Mixing
Explicit calculation of $int_0^{infty} frac{sin(2arctan x)}{(1+x^2)(exp(2pi x)-1)}mathrm{d}x$
$begingroup$
Is it possible to confirm the value of this integral using the methods of complex analysis or similar?
$$
int_0^{infty} frac{sin(2arctan x)}{(1+x^2)(exp(2pi x)-1)}mathrm{d}x=frac{pi^2-9}{12}
$$
Of course, one can reduce it to the definition of the polylogarithm and a $zeta$-function, but I was looking for an explicit calculation.
integration complex-analysis
edited Jan 13 at 17:30
amWhy
1
asked Jan 13 at 16:01
SchnarcoSchnarco
1618
$endgroup$
add a comment |
$begingroup$
Is it possible to confirm the value of this integral using the methods of complex analysis or similar?
$$
int_0^{infty} frac{sin(2arctan x)}{(1+x^2)(exp(2pi x)-1)}mathrm{d}x=frac{pi^2-9}{12}
$$
Of course, one can reduce it to the definition of the polylogarithm and a $zeta$-function, but I was looking for an explicit calculation.
integration complex-analysis
edited Jan 13 at 17:30
amWhy
1
asked Jan 13 at 16:01
SchnarcoSchnarco
1618
$endgroup$
$begingroup$
You have the value of the integral. What you're seeking is to confirm this answer.
$endgroup$
– David G. Stork
Jan 13 at 16:05
$begingroup$
Hint: $sin2theta=2sinthetacostheta=2tanthetacos^2theta=2tantheta/sec^2theta=2tantheta/(1+tan^2theta)$
$endgroup$
– Barry Cipra
Jan 13 at 16:13
add a comment |
$begingroup$
Is it possible to confirm the value of this integral using the methods of complex analysis or similar?
$$
int_0^{infty} frac{sin(2arctan x)}{(1+x^2)(exp(2pi x)-1)}mathrm{d}x=frac{pi^2-9}{12}
$$
Of course, one can reduce it to the definition of the polylogarithm and a $zeta$-function, but I was looking for an explicit calculation.
integration complex-analysis
edited Jan 13 at 17:30
amWhy
1
asked Jan 13 at 16:01
SchnarcoSchnarco
1618
$endgroup$
Is it possible to confirm the value of this integral using the methods of complex analysis or similar?
$$
int_0^{infty} frac{sin(2arctan x)}{(1+x^2)(exp(2pi x)-1)}mathrm{d}x=frac{pi^2-9}{12}
$$
Of course, one can reduce it to the definition of the polylogarithm and a $zeta$-function, but I was looking for an explicit calculation.
integration complex-analysis
integration complex-analysis
edited Jan 13 at 17:30
amWhy
1
asked Jan 13 at 16:01
SchnarcoSchnarco
1618
edited Jan 13 at 17:30
amWhy
1
asked Jan 13 at 16:01
SchnarcoSchnarco
1618
edited Jan 13 at 17:30
amWhy
1
edited Jan 13 at 17:30
amWhy
1
edited Jan 13 at 17:30
amWhy
1
1
asked Jan 13 at 16:01
SchnarcoSchnarco
1618
asked Jan 13 at 16:01
SchnarcoSchnarco
1618
asked Jan 13 at 16:01
SchnarcoSchnarco
1618
1618
$begingroup$
You have the value of the integral. What you're seeking is to confirm this answer.
$endgroup$
– David G. Stork
Jan 13 at 16:05
$begingroup$
Hint: $sin2theta=2sinthetacostheta=2tanthetacos^2theta=2tantheta/sec^2theta=2tantheta/(1+tan^2theta)$
$endgroup$
– Barry Cipra
Jan 13 at 16:13
add a comment |
$begingroup$
You have the value of the integral. What you're seeking is to confirm this answer.
$endgroup$
– David G. Stork
Jan 13 at 16:05
$begingroup$
Hint: $sin2theta=2sinthetacostheta=2tanthetacos^2theta=2tantheta/sec^2theta=2tantheta/(1+tan^2theta)$
$endgroup$
– Barry Cipra
Jan 13 at 16:13
$begingroup$
You have the value of the integral. What you're seeking is to confirm this answer.
$endgroup$
– David G. Stork
Jan 13 at 16:05
$begingroup$
You have the value of the integral. What you're seeking is to confirm this answer.
