrt{1+cos x}+(pm x^2+(pi-x)^2)sqrt{1-cos x}right],dx $$
and they can be tackled through the Fourier series of a periodic version of $x^2pm(pi-x)^2$. For instance
$$ x^2+(pi-x)^2 = frac{2pi^2}{3}+2sum_{ngeq 1}frac{cos(2nx)}{n^2}qquad forall xin[0,pi] $$
(yes, I am exploiting Bernoulli polynomials) and for any $ninmathbb{N}^+$
$$ int_{0}^{pi/2}frac{sqrt{1+cos x}+sqrt{1-cos x}}{sqrt{sin x}}cos(2nx),dx=frac{pi}{4^nsqrt{2}}binom{2n}{n}$$
so the whole question boils down to computing
$$ sum_{ngeq 1}frac{1}{n^2 4^n}binom{2n}{n} = frac{1}{2},phantom{}_4 F_3left(1,1,1,tfrac{3}{2};2,2,2;1right)=zeta(2)-2log^2(2).$$
The last equality follows by recalling $frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}(sin t)^{2n},dt$ and by tackling
$$ int_{0}^{1}frac{text{Li}_2(z)}{sqrt{z(1-z)}},dz $$
through Fourier-Legendre series or the reflection formula for $text{Li}_2$.
The remaining part is just related to the well-known
$$ int_{0}^{1}frac{log x}{sqrt{x(1-x)}},dx = 4int_{0}^{pi/2}logsintheta,dtheta = -2pilog 2.$$
Summarizing, all the involved integrals just depend on $left.frac{d^nu}{da^nu}Bleft(a,tfrac{1}{2}right)right|_{a=1/2}$ for $nuin{1,2}$.
answered Jan 12 at 18:38
Jack D'AurizioJack D'Aurizio
289k33282662
$endgroup$
3
$begingroup$
I don't know if this fits the requirements for 'elementary', but nice job anyway
$endgroup$
– clathratus
Jan 12 at 20:55
$begingroup$
Could you give me a tip on proving $$sum_{ngeq1}frac1{4^n n^2}{2nchoose n}=zeta(2)-2log^22$$?
$endgroup$
– clathratus
Jan 18 at 4:01
$begingroup$
The value of this series is (integrate first using variable t) begin{align}sum_{ngeq 1}^infty frac{1}{4^n n^2}binom{2n}{n}&=frac{2}{pi}int_{x=0}^1int_{t=0}^{frac{pi}{2}}frac{ln xsin^2 t }{xsin^2 t-1},dt,dxend{align}
$endgroup$
– FDP
Jan 18 at 21:40
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071027%2fintegral-int-0-frac-pi2-x2-sqrt-tan-x-mathrm-dx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
On the path of Zacky, the missing part...
Let,
begin{align}I&=int_0^{frac{pi}{2}}x^2sqrt{tan x},dx\
J&=int_0^{frac{pi}{2}}frac{x^2}{sqrt{tan x}},dx\
end{align}
Perform the change of variable $y=sqrt{tan x}$,
begin{align}I&=int_0^{infty}frac{2x^2arctan^2left(x^2right)}{1+x^4},dx\\
J&=int_0^{infty}frac{2x^2arctan^2left(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
begin{align}
text{I+J}&=int_0^{infty}frac{2x^2left(arctanleft(x^2right)+arctanleft(frac{1}{x^2}right)right)^2}{1+x^4},dx-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
&=frac{pi^2}{4}int_0^{infty}frac{2x^2}{1+x^4},dx-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
Perform the change of variable $y=dfrac{1}{x}$,
begin{align}
text{K}&=int_0^{infty}frac{2x^2}{1+x^4},dx\
&=int_0^{infty}frac{2}{1+x^4},dx\
end{align}
Therefore,
begin{align}
text{2K}=int_0^{infty}frac{2left(1+frac{1}{x^2}right)}{left(x-frac{1}{x}right)^2+2},dx
end{align}
Perform the change of variable $y=x-dfrac{1}{x}$,
begin{align}text{2K}&=2int_{-infty}^{+infty}frac{1}{2+x^2},dx\
&=2left[frac{1}{sqrt{2}}arctanleft(frac{x}{sqrt{2}}right)right]_{-infty}^{+infty}\
&=2times frac{pi}{sqrt{2}}
end{align}
therefore,
begin{align}
text{I+J}&=frac{pi^3}{4sqrt{2}}-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
Let $a>0$,
begin{align}
text{K}_1(a)&=int_0^{infty}frac{x^2}{a+x^4},dx\
&=frac{1}{a}int_0^{infty}frac{x^2}{1+left(a^{-frac{1}{4}}xright)^4},dx\
end{align}
Perform the change of variable $y=a^{-frac{1}{4}}x$,
begin{align}
text{K}_1(a)&=a^{-frac{1}{4}}int_0^{infty}frac{x^2}{1+x^4},dx\
&=frac{a^{-frac{1}{4}}pi}{2sqrt{2}}
end{align}
In the same manner,
begin{align}
text{K}_2(a)&=int_0^{infty}frac{x^2}{1+ax^4},dx\
&=frac{a^{-frac{3}{4}}pi}{2sqrt{2}}
end{align}
Since, for $a$ real,
begin{align}arctan a=int_0^1 frac{a}{1+a^2t^2},dtend{align}
then,
begin{align}text{L}&=int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
&=int_0^{infty}left(int_0^1 int_0^1 frac{x^2}{(1+u^2x^4)left(1+frac{v^2}{x^4}right)(1+x^4)},du,dvright),dx\
&=\
&int_0^{infty}left(int_0^1int_0^1 left(frac{x^2}{(1-u^2)(1-v^2)(1+x^4)}-frac{x^2}{1-u^2v^2}left(frac{u^2}{(1-u^2)(1+u^2x^4)}+frac{v^2}{(1-v^2)(v^2+x^4)}right)
right)dudvright)dx\
&=int_0^1int_0^1 left(frac{pi}{2sqrt{2}(1-u^2)(1-v^2)}-frac{1}{1-u^2v^2}left(frac{u^2text{K}_2(u^2)}{1-u^2}+frac{v^2text{K}_1(v^2)}{1-v^2}right)right)dudv\
&=frac{pi}{2sqrt{2}}int_0^1int_0^1 left(frac{1}{(1-u^2)(1-v^2)}-frac{1}{(1-u^2v^2)}left(frac{u^{frac{1}{2}}}{1-u^2}+frac{v^{frac{3}{2}}}{1-v^2}right)right)dudv\
&=piint_0^1left[frac{sqrt{v}left(text{ arctanh}left(sqrt{uv}right)-text{ arctan}left(sqrt{uv}right)-text{ arctanh}left(uvright)right)+arctanleft(sqrt{u}right)+lnleft(frac{sqrt{1+u}}{1+sqrt{u}}right)}{2sqrt{2}(1-v^2)}right]_{u=0}^{u=1},dv\