$endgroup$
– David G. Stork
Jan 13 at 16:05
$begingroup$
Hint: $sin2theta=2sinthetacostheta=2tanthetacos^2theta=2tantheta/sec^2theta=2tantheta/(1+tan^2theta)$
$endgroup$
– Barry Cipra
Jan 13 at 16:13
$begingroup$
Hint: $sin2theta=2sinthetacostheta=2tanthetacos^2theta=2tantheta/sec^2theta=2tantheta/(1+tan^2theta)$
$endgroup$
– Barry Cipra
Jan 13 at 16:13
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
This will require some calculations but no special functions. Making the change of variables $x = -x$ on $mathbb R^-$, we obtain
$$int_{mathbb R} frac x {(x^2 + 1)^2 (e^{2 pi x} - 1)} dx =
underbrace {int_{mathbb R^+} frac {2 x} {(x^2 + 1)^2 (e^{2 pi x} - 1)} dx}_
{= I} +
int_{mathbb R^+} frac x {(x^2 + 1)^2} dx, \
I = 2 pi i sum_{k geq 1}
operatorname*{Res}_{x = i k} frac x {(x^2 + 1)^2 (e^{2 pi x} - 1)} +
frac 1 {2 (x^2 + 1)} biggrvert_{x = 0}^infty = \
2 pi i frac {4 pi^2 + 3} {96 pi i} -
2 pi i sum_{k geq 2} frac k {2 pi i (k^2 - 1)^2} -
frac 1 2 = \
frac {4 pi^3 - 21} {48} - frac 1 4 sum_{k geq 2}
left( frac 1 {(k - 1)^2} - frac 1 {(k + 1)^2} right) = \
frac {4 pi^3 - 21} {48} - frac 1 4 left( 1 + frac 1 4 right) =
frac {pi^2 - 9} {12}.$$
answered Jan 13 at 20:36
MaximMaxim
5,5981219
$endgroup$
$begingroup$
How did you get the term $int_{mathbb R^+} frac x {(x^2 + 1)^2} dx$? What kind of integration contour did you use to be able to apply residue method?
$endgroup$
– user
Jan 13 at 23:41
1
$begingroup$
If $f(x) = x/((x^2 + 1)^2 (e^{2 pi x} - 1))$, $f(x) + f(-x) = 2 f(x) + x/(x^2 + 1)^2$. To evaluate $int_{mathbb R} f(x) dx$, take the sequence of contours consisting of $[-k - 1/2, k + 1/2]$ plus a half-circle on that diameter.
$endgroup$
– Maxim
Jan 14 at 1:23
$begingroup$
Could you demonstrate that the integral over the semicircle tends to 0, as $ktoinfty$? It should be enough to prove $|expleft(pi(2k+1)e^{iphi}right)-1|>M$ with a positive constant $M$.
$endgroup$
– user
Jan 14 at 15:05
1
$begingroup$
@user In fact, to make things simpler, take the polyline $$left[ k + frac 1 2, k + frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 right]$$ instead of a semicircle. On the vertical segments, $$|e^{2 pi x} - 1| geq ||e^{2 pi x}| - 1| = |e^{2 pi operatorname{Re} x} - 1|.$$ On the horizontal segment, $$|{e^{2 pi x} - 1}| = |{e^{2 pi(operatorname {Re} x + i(k + 1/2))} - 1}| = e^{2 pi operatorname{Re} x} + 1.$$
$endgroup$
– Maxim
Jan 15 at 10:49
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I am convinced (+1). Consider a possibility to add the clarifications you made in the answer.
$endgroup$
– user
Jan 15 at 13:28
add a comment |
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This approach is unnecessarily complicated, but quite fun:
As explained in the comments and the other answers, we may write
$$ I equiv int limits_0^infty frac{sin(2arctan x)}{(1+x^2)(exp(2pi x)-1)} , mathrm{d}x = int limits_0^infty frac{2 x}{(1+x^2)(mathrm{e}^{2pi x}-1)} , mathrm{d} x , . $$
We can then use the geometric series, integration by parts and this Laplace transform to obtain
begin{align}
I &= sum limits_{n=1}^infty int limits_0^infty frac{2 x}{(1+x^2)^2} mathrm{e}^{-2 pi n x} , mathrm{d} x = sum limits_{n=1}^infty left[1 - 2 pi n int limits_0^infty frac{mathrm{e}^{-2 pi n x}}{1+x^2} , mathrm{d} x right] \
&= sum limits_{n=1}^infty left[1 - 2 pi n left(frac{pi}{2} - operatorname{Si}(2 pi n)right)right] , ,
end{align}
where $operatorname{Si}$ is the sine integral . Plugging in its definition and integrating by parts twice more we arrive at
$$ I = sum limits_{n=1}^infty left[1 - 2 pi n int limits_1^infty frac{sin(2 pi n t)}{t} , mathrm{d} t right] = 2 sum limits_{n=1}^infty frac{1}{2 pi n} int limits_1^infty frac{sin(2 pi n t)}{t^3} , mathrm{d} t , .$$
Now the Fourier series
$$ 1 - 2 {t} = 4 sum limits_{n=1}^infty frac{sin(2 pi n t)}{2 pi n} , , , t in mathbb{R} , , $$
and the fractional part integral
$$ int limits_0^1 x left{frac{1}{x}right} , mathrm{d} t = int limits_1^infty frac{{t}}{t^3} , mathrm{d} t = 1 - frac{zeta(2)}{2} = 1 - frac{pi^2}{12} $$
are sufficient to derive the final result
$$ I = frac{1}{2} int limits_1^infty frac{1 - 2{t}}{t^3} , mathrm{d} t = frac{pi^2}{12} - frac{3}{4} = frac{pi^2 - 9}{12} $$
and thus prove $pi > 3$ in a rather convoluted manner.
answered Jan 13 at 18:59
ComplexYetTrivialComplexYetTrivial
4,5512631
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add a comment |
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Not an answer but I believe that the form below is better for contour integration $$int_0^infty frac{sin(2arctan(x))}{(1+x^2)(e^{2pi x} -1)}dx=2int_0^infty frac{sin(arctan(x))cos(arctan(x))}{(1+x^2)(e^{2 pi x} -1)}dx=2int_0^infty frac{x}{(1+x^2)^2 (e^{2 pi x} -1)}dx$$ Now all that remains is to find a suitable contour that includes $[0,infty)$, and apply residue theorem.
answered Jan 13 at 16:17
aledenaleden
2,352511
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add a comment |
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Here is an approach that first converts the integral to a double integral.
Let
$$I = int_0^infty frac{sin (2 tan^{-1} x)}{(1 + x^2) (e^{2 pi x} - 1)} , dx = 2 int_0^infty frac{x}{(1 + x^2)^2 (e^{2 pi x} - 1)} , dx.$$
Observe that
$$frac{1}{2} int_0^infty y e^{-y} sin (xy) , dy = frac{x}{(1 + x^2)^2}.$$
Thus the integral can be rewritten as
$$I = int_0^infty y e^{-y} int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx , dy, tag1$$
after a change of order has been made.
For the inner $x$-integral
begin{align}
int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx &= int_0^infty frac{e^{-2 pi x} sin (xy)}{1 - e^{-2 pi x}} , dx\
&= int_0^infty sum_{n = 0}^infty e^{-2 pi x} sin (xy) cdot e^{-2pi n x} , dx\
&= sum_{n = 1}^infty int_0^infty e^{-2pi n x} sin (xy) , dx,
end{align}
where a shift in the index of $n mapsto n - 1$ has been made. Now integrating by parts twice leads to
$$int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx = sum_{n = 1}^infty frac{y}{4 pi^2 n^2 + y^2} = frac{1}{2 pi} sum_{n = 1}^infty frac{frac{y}{2pi}}{left (frac{y}{2 pi} right )^2 + n^2}. tag2$$
Using the well-known result of
$$pi coth (pi z) = frac{1}{z} + 2 sum_{n = 1}^infty frac{z}{z^2 + n^2} , quad z neq 0,$$
the sum in (2) can be expressed as
$$int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx = frac{1}{4} coth left (frac{y}{2} right ) - frac{1}{2y}.$$
On returning to our integral in (1), we have
begin{align}
I &= -frac{1}{2} int_0^infty e^{-y} + frac{1}{4} int_0^infty y e^{-y} coth left (frac{y}{2} right ) , dy = -frac{1}{2} + frac{1}{4} J,
end{align}
where
$$J = int_0^infty y e^{-y} coth left (frac{y}{2} right ) , dy.$$
Finding $J$ we have
begin{align}
J &= int_0^infty y e^{-y} frac{e^{y/2} + e^{-y/2}}{e^{y/2} - e^{-y/2}} , dy\
&= int_0^infty y e^{-y} frac{1 + e^{-y}}{1 - e^{-y}} , dy\
&= int_0^infty y e^{-y} left (1 + frac{2 e^{-y}}{1 - e^{-y}} right ) , dy\
&= int_0^infty y e^{-y} , dy + 2 int_0^infty frac{y e^{-2y}}{1 - e^{-y}} , dy\
&= 1 + 2 sum_{n = 0}^infty int_0^infty y e^{-y(n + 2)} , dy\
&= 1 + 2 sum_{n = 0}^infty frac{1}{(n + 2)^2} qquad text{(by parts)}\
&= 1 + 2 sum_{n = 2}^infty frac{1}{n^2} \
&= 1 + 2 sum_{n = 1}^infty frac{1}{n^2} - 2\
&= -1 + 2 cdot frac{pi^2}{6}\
&= -1 + frac{pi^2}{3}.
end{align}
So finally we have
$$I = -frac{1}{2} + frac{1}{4} left (-1 + frac{pi^2}{3} right ),$$
or
$$int_0^infty frac{sin (2 tan^{-1} x)}{(1 + x^2) (e^{2 pi x} - 1)} , dx = frac{pi^2 - 9}{12},$$
as expected.