&=frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}big(text{ arctanh}left(sqrt{v}right)-text{ arctan}left(sqrt{v}right)-text{ arctanh}left(vright)big)+frac{pi}{4}-frac{1}{2}ln 2}{1-v^2},dv\
&=frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}arctanleft(frac{1-sqrt{v}}{1+sqrt{v}}right)}{1-v^2},dv+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right)int_0^1 frac{1-sqrt{v}}{1-v^2},dv+\
&frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}lnleft(frac{1+sqrt{v}}{2}right)}{1-v^2},dv-frac{pi}{4sqrt{2}}int_0^1frac{sqrt{v}lnleft(frac{1+v}{2}right)}{1-v^2},dv
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}text{R}_1&=int_0^1frac{sqrt{v}arctanleft(frac{1-sqrt{v}}{1+sqrt{v}}right)}{1-v^2},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2arctan v}{v(1+v^2)},dv\
&=frac{1}{2}int_0^1 frac{arctan v}{v},dv-int_0^1 frac{arctan v}{1+v^2},dv\
&=frac{1}{2}text{G}-frac{1}{2}Big[arctan^2 vBig]_0^1\
&=frac{1}{2}text{G}-frac{pi^2}{32}\
text{R}_2&=int_0^1 frac{1-sqrt{v}}{1-v^2},dv\
&=left[lnleft(frac{sqrt{1+v}}{1+sqrt{v}}right)+arctanleft(sqrt{v}right)right]_0^1\
&=frac{pi}{4}-frac{1}{2}ln 2\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}text{R}_3&=int_0^1frac{sqrt{v}lnleft(frac{1+sqrt{v}}{2}right)}{1-v^2},dv\
&=-frac{1}{2}int_0^1frac{(1-v)^2ln(1+v)}{v(1+v^2)},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{1}{2}int_0^1 frac{ln(1+v
)}{v},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{1}{4}int_0^1 frac{2vln(1-v^2)}{v^2},dv+frac{1}{2}int_0^1 frac{ln(1-v)}{v},dv\
end{align}
In the second integral perform the change of variable $y=v^2$,
begin{align}text{R}_3&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}int_0^1 frac{ln(1-v)}{v},dv\
end{align}
In the second integral perform the change of variable $y=1-v$,
begin{align}text{R}_3&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}int_0^1 frac{ln v}{1-v},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}times -zeta(2)\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{pi^2}{24}\
end{align}
Perform the change of variable $y=dfrac{1-v}{1+v}$,
begin{align}
text{S}_1&=int_0^1frac{ln(1+v)}{1+v^2},dv\
&=int_0^1frac{ln(frac{2}{1+v})}{1+v^2},dv\
&=ln 2int_0^1 frac{1}{1+v^2},dv-text{S}_1\
&=frac{pi}{4}ln 2-text{S}_1
end{align}
Therefore,
begin{align}
text{S}_1&=frac{pi}{8}ln 2\
text{R}_3&=frac{pi}{8}ln 2-frac{pi^2}{24}\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}
text{R}_4&=int_0^1frac{sqrt{v}lnleft(frac{1+v}{2}right)}{1-v^2},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2lnleft(frac{1+v^2}{(1+v)^2}right)}{v(1+v^2)},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2lnleft(1+v^2right)}{v(1+v^2)},dv+2text{R}_3\
&=frac{1}{2}int_0^1frac{ln(1+v^2)}{v},dv-int_0^1frac{ln(1+v^2)}{1+v^2},dv+frac{pi}{4}ln 2-frac{pi^2}{12}\
&=frac{1}{2}times frac{1}{4}zeta(2)-int_0^1frac{ln(1+v^2)}{1+v^2},dv+frac{pi}{4}ln 2-frac{pi^2}{12}\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1frac{ln(1+v^2)}{1+v^2},dv\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1int_0^1frac{v^2}{(1+v^2)(1+v^2t)},dt,dv\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1 left[frac{arctanleft(vright)sqrt{t}-arctanleft(vsqrt{t}right)}{(t-1)sqrt{t}}right]_{v=0}^{v=1},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1 frac{frac{pisqrt{t}}{4}-arctanleft(sqrt{t}right)}{(t-1)sqrt{t}},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}+int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}int_0^1 frac{sqrt{t}-1}{(t-1)sqrt{t}},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}+int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}Big[2lnleft(1+sqrt{t}right)Big]_0^1\
&=int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}ln 2-frac{pi^2}{16}\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{t}}{1+sqrt{t}}$,
begin{align}
text{R}_4&=int_0^1 frac{arctan t}{t},dt-frac{pi}{4}ln 2-frac{pi^2}{16}\
&=text{G}-frac{pi}{4}ln 2-frac{pi^2}{16}\
end{align}
Therefore,
begin{align}L&=frac{pi}{2sqrt{2}}text{R}_1+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right) text{R}_2+frac{pi}{2sqrt{2}}text{R}_3-frac{pi}{4sqrt{2}}text{R}_4\
&=frac{pi}{2sqrt{2}}left(frac{text{G}}{2}-frac{pi^2}{32}right)+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right)^2+frac{pi}{2sqrt{2}}left(frac{pi}{8}ln 2-frac{pi^2}{24}right)-\
&frac{pi}{4sqrt{2}}left(text{G}-frac{pi}{4}ln 2-frac{pi^2}{16}right)\
&=frac{pi^3}{96sqrt{2}}+frac{piln^2 2}{8sqrt{2}}
end{align}
Thus,
begin{align}text{I+J}&=frac{pi^3}{4sqrt{2}}-4text{L}\
&=frac{pi^3}{4sqrt{2}}-4left(frac{pi^3}{96sqrt{2}}+frac{piln^2 2}{8sqrt{2}}right)\
&=boxed{frac{5pi^3}{24sqrt{2}}-frac{piln^2 2}{2sqrt{2}}}
end{align}
answered Jan 18 at 14:45
FDPFDP
5,35711524
$endgroup$
$begingroup$
I wish I could up-vote this multiple times. Excellent work
$endgroup$
– clathratus
yesterday
add a comment |
$begingroup$
On the path of Zacky, the missing part...
Let,
begin{align}I&=int_0^{frac{pi}{2}}x^2sqrt{tan x},dx\
J&=int_0^{frac{pi}{2}}frac{x^2}{sqrt{tan x}},dx\
end{align}
Perform the change of variable $y=sqrt{tan x}$,
begin{align}I&=int_0^{infty}frac{2x^2arctan^2left(x^2right)}{1+x^4},dx\\
J&=int_0^{infty}frac{2x^2arctan^2left(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
begin{align}
text{I+J}&=int_0^{infty}frac{2x^2left(arctanleft(x^2right)+arctanleft(frac{1}{x^2}right)right)^2}{1+x^4},dx-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
&=frac{pi^2}{4}int_0^{infty}frac{2x^2}{1+x^4},dx-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
Perform the change of variable $y=dfrac{1}{x}$,
begin{align}
text{K}&=int_0^{infty}frac{2x^2}{1+x^4},dx\
&=int_0^{infty}frac{2}{1+x^4},dx\
end{align}
Therefore,
begin{align}
text{2K}=int_0^{infty}frac{2left(1+frac{1}{x^2}right)}{left(x-frac{1}{x}right)^2+2},dx
end{align}
Perform the change of variable $y=x-dfrac{1}{x}$,
begin{align}text{2K}&=2int_{-infty}^{+infty}frac{1}{2+x^2},dx\
&=2left[frac{1}{sqrt{2}}arctanleft(frac{x}{sqrt{2}}right)right]_{-infty}^{+infty}\
&=2times frac{pi}{sqrt{2}}
end{align}
therefore,
begin{align}
text{I+J}&=frac{pi^3}{4sqrt{2}}-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
Let $a>0$,
begin{align}
text{K}_1(a)&=int_0^{infty}frac{x^2}{a+x^4},dx\
&=frac{1}{a}int_0^{infty}frac{x^2}{1+left(a^{-frac{1}{4}}xright)^4},dx\
end{align}
Perform the change of variable $y=a^{-frac{1}{4}}x$,
begin{align}
text{K}_1(a)&=a^{-frac{1}{4}}int_0^{infty}frac{x^2}{1+x^4},dx\
&=frac{a^{-frac{1}{4}}pi}{2sqrt{2}}
end{align}
In the same manner,
begin{align}
text{K}_2(a)&=int_0^{infty}frac{x^2}{1+ax^4},dx\
&=frac{a^{-frac{3}{4}}pi}{2sqrt{2}}
end{align}
Since, for $a$ real,
begin{align}arctan a=int_0^1 frac{a}{1+a^2t^2},dtend{align}
then,
begin{align}text{L}&=int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
&=int_0^{infty}left(int_0^1 int_0^1 frac{x^2}{(1+u^2x^4)left(1+frac{v^2}{x^4}right)(1+x^4)},du,dvright),dx\
&=\
&int_0^{infty}left(int_0^1int_0^1 left(frac{x^2}{(1-u^2)(1-v^2)(1+x^4)}-frac{x^2}{1-u^2v^2}left(frac{u^2}{(1-u^2)(1+u^2x^4)}+frac{v^2}{(1-v^2)(v^2+x^4)}right)
right)dudvright)dx\
&=int_0^1int_0^1 left(frac{pi}{2sqrt{2}(1-u^2)(1-v^2)}-frac{1}{1-u^2v^2}left(frac{u^2text{K}_2(u^2)}{1-u^2}+frac{v^2text{K}_1(v^2)}{1-v^2}right)right)dudv\
&=frac{pi}{2sqrt{2}}int_0^1int_0^1 left(frac{1}{(1-u^2)(1-v^2)}-frac{1}{(1-u^2v^2)}left(frac{u^{frac{1}{2}}}{1-u^2}+frac{v^{frac{3}{2}}}{1-v^2}right)right)dudv\
&=piint_0^1left[frac{sqrt{v}left(text{ arctanh}left(sqrt{uv}right)-text{ arctan}left(sqrt{uv}right)-text{ arctanh}left(uvright)right)+arctanleft(sqrt{u}right)+lnleft(frac{sqrt{1+u}}{1+sqrt{u}}right)}{2sqrt{2}(1-v^2)}right]_{u=0}^{u=1},dv\
&=frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}big(text{ arctanh}left(sqrt{v}right)-text{ arctan}left(sqrt{v}right)-text{ arctanh}left(vright)big)+frac{pi}{4}-frac{1}{2}ln 2}{1-v^2},dv\
&=frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}arctanleft(frac{1-sqrt{v}}{1+sqrt{v}}right)}{1-v^2},dv+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right)int_0^1 frac{1-sqrt{v}}{1-v^2},dv+\
&frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}lnleft(frac{1+sqrt{v}}{2}right)}{1-v^2},dv-frac{pi}{4sqrt{2}}int_0^1frac{sqrt{v}lnleft(frac{1+v}{2}right)}{1-v^2},dv
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}text{R}_1&=int_0^1frac{sqrt{v}arctanleft(frac{1-sqrt{v}}{1+sqrt{v}}right)}{1-v^2},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2arctan v}{v(1+v^2)},dv\
&=frac{1}{2}int_0^1 frac{arctan v}{v},dv-int_0^1 frac{arctan v}{1+v^2},dv\
&=frac{1}{2}text{G}-frac{1}{2}Big[arctan^2 vBig]_0^1\
&=frac{1}{2}text{G}-frac{pi^2}{32}\
text{R}_2&=int_0^1 frac{1-sqrt{v}}{1-v^2},dv\
&=left[lnleft(frac{sqrt{1+v}}{1+sqrt{v}}right)+arctanleft(sqrt{v}right)right]_0^1\
&=frac{pi}{4}-frac{1}{2}ln 2\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}text{R}_3&=int_0^1frac{sqrt{v}lnleft(frac{1+sqrt{v}}{2}right)}{1-v^2},dv\
&=-frac{1}{2}int_0^1frac{(1-v)^2ln(1+v)}{v(1+v^2)},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{1}{2}int_0^1 frac{ln(1+v
)}{v},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{1}{4}int_0^1 frac{2vln(1-v^2)}{v^2},dv+frac{1}{2}int_0^1 frac{ln(1-v)}{v},dv\
end{align}
In the second integral perform the change of variable $y=v^2$,
begin{align}text{R}_3&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}int_0^1 frac{ln(1-v)}{v},dv\
end{align}
In the second integral perform the change of variable $y=1-v$,
begin{align}text{R}_3&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}int_0^1 frac{ln v}{1-v},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}times -zeta(2)\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{pi^2}{24}\
end{align}
Perform the change of variable $y=dfrac{1-v}{1+v}$,
begin{align}
text{S}_1&=int_0^1frac{ln(1+v)}{1+v^2},dv\
&=int_0^1frac{ln(frac{2}{1+v})}{1+v^2},dv\
&=ln 2int_0^1 frac{1}{1+v^2},dv-text{S}_1\
&=frac{pi}{4}ln 2-text{S}_1
end{align}
Therefore,
begin{align}
text{S}_1&=frac{pi}{8}ln 2\
text{R}_3&=frac{pi}{8}ln 2-frac{pi^2}{24}\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}
text{R}_4&=int_0^1frac{sqrt{v}lnleft(frac{1+v}{2}right)}{1-v^2},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2lnleft(frac{1+v^2}{(1+v)^2}right)}{v(1+v^2)},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2lnleft(1+v^2right)}{v(1+v^2)},dv+2text{R}_3\
&=frac{1}{2}int_0^1frac{ln(1+v^2)}{v},dv-int_0^1frac{ln(1+v^2)}{1+v^2},dv+frac{pi}{4}ln 2-frac{pi^2}{12}\
&=frac{1}{2}times frac{1}{4}zeta(2)-int_0^1frac{ln(1+v^2)}{1+v^2},dv+frac{pi}{4}ln 2-frac{pi^2}{12}\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1frac{ln(1+v^2)}{1+v^2},dv\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1int_0^1frac{v^2}{(1+v^2)(1+v^2t)},dt,dv\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1 left[frac{arctanleft(vright)sqrt{t}-arctanleft(vsqrt{t}right)}{(t-1)sqrt{t}}right]_{v=0}^{v=1},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1 frac{frac{pisqrt{t}}{4}-arctanleft(sqrt{t}right)}{(t-1)sqrt{t}},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}+int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}int_0^1 frac{sqrt{t}-1}{(t-1)sqrt{t}},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}+int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}Big[2lnleft(1+sqrt{t}right)Big]_0^1\
&=int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}ln 2-frac{pi^2}{16}\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{t}}{1+sqrt{t}}$,
begin{align}
text{R}_4&=int_0^1 frac{arctan t}{t},dt-frac{pi}{4}ln 2-frac{pi^2}{16}\
&=text{G}-frac{pi}{4}ln 2-frac{pi^2}{16}\
end{align}
Therefore,
begin{align}L&=frac{pi}{2sqrt{2}}text{R}_1+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right) text{R}_2+frac{pi}{2sqrt{2}}text{R}_3-frac{pi}{4sqrt{2}}text{R}_4\
&=frac{pi}{2sqrt{2}}left(frac{text{G}}{2}-frac{pi^2}{32}right)+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right)^2+frac{pi}{2sqrt{2}}left(frac{pi}{8}ln 2-frac{pi^2}{24}right)-\
&frac{pi}{4sqrt{2}}left(text{G}-frac{pi}{4}ln 2-frac{pi^2}{16}right)\
&=frac{pi^3}{96sqrt{2}}+frac{piln^2 2}{8sqrt{2}}
end{align}
Thus,
begin{align}text{I+J}&=frac{pi^3}{4sqrt{2}}-4text{L}\
&=frac{pi^3}{4sqrt{2}}-4left(frac{pi^3}{96sqrt{2}}+frac{piln^2 2}{8sqrt{2}}right)\
&=boxed{frac{5pi^3}{24sqrt{2}}-frac{piln^2 2}{2sqrt{2}}}
end{align}
answered Jan 18 at 14:45
FDPFDP
5,35711524
$endgroup$
$begingroup$
I wish I could up-vote this multiple times. Excellent work
$endgroup$
– clathratus
yesterday
add a comment |
$begingroup$
On the path of Zacky, the missing part...