answered Jan 16 at 5:53
omegadotomegadot
5,9122728
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4 Answers
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4 Answers
4
active
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$begingroup$
This will require some calculations but no special functions. Making the change of variables $x = -x$ on $mathbb R^-$, we obtain
$$int_{mathbb R} frac x {(x^2 + 1)^2 (e^{2 pi x} - 1)} dx =
underbrace {int_{mathbb R^+} frac {2 x} {(x^2 + 1)^2 (e^{2 pi x} - 1)} dx}_
{= I} +
int_{mathbb R^+} frac x {(x^2 + 1)^2} dx, \
I = 2 pi i sum_{k geq 1}
operatorname*{Res}_{x = i k} frac x {(x^2 + 1)^2 (e^{2 pi x} - 1)} +
frac 1 {2 (x^2 + 1)} biggrvert_{x = 0}^infty = \
2 pi i frac {4 pi^2 + 3} {96 pi i} -
2 pi i sum_{k geq 2} frac k {2 pi i (k^2 - 1)^2} -
frac 1 2 = \
frac {4 pi^3 - 21} {48} - frac 1 4 sum_{k geq 2}
left( frac 1 {(k - 1)^2} - frac 1 {(k + 1)^2} right) = \
frac {4 pi^3 - 21} {48} - frac 1 4 left( 1 + frac 1 4 right) =
frac {pi^2 - 9} {12}.$$
answered Jan 13 at 20:36
MaximMaxim
5,5981219
$endgroup$
$begingroup$
How did you get the term $int_{mathbb R^+} frac x {(x^2 + 1)^2} dx$? What kind of integration contour did you use to be able to apply residue method?
$endgroup$
– user
Jan 13 at 23:41
1
$begingroup$
If $f(x) = x/((x^2 + 1)^2 (e^{2 pi x} - 1))$, $f(x) + f(-x) = 2 f(x) + x/(x^2 + 1)^2$. To evaluate $int_{mathbb R} f(x) dx$, take the sequence of contours consisting of $[-k - 1/2, k + 1/2]$ plus a half-circle on that diameter.
$endgroup$
– Maxim
Jan 14 at 1:23
$begingroup$
Could you demonstrate that the integral over the semicircle tends to 0, as $ktoinfty$? It should be enough to prove $|expleft(pi(2k+1)e^{iphi}right)-1|>M$ with a positive constant $M$.
$endgroup$
– user
Jan 14 at 15:05
1
$begingroup$
@user In fact, to make things simpler, take the polyline $$left[ k + frac 1 2, k + frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 right]$$ instead of a semicircle. On the vertical segments, $$|e^{2 pi x} - 1| geq ||e^{2 pi x}| - 1| = |e^{2 pi operatorname{Re} x} - 1|.$$ On the horizontal segment, $$|{e^{2 pi x} - 1}| = |{e^{2 pi(operatorname {Re} x + i(k + 1/2))} - 1}| = e^{2 pi operatorname{Re} x} + 1.$$
$endgroup$
– Maxim
Jan 15 at 10:49
$begingroup$
I am convinced (+1). Consider a possibility to add the clarifications you made in the answer.
$endgroup$
– user
Jan 15 at 13:28
add a comment |
$begingroup$
This will require some calculations but no special functions. Making the change of variables $x = -x$ on $mathbb R^-$, we obtain
$$int_{mathbb R} frac x {(x^2 + 1)^2 (e^{2 pi x} - 1)} dx =
underbrace {int_{mathbb R^+} frac {2 x} {(x^2 + 1)^2 (e^{2 pi x} - 1)} dx}_
{= I} +
int_{mathbb R^+} frac x {(x^2 + 1)^2} dx, \
I = 2 pi i sum_{k geq 1}
operatorname*{Res}_{x = i k} frac x {(x^2 + 1)^2 (e^{2 pi x} - 1)} +
frac 1 {2 (x^2 + 1)} biggrvert_{x = 0}^infty = \
2 pi i frac {4 pi^2 + 3} {96 pi i} -
2 pi i sum_{k geq 2} frac k {2 pi i (k^2 - 1)^2} -
frac 1 2 = \
frac {4 pi^3 - 21} {48} - frac 1 4 sum_{k geq 2}
left( frac 1 {(k - 1)^2} - frac 1 {(k + 1)^2} right) = \
frac {4 pi^3 - 21} {48} - frac 1 4 left( 1 + frac 1 4 right) =
frac {pi^2 - 9} {12}.$$
answered Jan 13 at 20:36
MaximMaxim
5,5981219
$endgroup$
$begingroup$
How did you get the term $int_{mathbb R^+} frac x {(x^2 + 1)^2} dx$? What kind of integration contour did you use to be able to apply residue method?
$endgroup$
– user
Jan 13 at 23:41
1
$begingroup$
If $f(x) = x/((x^2 + 1)^2 (e^{2 pi x} - 1))$, $f(x) + f(-x) = 2 f(x) + x/(x^2 + 1)^2$. To evaluate $int_{mathbb R} f(x) dx$, take the sequence of contours consisting of $[-k - 1/2, k + 1/2]$ plus a half-circle on that diameter.
$endgroup$
– Maxim
Jan 14 at 1:23
$begingroup$
Could you demonstrate that the integral over the semicircle tends to 0, as $ktoinfty$? It should be enough to prove $|expleft(pi(2k+1)e^{iphi}right)-1|>M$ with a positive constant $M$.
$endgroup$
– user
Jan 14 at 15:05
1
$begingroup$
@user In fact, to make things simpler, take the polyline $$left[ k + frac 1 2, k + frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 right]$$ instead of a semicircle. On the vertical segments, $$|e^{2 pi x} - 1| geq ||e^{2 pi x}| - 1| = |e^{2 pi operatorname{Re} x} - 1|.$$ On the horizontal segment, $$|{e^{2 pi x} - 1}| = |{e^{2 pi(operatorname {Re} x + i(k + 1/2))} - 1}| = e^{2 pi operatorname{Re} x} + 1.$$
$endgroup$
– Maxim
Jan 15 at 10:49
$begingroup$
I am convinced (+1). Consider a possibility to add the clarifications you made in the answer.