Let,
begin{align}I&=int_0^{frac{pi}{2}}x^2sqrt{tan x},dx\
J&=int_0^{frac{pi}{2}}frac{x^2}{sqrt{tan x}},dx\
end{align}
Perform the change of variable $y=sqrt{tan x}$,
begin{align}I&=int_0^{infty}frac{2x^2arctan^2left(x^2right)}{1+x^4},dx\\
J&=int_0^{infty}frac{2x^2arctan^2left(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
begin{align}
text{I+J}&=int_0^{infty}frac{2x^2left(arctanleft(x^2right)+arctanleft(frac{1}{x^2}right)right)^2}{1+x^4},dx-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
&=frac{pi^2}{4}int_0^{infty}frac{2x^2}{1+x^4},dx-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
Perform the change of variable $y=dfrac{1}{x}$,
begin{align}
text{K}&=int_0^{infty}frac{2x^2}{1+x^4},dx\
&=int_0^{infty}frac{2}{1+x^4},dx\
end{align}
Therefore,
begin{align}
text{2K}=int_0^{infty}frac{2left(1+frac{1}{x^2}right)}{left(x-frac{1}{x}right)^2+2},dx
end{align}
Perform the change of variable $y=x-dfrac{1}{x}$,
begin{align}text{2K}&=2int_{-infty}^{+infty}frac{1}{2+x^2},dx\
&=2left[frac{1}{sqrt{2}}arctanleft(frac{x}{sqrt{2}}right)right]_{-infty}^{+infty}\
&=2times frac{pi}{sqrt{2}}
end{align}
therefore,
begin{align}
text{I+J}&=frac{pi^3}{4sqrt{2}}-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
Let $a>0$,
begin{align}
text{K}_1(a)&=int_0^{infty}frac{x^2}{a+x^4},dx\
&=frac{1}{a}int_0^{infty}frac{x^2}{1+left(a^{-frac{1}{4}}xright)^4},dx\
end{align}
Perform the change of variable $y=a^{-frac{1}{4}}x$,
begin{align}
text{K}_1(a)&=a^{-frac{1}{4}}int_0^{infty}frac{x^2}{1+x^4},dx\
&=frac{a^{-frac{1}{4}}pi}{2sqrt{2}}
end{align}
In the same manner,
begin{align}
text{K}_2(a)&=int_0^{infty}frac{x^2}{1+ax^4},dx\
&=frac{a^{-frac{3}{4}}pi}{2sqrt{2}}
end{align}
Since, for $a$ real,
begin{align}arctan a=int_0^1 frac{a}{1+a^2t^2},dtend{align}
then,
begin{align}text{L}&=int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
&=int_0^{infty}left(int_0^1 int_0^1 frac{x^2}{(1+u^2x^4)left(1+frac{v^2}{x^4}right)(1+x^4)},du,dvright),dx\
&=\
&int_0^{infty}left(int_0^1int_0^1 left(frac{x^2}{(1-u^2)(1-v^2)(1+x^4)}-frac{x^2}{1-u^2v^2}left(frac{u^2}{(1-u^2)(1+u^2x^4)}+frac{v^2}{(1-v^2)(v^2+x^4)}right)
right)dudvright)dx\
&=int_0^1int_0^1 left(frac{pi}{2sqrt{2}(1-u^2)(1-v^2)}-frac{1}{1-u^2v^2}left(frac{u^2text{K}_2(u^2)}{1-u^2}+frac{v^2text{K}_1(v^2)}{1-v^2}right)right)dudv\
&=frac{pi}{2sqrt{2}}int_0^1int_0^1 left(frac{1}{(1-u^2)(1-v^2)}-frac{1}{(1-u^2v^2)}left(frac{u^{frac{1}{2}}}{1-u^2}+frac{v^{frac{3}{2}}}{1-v^2}right)right)dudv\
&=piint_0^1left[frac{sqrt{v}left(text{ arctanh}left(sqrt{uv}right)-text{ arctan}left(sqrt{uv}right)-text{ arctanh}left(uvright)right)+arctanleft(sqrt{u}right)+lnleft(frac{sqrt{1+u}}{1+sqrt{u}}right)}{2sqrt{2}(1-v^2)}right]_{u=0}^{u=1},dv\
&=frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}big(text{ arctanh}left(sqrt{v}right)-text{ arctan}left(sqrt{v}right)-text{ arctanh}left(vright)big)+frac{pi}{4}-frac{1}{2}ln 2}{1-v^2},dv\
&=frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}arctanleft(frac{1-sqrt{v}}{1+sqrt{v}}right)}{1-v^2},dv+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right)int_0^1 frac{1-sqrt{v}}{1-v^2},dv+\
&frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}lnleft(frac{1+sqrt{v}}{2}right)}{1-v^2},dv-frac{pi}{4sqrt{2}}int_0^1frac{sqrt{v}lnleft(frac{1+v}{2}right)}{1-v^2},dv
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}text{R}_1&=int_0^1frac{sqrt{v}arctanleft(frac{1-sqrt{v}}{1+sqrt{v}}right)}{1-v^2},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2arctan v}{v(1+v^2)},dv\
&=frac{1}{2}int_0^1 frac{arctan v}{v},dv-int_0^1 frac{arctan v}{1+v^2},dv\
&=frac{1}{2}text{G}-frac{1}{2}Big[arctan^2 vBig]_0^1\
&=frac{1}{2}text{G}-frac{pi^2}{32}\
text{R}_2&=int_0^1 frac{1-sqrt{v}}{1-v^2},dv\
&=left[lnleft(frac{sqrt{1+v}}{1+sqrt{v}}right)+arctanleft(sqrt{v}right)right]_0^1\
&=frac{pi}{4}-frac{1}{2}ln 2\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}text{R}_3&=int_0^1frac{sqrt{v}lnleft(frac{1+sqrt{v}}{2}right)}{1-v^2},dv\
&=-frac{1}{2}int_0^1frac{(1-v)^2ln(1+v)}{v(1+v^2)},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{1}{2}int_0^1 frac{ln(1+v
)}{v},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{1}{4}int_0^1 frac{2vln(1-v^2)}{v^2},dv+frac{1}{2}int_0^1 frac{ln(1-v)}{v},dv\
end{align}
In the second integral perform the change of variable $y=v^2$,
begin{align}text{R}_3&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}int_0^1 frac{ln(1-v)}{v},dv\
end{align}
In the second integral perform the change of variable $y=1-v$,
begin{align}text{R}_3&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}int_0^1 frac{ln v}{1-v},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}times -zeta(2)\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{pi^2}{24}\
end{align}
Perform the change of variable $y=dfrac{1-v}{1+v}$,
begin{align}
text{S}_1&=int_0^1frac{ln(1+v)}{1+v^2},dv\
&=int_0^1frac{ln(frac{2}{1+v})}{1+v^2},dv\
&=ln 2int_0^1 frac{1}{1+v^2},dv-text{S}_1\
&=frac{pi}{4}ln 2-text{S}_1
end{align}
Therefore,
begin{align}
text{S}_1&=frac{pi}{8}ln 2\
text{R}_3&=frac{pi}{8}ln 2-frac{pi^2}{24}\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}
text{R}_4&=int_0^1frac{sqrt{v}lnleft(frac{1+v}{2}right)}{1-v^2},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2lnleft(frac{1+v^2}{(1+v)^2}right)}{v(1+v^2)},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2lnleft(1+v^2right)}{v(1+v^2)},dv+2text{R}_3\
&=frac{1}{2}int_0^1frac{ln(1+v^2)}{v},dv-int_0^1frac{ln(1+v^2)}{1+v^2},dv+frac{pi}{4}ln 2-frac{pi^2}{12}\
&=frac{1}{2}times frac{1}{4}zeta(2)-int_0^1frac{ln(1+v^2)}{1+v^2},dv+frac{pi}{4}ln 2-frac{pi^2}{12}\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1frac{ln(1+v^2)}{1+v^2},dv\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1int_0^1frac{v^2}{(1+v^2)(1+v^2t)},dt,dv\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1 left[frac{arctanleft(vright)sqrt{t}-arctanleft(vsqrt{t}right)}{(t-1)sqrt{t}}right]_{v=0}^{v=1},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1 frac{frac{pisqrt{t}}{4}-arctanleft(sqrt{t}right)}{(t-1)sqrt{t}},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}+int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}int_0^1 frac{sqrt{t}-1}{(t-1)sqrt{t}},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}+int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}Big[2lnleft(1+sqrt{t}right)Big]_0^1\
&=int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}ln 2-frac{pi^2}{16}\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{t}}{1+sqrt{t}}$,
begin{align}
text{R}_4&=int_0^1 frac{arctan t}{t},dt-frac{pi}{4}ln 2-frac{pi^2}{16}\
&=text{G}-frac{pi}{4}ln 2-frac{pi^2}{16}\
end{align}
Therefore,
begin{align}L&=frac{pi}{2sqrt{2}}text{R}_1+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right) text{R}_2+frac{pi}{2sqrt{2}}text{R}_3-frac{pi}{4sqrt{2}}text{R}_4\
&=frac{pi}{2sqrt{2}}left(frac{text{G}}{2}-frac{pi^2}{32}right)+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right)^2+frac{pi}{2sqrt{2}}left(frac{pi}{8}ln 2-frac{pi^2}{24}right)-\
&frac{pi}{4sqrt{2}}left(text{G}-frac{pi}{4}ln 2-frac{pi^2}{16}right)\
&=frac{pi^3}{96sqrt{2}}+frac{piln^2 2}{8sqrt{2}}
end{align}
Thus,
begin{align}text{I+J}&=frac{pi^3}{4sqrt{2}}-4text{L}\
&=frac{pi^3}{4sqrt{2}}-4left(frac{pi^3}{96sqrt{2}}+frac{piln^2 2}{8sqrt{2}}right)\
&=boxed{frac{5pi^3}{24sqrt{2}}-frac{piln^2 2}{2sqrt{2}}}
end{align}
answered Jan 18 at 14:45
FDPFDP
5,35711524
$endgroup$
On the path of Zacky, the missing part...
Let,
begin{align}I&=int_0^{frac{pi}{2}}x^2sqrt{tan x},dx\
J&=int_0^{frac{pi}{2}}frac{x^2}{sqrt{tan x}},dx\
end{align}
Perform the change of variable $y=sqrt{tan x}$,
begin{align}I&=int_0^{infty}frac{2x^2arctan^2left(x^2right)}{1+x^4},dx\\
J&=int_0^{infty}frac{2x^2arctan^2left(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
begin{align}
text{I+J}&=int_0^{infty}frac{2x^2left(arctanleft(x^2right)+arctanleft(frac{1}{x^2}right)right)^2}{1+x^4},dx-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
&=frac{pi^2}{4}int_0^{infty}frac{2x^2}{1+x^4},dx-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
Perform the change of variable $y=dfrac{1}{x}$,
begin{align}
text{K}&=int_0^{infty}frac{2x^2}{1+x^4},dx\
&=int_0^{infty}frac{2}{1+x^4},dx\
end{align}
Therefore,
begin{align}
text{2K}=int_0^{infty}frac{2left(1+frac{1}{x^2}right)}{left(x-frac{1}{x}right)^2+2},dx
end{align}
Perform the change of variable $y=x-dfrac{1}{x}$,
begin{align}text{2K}&=2int_{-infty}^{+infty}frac{1}{2+x^2},dx\
&=2left[frac{1}{sqrt{2}}arctanleft(frac{x}{sqrt{2}}right)right]_{-infty}^{+infty}\
&=2times frac{pi}{sqrt{2}}
end{align}
therefore,
begin{align}
text{I+J}&=frac{pi^3}{4sqrt{2}}-4int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
end{align}
Let $a>0$,
begin{align}
text{K}_1(a)&=int_0^{infty}frac{x^2}{a+x^4},dx\
&=frac{1}{a}int_0^{infty}frac{x^2}{1+left(a^{-frac{1}{4}}xright)^4},dx\
end{align}
Perform the change of variable $y=a^{-frac{1}{4}}x$,
begin{align}
text{K}_1(a)&=a^{-frac{1}{4}}int_0^{infty}frac{x^2}{1+x^4},dx\
&=frac{a^{-frac{1}{4}}pi}{2sqrt{2}}
end{align}
In the same manner,
begin{align}
text{K}_2(a)&=int_0^{infty}frac{x^2}{1+ax^4},dx\
&=frac{a^{-frac{3}{4}}pi}{2sqrt{2}}
end{align}
Since, for $a$ real,
begin{align}arctan a=int_0^1 frac{a}{1+a^2t^2},dtend{align}
then,
begin{align}text{L}&=int_0^{infty}frac{x^2arctanleft(x^2right)arctanleft(frac{1}{x^2}right)}{1+x^4},dx\
&=int_0^{infty}left(int_0^1 int_0^1 frac{x^2}{(1+u^2x^4)left(1+frac{v^2}{x^4}right)(1+x^4)},du,dvright),dx\
&=\
&int_0^{infty}left(int_0^1int_0^1 left(frac{x^2}{(1-u^2)(1-v^2)(1+x^4)}-frac{x^2}{1-u^2v^2}left(frac{u^2}{(1-u^2)(1+u^2x^4)}+frac{v^2}{(1-v^2)(v^2+x^4)}right)