$endgroup$
– user
Jan 15 at 13:28
add a comment |
$begingroup$
This will require some calculations but no special functions. Making the change of variables $x = -x$ on $mathbb R^-$, we obtain
$$int_{mathbb R} frac x {(x^2 + 1)^2 (e^{2 pi x} - 1)} dx =
underbrace {int_{mathbb R^+} frac {2 x} {(x^2 + 1)^2 (e^{2 pi x} - 1)} dx}_
{= I} +
int_{mathbb R^+} frac x {(x^2 + 1)^2} dx, \
I = 2 pi i sum_{k geq 1}
operatorname*{Res}_{x = i k} frac x {(x^2 + 1)^2 (e^{2 pi x} - 1)} +
frac 1 {2 (x^2 + 1)} biggrvert_{x = 0}^infty = \
2 pi i frac {4 pi^2 + 3} {96 pi i} -
2 pi i sum_{k geq 2} frac k {2 pi i (k^2 - 1)^2} -
frac 1 2 = \
frac {4 pi^3 - 21} {48} - frac 1 4 sum_{k geq 2}
left( frac 1 {(k - 1)^2} - frac 1 {(k + 1)^2} right) = \
frac {4 pi^3 - 21} {48} - frac 1 4 left( 1 + frac 1 4 right) =
frac {pi^2 - 9} {12}.$$
answered Jan 13 at 20:36
MaximMaxim
5,5981219
$endgroup$
This will require some calculations but no special functions. Making the change of variables $x = -x$ on $mathbb R^-$, we obtain
$$int_{mathbb R} frac x {(x^2 + 1)^2 (e^{2 pi x} - 1)} dx =
underbrace {int_{mathbb R^+} frac {2 x} {(x^2 + 1)^2 (e^{2 pi x} - 1)} dx}_
{= I} +
int_{mathbb R^+} frac x {(x^2 + 1)^2} dx, \
I = 2 pi i sum_{k geq 1}
operatorname*{Res}_{x = i k} frac x {(x^2 + 1)^2 (e^{2 pi x} - 1)} +
frac 1 {2 (x^2 + 1)} biggrvert_{x = 0}^infty = \
2 pi i frac {4 pi^2 + 3} {96 pi i} -
2 pi i sum_{k geq 2} frac k {2 pi i (k^2 - 1)^2} -
frac 1 2 = \
frac {4 pi^3 - 21} {48} - frac 1 4 sum_{k geq 2}
left( frac 1 {(k - 1)^2} - frac 1 {(k + 1)^2} right) = \
frac {4 pi^3 - 21} {48} - frac 1 4 left( 1 + frac 1 4 right) =
frac {pi^2 - 9} {12}.$$
answered Jan 13 at 20:36
MaximMaxim
5,5981219
edited Jan 13 at 20:41
answered Jan 13 at 20:36
MaximMaxim
5,5981219
answered Jan 13 at 20:36
MaximMaxim
5,5981219
answered Jan 13 at 20:36
MaximMaxim
5,5981219
5,5981219
$begingroup$
How did you get the term $int_{mathbb R^+} frac x {(x^2 + 1)^2} dx$? What kind of integration contour did you use to be able to apply residue method?
$endgroup$
– user
Jan 13 at 23:41
1
$begingroup$
If $f(x) = x/((x^2 + 1)^2 (e^{2 pi x} - 1))$, $f(x) + f(-x) = 2 f(x) + x/(x^2 + 1)^2$. To evaluate $int_{mathbb R} f(x) dx$, take the sequence of contours consisting of $[-k - 1/2, k + 1/2]$ plus a half-circle on that diameter.
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– Maxim
Jan 14 at 1:23
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Could you demonstrate that the integral over the semicircle tends to 0, as $ktoinfty$? It should be enough to prove $|expleft(pi(2k+1)e^{iphi}right)-1|>M$ with a positive constant $M$.
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– user
Jan 14 at 15:05
1
$begingroup$
@user In fact, to make things simpler, take the polyline $$left[ k + frac 1 2, k + frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 right]$$ instead of a semicircle. On the vertical segments, $$|e^{2 pi x} - 1| geq ||e^{2 pi x}| - 1| = |e^{2 pi operatorname{Re} x} - 1|.$$ On the horizontal segment, $$|{e^{2 pi x} - 1}| = |{e^{2 pi(operatorname {Re} x + i(k + 1/2))} - 1}| = e^{2 pi operatorname{Re} x} + 1.$$
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– Maxim
Jan 15 at 10:49
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I am convinced (+1). Consider a possibility to add the clarifications you made in the answer.