right)dudvright)dx\
&=int_0^1int_0^1 left(frac{pi}{2sqrt{2}(1-u^2)(1-v^2)}-frac{1}{1-u^2v^2}left(frac{u^2text{K}_2(u^2)}{1-u^2}+frac{v^2text{K}_1(v^2)}{1-v^2}right)right)dudv\
&=frac{pi}{2sqrt{2}}int_0^1int_0^1 left(frac{1}{(1-u^2)(1-v^2)}-frac{1}{(1-u^2v^2)}left(frac{u^{frac{1}{2}}}{1-u^2}+frac{v^{frac{3}{2}}}{1-v^2}right)right)dudv\
&=piint_0^1left[frac{sqrt{v}left(text{ arctanh}left(sqrt{uv}right)-text{ arctan}left(sqrt{uv}right)-text{ arctanh}left(uvright)right)+arctanleft(sqrt{u}right)+lnleft(frac{sqrt{1+u}}{1+sqrt{u}}right)}{2sqrt{2}(1-v^2)}right]_{u=0}^{u=1},dv\
&=frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}big(text{ arctanh}left(sqrt{v}right)-text{ arctan}left(sqrt{v}right)-text{ arctanh}left(vright)big)+frac{pi}{4}-frac{1}{2}ln 2}{1-v^2},dv\
&=frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}arctanleft(frac{1-sqrt{v}}{1+sqrt{v}}right)}{1-v^2},dv+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right)int_0^1 frac{1-sqrt{v}}{1-v^2},dv+\
&frac{pi}{2sqrt{2}}int_0^1frac{sqrt{v}lnleft(frac{1+sqrt{v}}{2}right)}{1-v^2},dv-frac{pi}{4sqrt{2}}int_0^1frac{sqrt{v}lnleft(frac{1+v}{2}right)}{1-v^2},dv
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}text{R}_1&=int_0^1frac{sqrt{v}arctanleft(frac{1-sqrt{v}}{1+sqrt{v}}right)}{1-v^2},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2arctan v}{v(1+v^2)},dv\
&=frac{1}{2}int_0^1 frac{arctan v}{v},dv-int_0^1 frac{arctan v}{1+v^2},dv\
&=frac{1}{2}text{G}-frac{1}{2}Big[arctan^2 vBig]_0^1\
&=frac{1}{2}text{G}-frac{pi^2}{32}\
text{R}_2&=int_0^1 frac{1-sqrt{v}}{1-v^2},dv\
&=left[lnleft(frac{sqrt{1+v}}{1+sqrt{v}}right)+arctanleft(sqrt{v}right)right]_0^1\
&=frac{pi}{4}-frac{1}{2}ln 2\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}text{R}_3&=int_0^1frac{sqrt{v}lnleft(frac{1+sqrt{v}}{2}right)}{1-v^2},dv\
&=-frac{1}{2}int_0^1frac{(1-v)^2ln(1+v)}{v(1+v^2)},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{1}{2}int_0^1 frac{ln(1+v
)}{v},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{1}{4}int_0^1 frac{2vln(1-v^2)}{v^2},dv+frac{1}{2}int_0^1 frac{ln(1-v)}{v},dv\
end{align}
In the second integral perform the change of variable $y=v^2$,
begin{align}text{R}_3&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}int_0^1 frac{ln(1-v)}{v},dv\
end{align}
In the second integral perform the change of variable $y=1-v$,
begin{align}text{R}_3&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}int_0^1 frac{ln v}{1-v},dv\
&=int_0^1frac{ln(1+v)}{1+v^2},dv+frac{1}{4}times -zeta(2)\
&=int_0^1frac{ln(1+v)}{1+v^2},dv-frac{pi^2}{24}\
end{align}
Perform the change of variable $y=dfrac{1-v}{1+v}$,
begin{align}
text{S}_1&=int_0^1frac{ln(1+v)}{1+v^2},dv\
&=int_0^1frac{ln(frac{2}{1+v})}{1+v^2},dv\
&=ln 2int_0^1 frac{1}{1+v^2},dv-text{S}_1\
&=frac{pi}{4}ln 2-text{S}_1
end{align}
Therefore,
begin{align}
text{S}_1&=frac{pi}{8}ln 2\
text{R}_3&=frac{pi}{8}ln 2-frac{pi^2}{24}\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{v}}{1+sqrt{v}}$,
begin{align}
text{R}_4&=int_0^1frac{sqrt{v}lnleft(frac{1+v}{2}right)}{1-v^2},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2lnleft(frac{1+v^2}{(1+v)^2}right)}{v(1+v^2)},dv\
&=frac{1}{2}int_0^1 frac{(1-v)^2lnleft(1+v^2right)}{v(1+v^2)},dv+2text{R}_3\
&=frac{1}{2}int_0^1frac{ln(1+v^2)}{v},dv-int_0^1frac{ln(1+v^2)}{1+v^2},dv+frac{pi}{4}ln 2-frac{pi^2}{12}\
&=frac{1}{2}times frac{1}{4}zeta(2)-int_0^1frac{ln(1+v^2)}{1+v^2},dv+frac{pi}{4}ln 2-frac{pi^2}{12}\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1frac{ln(1+v^2)}{1+v^2},dv\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1int_0^1frac{v^2}{(1+v^2)(1+v^2t)},dt,dv\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1 left[frac{arctanleft(vright)sqrt{t}-arctanleft(vsqrt{t}right)}{(t-1)sqrt{t}}right]_{v=0}^{v=1},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}-int_0^1 frac{frac{pisqrt{t}}{4}-arctanleft(sqrt{t}right)}{(t-1)sqrt{t}},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}+int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}int_0^1 frac{sqrt{t}-1}{(t-1)sqrt{t}},dt\
&=frac{pi}{4}ln 2-frac{pi^2}{16}+int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}Big[2lnleft(1+sqrt{t}right)Big]_0^1\
&=int_0^1 frac{arctanleft(frac{1-sqrt{t}}{1+sqrt{t}}right)}{(1-t)sqrt{t}},dt-frac{pi}{4}ln 2-frac{pi^2}{16}\
end{align}
Perform the change of variable $y=dfrac{1-sqrt{t}}{1+sqrt{t}}$,
begin{align}
text{R}_4&=int_0^1 frac{arctan t}{t},dt-frac{pi}{4}ln 2-frac{pi^2}{16}\
&=text{G}-frac{pi}{4}ln 2-frac{pi^2}{16}\
end{align}
Therefore,
begin{align}L&=frac{pi}{2sqrt{2}}text{R}_1+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right) text{R}_2+frac{pi}{2sqrt{2}}text{R}_3-frac{pi}{4sqrt{2}}text{R}_4\
&=frac{pi}{2sqrt{2}}left(frac{text{G}}{2}-frac{pi^2}{32}right)+frac{pi}{2sqrt{2}}left(frac{pi}{4}-frac{1}{2}ln 2right)^2+frac{pi}{2sqrt{2}}left(frac{pi}{8}ln 2-frac{pi^2}{24}right)-\
&frac{pi}{4sqrt{2}}left(text{G}-frac{pi}{4}ln 2-frac{pi^2}{16}right)\
&=frac{pi^3}{96sqrt{2}}+frac{piln^2 2}{8sqrt{2}}
end{align}
Thus,
begin{align}text{I+J}&=frac{pi^3}{4sqrt{2}}-4text{L}\
&=frac{pi^3}{4sqrt{2}}-4left(frac{pi^3}{96sqrt{2}}+frac{piln^2 2}{8sqrt{2}}right)\
&=boxed{frac{5pi^3}{24sqrt{2}}-frac{piln^2 2}{2sqrt{2}}}
end{align}
answered Jan 18 at 14:45
FDPFDP
5,35711524
edited Jan 18 at 14:53
answered Jan 18 at 14:45
FDPFDP
5,35711524
answered Jan 18 at 14:45
FDPFDP
5,35711524
answered Jan 18 at 14:45
FDPFDP
5,35711524
5,35711524
$begingroup$
I wish I could up-vote this multiple times. Excellent work
$endgroup$
– clathratus
yesterday
add a comment |
$begingroup$
I wish I could up-vote this multiple times. Excellent work
$endgroup$
– clathratus
yesterday
$begingroup$
I wish I could up-vote this multiple times. Excellent work
$endgroup$
– clathratus
yesterday
$begingroup$
I wish I could up-vote this multiple times. Excellent work
$endgroup$
– clathratus
yesterday
add a comment |
$begingroup$
Both integrals depend on
$$int_{0}^{pi} x^2 frac{sqrt{1+cos x}pm sqrt{1-cos x}}{sqrt{sin x}},dx=\int_{0}^{pi/2}frac{dx}{sqrt{sin x}}left[(x^2pm(pi-x)^2)sqrt{1+cos x}+(pm x^2+(pi-x)^2)sqrt{1-cos x}right],dx $$
and they can be tackled through the Fourier series of a periodic version of $x^2pm(pi-x)^2$. For instance
$$ x^2+(pi-x)^2 = frac{2pi^2}{3}+2sum_{ngeq 1}frac{cos(2nx)}{n^2}qquad forall xin[0,pi] $$
(yes, I am exploiting Bernoulli polynomials) and for any $ninmathbb{N}^+$
$$ int_{0}^{pi/2}frac{sqrt{1+cos x}+sqrt{1-cos x}}{sqrt{sin x}}cos(2nx),dx=frac{pi}{4^nsqrt{2}}binom{2n}{n}$$
so the whole question boils down to computing
$$ sum_{ngeq 1}frac{1}{n^2 4^n}binom{2n}{n} = frac{1}{2},phantom{}_4 F_3left(1,1,1,tfrac{3}{2};2,2,2;1right)=zeta(2)-2log^2(2).$$
The last equality follows by recalling $frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}(sin t)^{2n},dt$ and by tackling
$$ int_{0}^{1}frac{text{Li}_2(z)}{sqrt{z(1-z)}},dz $$
through Fourier-Legendre series or the reflection formula for $text{Li}_2$.
The remaining part is just related to the well-known
$$ int_{0}^{1}frac{log x}{sqrt{x(1-x)}},dx = 4int_{0}^{pi/2}logsintheta,dtheta = -2pilog 2.$$
Summarizing, all the involved integrals just depend on $left.frac{d^nu}{da^nu}Bleft(a,tfrac{1}{2}right)right|_{a=1/2}$ for $nuin{1,2}$.
answered Jan 12 at 18:38
Jack D'AurizioJack D'Aurizio
289k33282662
$endgroup$
3
$begingroup$
I don't know if this fits the requirements for 'elementary', but nice job anyway
$endgroup$
– clathratus
Jan 12 at 20:55
$begingroup$
Could you give me a tip on proving $$sum_{ngeq1}frac1{4^n n^2}{2nchoose n}=zeta(2)-2log^22$$?
$endgroup$
– clathratus
Jan 18 at 4:01
$begingroup$
The value of this series is (integrate first using variable t) begin{align}sum_{ngeq 1}^infty frac{1}{4^n n^2}binom{2n}{n}&=frac{2}{pi}int_{x=0}^1int_{t=0}^{frac{pi}{2}}frac{ln xsin^2 t }{xsin^2 t-1},dt,dxend{align}
$endgroup$
– FDP
Jan 18 at 21:40
add a comment |
$begingroup$
Both integrals depend on
$$int_{0}^{pi} x^2 frac{sqrt{1+cos x}pm sqrt{1-cos x}}{sqrt{sin x}},dx=\int_{0}^{pi/2}frac{dx}{sqrt{sin x}}left[(x^2pm(pi-x)^2)sqrt{1+cos x}+(pm x^2+(pi-x)^2)sqrt{1-cos x}right],dx $$
and they can be tackled through the Fourier series of a periodic version of $x^2pm(pi-x)^2$. For instance
$$ x^2+(pi-x)^2 = frac{2pi^2}{3}+2sum_{ngeq 1}frac{cos(2nx)}{n^2}qquad forall xin[0,pi] $$
(yes, I am exploiting Bernoulli polynomials) and for any $ninmathbb{N}^+$
$$ int_{0}^{pi/2}frac{sqrt{1+cos x}+sqrt{1-cos x}}{sqrt{sin x}}cos(2nx),dx=frac{pi}{4^nsqrt{2}}binom{2n}{n}$$
so the whole question boils down to computing
$$ sum_{ngeq 1}frac{1}{n^2 4^n}binom{2n}{n} = frac{1}{2},phantom{}_4 F_3left(1,1,1,tfrac{3}{2};2,2,2;1right)=zeta(2)-2log^2(2).$$
The last equality follows by recalling $frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}(sin t)^{2n},dt$ and by tackling
$$ int_{0}^{1}frac{text{Li}_2(z)}{sqrt{z(1-z)}},dz $$
through Fourier-Legendre series or the reflection formula for $text{Li}_2$.
The remaining part is just related to the well-known
$$ int_{0}^{1}frac{log x}{sqrt{x(1-x)}},dx = 4int_{0}^{pi/2}logsintheta,dtheta = -2pilog 2.$$
Summarizing, all the involved integrals just depend on $left.frac{d^nu}{da^nu}Bleft(a,tfrac{1}{2}right)right|_{a=1/2}$ for $nuin{1,2}$.
answered Jan 12 at 18:38
Jack D'AurizioJack D'Aurizio
289k33282662
$endgroup$
3
$begingroup$
I don't know if this fits the requirements for 'elementary', but nice job anyway
$endgroup$
– clathratus
Jan 12 at 20:55
$begingroup$
Could you give me a tip on proving $$sum_{ngeq1}frac1{4^n n^2}{2nchoose n}=zeta(2)-2log^22$$?