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– user
Jan 15 at 13:28
add a comment |
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How did you get the term $int_{mathbb R^+} frac x {(x^2 + 1)^2} dx$? What kind of integration contour did you use to be able to apply residue method?
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– user
Jan 13 at 23:41
1
$begingroup$
If $f(x) = x/((x^2 + 1)^2 (e^{2 pi x} - 1))$, $f(x) + f(-x) = 2 f(x) + x/(x^2 + 1)^2$. To evaluate $int_{mathbb R} f(x) dx$, take the sequence of contours consisting of $[-k - 1/2, k + 1/2]$ plus a half-circle on that diameter.
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– Maxim
Jan 14 at 1:23
$begingroup$
Could you demonstrate that the integral over the semicircle tends to 0, as $ktoinfty$? It should be enough to prove $|expleft(pi(2k+1)e^{iphi}right)-1|>M$ with a positive constant $M$.
$endgroup$
– user
Jan 14 at 15:05
1
$begingroup$
@user In fact, to make things simpler, take the polyline $$left[ k + frac 1 2, k + frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 right]$$ instead of a semicircle. On the vertical segments, $$|e^{2 pi x} - 1| geq ||e^{2 pi x}| - 1| = |e^{2 pi operatorname{Re} x} - 1|.$$ On the horizontal segment, $$|{e^{2 pi x} - 1}| = |{e^{2 pi(operatorname {Re} x + i(k + 1/2))} - 1}| = e^{2 pi operatorname{Re} x} + 1.$$
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– Maxim
Jan 15 at 10:49
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I am convinced (+1). Consider a possibility to add the clarifications you made in the answer.
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– user
Jan 15 at 13:28
$begingroup$
How did you get the term $int_{mathbb R^+} frac x {(x^2 + 1)^2} dx$? What kind of integration contour did you use to be able to apply residue method?
$endgroup$
– user
Jan 13 at 23:41
$begingroup$
How did you get the term $int_{mathbb R^+} frac x {(x^2 + 1)^2} dx$? What kind of integration contour did you use to be able to apply residue method?
$endgroup$
– user
Jan 13 at 23:41
1
1
$begingroup$
If $f(x) = x/((x^2 + 1)^2 (e^{2 pi x} - 1))$, $f(x) + f(-x) = 2 f(x) + x/(x^2 + 1)^2$. To evaluate $int_{mathbb R} f(x) dx$, take the sequence of contours consisting of $[-k - 1/2, k + 1/2]$ plus a half-circle on that diameter.
$endgroup$
– Maxim
Jan 14 at 1:23
$begingroup$
If $f(x) = x/((x^2 + 1)^2 (e^{2 pi x} - 1))$, $f(x) + f(-x) = 2 f(x) + x/(x^2 + 1)^2$. To evaluate $int_{mathbb R} f(x) dx$, take the sequence of contours consisting of $[-k - 1/2, k + 1/2]$ plus a half-circle on that diameter.
$endgroup$
– Maxim
Jan 14 at 1:23
$begingroup$
Could you demonstrate that the integral over the semicircle tends to 0, as $ktoinfty$? It should be enough to prove $|expleft(pi(2k+1)e^{iphi}right)-1|>M$ with a positive constant $M$.
$endgroup$
– user
Jan 14 at 15:05
$begingroup$
Could you demonstrate that the integral over the semicircle tends to 0, as $ktoinfty$? It should be enough to prove $|expleft(pi(2k+1)e^{iphi}right)-1|>M$ with a positive constant $M$.
$endgroup$
– user
Jan 14 at 15:05
1
1
$begingroup$
@user In fact, to make things simpler, take the polyline $$left[ k + frac 1 2, k + frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 right]$$ instead of a semicircle. On the vertical segments, $$|e^{2 pi x} - 1| geq ||e^{2 pi x}| - 1| = |e^{2 pi operatorname{Re} x} - 1|.$$ On the horizontal segment, $$|{e^{2 pi x} - 1}| = |{e^{2 pi(operatorname {Re} x + i(k + 1/2))} - 1}| = e^{2 pi operatorname{Re} x} + 1.$$
$endgroup$
– Maxim
Jan 15 at 10:49
$begingroup$
@user In fact, to make things simpler, take the polyline $$left[ k + frac 1 2, k + frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 + i left( k + frac 1 2 right), -k - frac 1 2 right]$$ instead of a semicircle. On the vertical segments, $$|e^{2 pi x} - 1| geq ||e^{2 pi x}| - 1| = |e^{2 pi operatorname{Re} x} - 1|.$$ On the horizontal segment, $$|{e^{2 pi x} - 1}| = |{e^{2 pi(operatorname {Re} x + i(k + 1/2))} - 1}| = e^{2 pi operatorname{Re} x} + 1.$$
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– Maxim
Jan 15 at 10:49
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I am convinced (+1). Consider a possibility to add the clarifications you made in the answer.
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– user
Jan 15 at 13:28
$begingroup$
I am convinced (+1). Consider a possibility to add the clarifications you made in the answer.
$endgroup$
– user
Jan 15 at 13:28
add a comment |
$begingroup$
This approach is unnecessarily complicated, but quite fun:
As explained in the comments and the other answers, we may write
$$ I equiv int limits_0^infty frac{sin(2arctan x)}{(1+x^2)(exp(2pi x)-1)} , mathrm{d}x = int limits_0^infty frac{2 x}{(1+x^2)(mathrm{e}^{2pi x}-1)} , mathrm{d} x , . $$
We can then use the geometric series, integration by parts and this Laplace transform to obtain
begin{align}
I &= sum limits_{n=1}^infty int limits_0^infty frac{2 x}{(1+x^2)^2} mathrm{e}^{-2 pi n x} , mathrm{d} x = sum limits_{n=1}^infty left[1 - 2 pi n int limits_0^infty frac{mathrm{e}^{-2 pi n x}}{1+x^2} , mathrm{d} x right] \
&= sum limits_{n=1}^infty left[1 - 2 pi n left(frac{pi}{2} - operatorname{Si}(2 pi n)right)right] , ,
end{align}
where $operatorname{Si}$ is the sine integral . Plugging in its definition and integrating by parts twice more we arrive at
$$ I = sum limits_{n=1}^infty left[1 - 2 pi n int limits_1^infty frac{sin(2 pi n t)}{t} , mathrm{d} t right] = 2 sum limits_{n=1}^infty frac{1}{2 pi n} int limits_1^infty frac{sin(2 pi n t)}{t^3} , mathrm{d} t , .$$
Now the Fourier series
$$ 1 - 2 {t} = 4 sum limits_{n=1}^infty frac{sin(2 pi n t)}{2 pi n} , , , t in mathbb{R} , , $$
and the fractional part integral
$$ int limits_0^1 x left{frac{1}{x}right} , mathrm{d} t = int limits_1^infty frac{{t}}{t^3} , mathrm{d} t = 1 - frac{zeta(2)}{2} = 1 - frac{pi^2}{12} $$
are sufficient to derive the final result
$$ I = frac{1}{2} int limits_1^infty frac{1 - 2{t}}{t^3} , mathrm{d} t = frac{pi^2}{12} - frac{3}{4} = frac{pi^2 - 9}{12} $$
and thus prove $pi > 3$ in a rather convoluted manner.
answered Jan 13 at 18:59
ComplexYetTrivialComplexYetTrivial
4,5512631
$endgroup$
add a comment |
$begingroup$
This approach is unnecessarily complicated, but quite fun:
As explained in the comments and the other answers, we may write
$$ I equiv int limits_0^infty frac{sin(2arctan x)}{(1+x^2)(exp(2pi x)-1)} , mathrm{d}x = int limits_0^infty frac{2 x}{(1+x^2)(mathrm{e}^{2pi x}-1)} , mathrm{d} x , . $$
We can then use the geometric series, integration by parts and this Laplace transform to obtain
begin{align}
I &= sum limits_{n=1}^infty int limits_0^infty frac{2 x}{(1+x^2)^2} mathrm{e}^{-2 pi n x} , mathrm{d} x = sum limits_{n=1}^infty left[1 - 2 pi n int limits_0^infty frac{mathrm{e}^{-2 pi n x}}{1+x^2} , mathrm{d} x right] \
&= sum limits_{n=1}^infty left[1 - 2 pi n left(frac{pi}{2} - operatorname{Si}(2 pi n)right)right] , ,
end{align}
where $operatorname{Si}$ is the sine integral . Plugging in its definition and integrating by parts twice more we arrive at
$$ I = sum limits_{n=1}^infty left[1 - 2 pi n int limits_1^infty frac{sin(2 pi n t)}{t} , mathrm{d} t right] = 2 sum limits_{n=1}^infty frac{1}{2 pi n} int limits_1^infty frac{sin(2 pi n t)}{t^3} , mathrm{d} t , .$$
Now the Fourier series
$$ 1 - 2 {t} = 4 sum limits_{n=1}^infty frac{sin(2 pi n t)}{2 pi n} , , , t in mathbb{R} , , $$
and the fractional part integral
$$ int limits_0^1 x left{frac{1}{x}right} , mathrm{d} t = int limits_1^infty frac{{t}}{t^3} , mathrm{d} t = 1 - frac{zeta(2)}{2} = 1 - frac{pi^2}{12} $$
are sufficient to derive the final result
$$ I = frac{1}{2} int limits_1^infty frac{1 - 2{t}}{t^3} , mathrm{d} t = frac{pi^2}{12} - frac{3}{4} = frac{pi^2 - 9}{12} $$
and thus prove $pi > 3$ in a rather convoluted manner.