$endgroup$
– clathratus
Jan 18 at 4:01
$begingroup$
The value of this series is (integrate first using variable t) begin{align}sum_{ngeq 1}^infty frac{1}{4^n n^2}binom{2n}{n}&=frac{2}{pi}int_{x=0}^1int_{t=0}^{frac{pi}{2}}frac{ln xsin^2 t }{xsin^2 t-1},dt,dxend{align}
$endgroup$
– FDP
Jan 18 at 21:40
add a comment |
$begingroup$
Both integrals depend on
$$int_{0}^{pi} x^2 frac{sqrt{1+cos x}pm sqrt{1-cos x}}{sqrt{sin x}},dx=\int_{0}^{pi/2}frac{dx}{sqrt{sin x}}left[(x^2pm(pi-x)^2)sqrt{1+cos x}+(pm x^2+(pi-x)^2)sqrt{1-cos x}right],dx $$
and they can be tackled through the Fourier series of a periodic version of $x^2pm(pi-x)^2$. For instance
$$ x^2+(pi-x)^2 = frac{2pi^2}{3}+2sum_{ngeq 1}frac{cos(2nx)}{n^2}qquad forall xin[0,pi] $$
(yes, I am exploiting Bernoulli polynomials) and for any $ninmathbb{N}^+$
$$ int_{0}^{pi/2}frac{sqrt{1+cos x}+sqrt{1-cos x}}{sqrt{sin x}}cos(2nx),dx=frac{pi}{4^nsqrt{2}}binom{2n}{n}$$
so the whole question boils down to computing
$$ sum_{ngeq 1}frac{1}{n^2 4^n}binom{2n}{n} = frac{1}{2},phantom{}_4 F_3left(1,1,1,tfrac{3}{2};2,2,2;1right)=zeta(2)-2log^2(2).$$
The last equality follows by recalling $frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}(sin t)^{2n},dt$ and by tackling
$$ int_{0}^{1}frac{text{Li}_2(z)}{sqrt{z(1-z)}},dz $$
through Fourier-Legendre series or the reflection formula for $text{Li}_2$.
The remaining part is just related to the well-known
$$ int_{0}^{1}frac{log x}{sqrt{x(1-x)}},dx = 4int_{0}^{pi/2}logsintheta,dtheta = -2pilog 2.$$
Summarizing, all the involved integrals just depend on $left.frac{d^nu}{da^nu}Bleft(a,tfrac{1}{2}right)right|_{a=1/2}$ for $nuin{1,2}$.
answered Jan 12 at 18:38
Jack D'AurizioJack D'Aurizio
289k33282662
$endgroup$
Both integrals depend on
$$int_{0}^{pi} x^2 frac{sqrt{1+cos x}pm sqrt{1-cos x}}{sqrt{sin x}},dx=\int_{0}^{pi/2}frac{dx}{sqrt{sin x}}left[(x^2pm(pi-x)^2)sqrt{1+cos x}+(pm x^2+(pi-x)^2)sqrt{1-cos x}right],dx $$
and they can be tackled through the Fourier series of a periodic version of $x^2pm(pi-x)^2$. For instance
$$ x^2+(pi-x)^2 = frac{2pi^2}{3}+2sum_{ngeq 1}frac{cos(2nx)}{n^2}qquad forall xin[0,pi] $$
(yes, I am exploiting Bernoulli polynomials) and for any $ninmathbb{N}^+$
$$ int_{0}^{pi/2}frac{sqrt{1+cos x}+sqrt{1-cos x}}{sqrt{sin x}}cos(2nx),dx=frac{pi}{4^nsqrt{2}}binom{2n}{n}$$
so the whole question boils down to computing
$$ sum_{ngeq 1}frac{1}{n^2 4^n}binom{2n}{n} = frac{1}{2},phantom{}_4 F_3left(1,1,1,tfrac{3}{2};2,2,2;1right)=zeta(2)-2log^2(2).$$
The last equality follows by recalling $frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}(sin t)^{2n},dt$ and by tackling
$$ int_{0}^{1}frac{text{Li}_2(z)}{sqrt{z(1-z)}},dz $$
through Fourier-Legendre series or the reflection formula for $text{Li}_2$.
The remaining part is just related to the well-known
$$ int_{0}^{1}frac{log x}{sqrt{x(1-x)}},dx = 4int_{0}^{pi/2}logsintheta,dtheta = -2pilog 2.$$
Summarizing, all the involved integrals just depend on $left.frac{d^nu}{da^nu}Bleft(a,tfrac{1}{2}right)right|_{a=1/2}$ for $nuin{1,2}$.
answered Jan 12 at 18:38
Jack D'AurizioJack D'Aurizio
289k33282662
edited Jan 12 at 18:52
answered Jan 12 at 18:38
Jack D'AurizioJack D'Aurizio
289k33282662
answered Jan 12 at 18:38
Jack D'AurizioJack D'Aurizio
289k33282662
answered Jan 12 at 18:38
Jack D'AurizioJack D'Aurizio
289k33282662
289k33282662
3
$begingroup$
I don't know if this fits the requirements for 'elementary', but nice job anyway
$endgroup$
– clathratus
Jan 12 at 20:55
$begingroup$
Could you give me a tip on proving $$sum_{ngeq1}frac1{4^n n^2}{2nchoose n}=zeta(2)-2log^22$$?
$endgroup$
– clathratus
Jan 18 at 4:01
$begingroup$
The value of this series is (integrate first using variable t) begin{align}sum_{ngeq 1}^infty frac{1}{4^n n^2}binom{2n}{n}&=frac{2}{pi}int_{x=0}^1int_{t=0}^{frac{pi}{2}}frac{ln xsin^2 t }{xsin^2 t-1},dt,dxend{align}
$endgroup$
– FDP
Jan 18 at 21:40
add a comment |
3
$begingroup$
I don't know if this fits the requirements for 'elementary', but nice job anyway
$endgroup$
– clathratus
Jan 12 at 20:55
$begingroup$
Could you give me a tip on proving $$sum_{ngeq1}frac1{4^n n^2}{2nchoose n}=zeta(2)-2log^22$$?
$endgroup$
– clathratus
Jan 18 at 4:01
$begingroup$
The value of this series is (integrate first using variable t) begin{align}sum_{ngeq 1}^infty frac{1}{4^n n^2}binom{2n}{n}&=frac{2}{pi}int_{x=0}^1int_{t=0}^{frac{pi}{2}}frac{ln xsin^2 t }{xsin^2 t-1},dt,dxend{align}
$endgroup$
– FDP
Jan 18 at 21:40
3
3
$begingroup$
I don't know if this fits the requirements for 'elementary', but nice job anyway
$endgroup$
– clathratus
Jan 12 at 20:55
$begingroup$
I don't know if this fits the requirements for 'elementary', but nice job anyway
$endgroup$
– clathratus
Jan 12 at 20:55
$begingroup$
Could you give me a tip on proving $$sum_{ngeq1}frac1{4^n n^2}{2nchoose n}=zeta(2)-2log^22$$?
$endgroup$
– clathratus
Jan 18 at 4:01
$begingroup$
Could you give me a tip on proving $$sum_{ngeq1}frac1{4^n n^2}{2nchoose n}=zeta(2)-2log^22$$?
$endgroup$
– clathratus
Jan 18 at 4:01
$begingroup$
The value of this series is (integrate first using variable t) begin{align}sum_{ngeq 1}^infty frac{1}{4^n n^2}binom{2n}{n}&=frac{2}{pi}int_{x=0}^1int_{t=0}^{frac{pi}{2}}frac{ln xsin^2 t }{xsin^2 t-1},dt,dxend{align}
$endgroup$
– FDP
Jan 18 at 21:40
$begingroup$
The value of this series is (integrate first using variable t) begin{align}sum_{ngeq 1}^infty frac{1}{4^n n^2}binom{2n}{n}&=frac{2}{pi}int_{x=0}^1int_{t=0}^{frac{pi}{2}}frac{ln xsin^2 t }{xsin^2 t-1},dt,dxend{align}
$endgroup$
– FDP
Jan 18 at 21:40
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071027%2fintegral-int-0-frac-pi2-x2-sqrt-tan-x-mathrm-dx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
I look forward to being about to integrate as you do. Very nice work here.
$endgroup$
– DavidG
Jan 13 at 6:42