answered Jan 13 at 18:59
ComplexYetTrivialComplexYetTrivial
4,5512631
$endgroup$
add a comment |
$begingroup$
This approach is unnecessarily complicated, but quite fun:
As explained in the comments and the other answers, we may write
$$ I equiv int limits_0^infty frac{sin(2arctan x)}{(1+x^2)(exp(2pi x)-1)} , mathrm{d}x = int limits_0^infty frac{2 x}{(1+x^2)(mathrm{e}^{2pi x}-1)} , mathrm{d} x , . $$
We can then use the geometric series, integration by parts and this Laplace transform to obtain
begin{align}
I &= sum limits_{n=1}^infty int limits_0^infty frac{2 x}{(1+x^2)^2} mathrm{e}^{-2 pi n x} , mathrm{d} x = sum limits_{n=1}^infty left[1 - 2 pi n int limits_0^infty frac{mathrm{e}^{-2 pi n x}}{1+x^2} , mathrm{d} x right] \
&= sum limits_{n=1}^infty left[1 - 2 pi n left(frac{pi}{2} - operatorname{Si}(2 pi n)right)right] , ,
end{align}
where $operatorname{Si}$ is the sine integral . Plugging in its definition and integrating by parts twice more we arrive at
$$ I = sum limits_{n=1}^infty left[1 - 2 pi n int limits_1^infty frac{sin(2 pi n t)}{t} , mathrm{d} t right] = 2 sum limits_{n=1}^infty frac{1}{2 pi n} int limits_1^infty frac{sin(2 pi n t)}{t^3} , mathrm{d} t , .$$
Now the Fourier series
$$ 1 - 2 {t} = 4 sum limits_{n=1}^infty frac{sin(2 pi n t)}{2 pi n} , , , t in mathbb{R} , , $$
and the fractional part integral
$$ int limits_0^1 x left{frac{1}{x}right} , mathrm{d} t = int limits_1^infty frac{{t}}{t^3} , mathrm{d} t = 1 - frac{zeta(2)}{2} = 1 - frac{pi^2}{12} $$
are sufficient to derive the final result
$$ I = frac{1}{2} int limits_1^infty frac{1 - 2{t}}{t^3} , mathrm{d} t = frac{pi^2}{12} - frac{3}{4} = frac{pi^2 - 9}{12} $$
and thus prove $pi > 3$ in a rather convoluted manner.
answered Jan 13 at 18:59
ComplexYetTrivialComplexYetTrivial
4,5512631
$endgroup$
This approach is unnecessarily complicated, but quite fun:
As explained in the comments and the other answers, we may write
$$ I equiv int limits_0^infty frac{sin(2arctan x)}{(1+x^2)(exp(2pi x)-1)} , mathrm{d}x = int limits_0^infty frac{2 x}{(1+x^2)(mathrm{e}^{2pi x}-1)} , mathrm{d} x , . $$
We can then use the geometric series, integration by parts and this Laplace transform to obtain
begin{align}
I &= sum limits_{n=1}^infty int limits_0^infty frac{2 x}{(1+x^2)^2} mathrm{e}^{-2 pi n x} , mathrm{d} x = sum limits_{n=1}^infty left[1 - 2 pi n int limits_0^infty frac{mathrm{e}^{-2 pi n x}}{1+x^2} , mathrm{d} x right] \
&= sum limits_{n=1}^infty left[1 - 2 pi n left(frac{pi}{2} - operatorname{Si}(2 pi n)right)right] , ,
end{align}
where $operatorname{Si}$ is the sine integral . Plugging in its definition and integrating by parts twice more we arrive at
$$ I = sum limits_{n=1}^infty left[1 - 2 pi n int limits_1^infty frac{sin(2 pi n t)}{t} , mathrm{d} t right] = 2 sum limits_{n=1}^infty frac{1}{2 pi n} int limits_1^infty frac{sin(2 pi n t)}{t^3} , mathrm{d} t , .$$
Now the Fourier series
$$ 1 - 2 {t} = 4 sum limits_{n=1}^infty frac{sin(2 pi n t)}{2 pi n} , , , t in mathbb{R} , , $$
and the fractional part integral
$$ int limits_0^1 x left{frac{1}{x}right} , mathrm{d} t = int limits_1^infty frac{{t}}{t^3} , mathrm{d} t = 1 - frac{zeta(2)}{2} = 1 - frac{pi^2}{12} $$
are sufficient to derive the final result
$$ I = frac{1}{2} int limits_1^infty frac{1 - 2{t}}{t^3} , mathrm{d} t = frac{pi^2}{12} - frac{3}{4} = frac{pi^2 - 9}{12} $$
and thus prove $pi > 3$ in a rather convoluted manner.
answered Jan 13 at 18:59
ComplexYetTrivialComplexYetTrivial
4,5512631
answered Jan 13 at 18:59
ComplexYetTrivialComplexYetTrivial
4,5512631
answered Jan 13 at 18:59
ComplexYetTrivialComplexYetTrivial
4,5512631
answered Jan 13 at 18:59
ComplexYetTrivialComplexYetTrivial
4,5512631
4,5512631
add a comment |
add a comment |
$begingroup$
Not an answer but I believe that the form below is better for contour integration $$int_0^infty frac{sin(2arctan(x))}{(1+x^2)(e^{2pi x} -1)}dx=2int_0^infty frac{sin(arctan(x))cos(arctan(x))}{(1+x^2)(e^{2 pi x} -1)}dx=2int_0^infty frac{x}{(1+x^2)^2 (e^{2 pi x} -1)}dx$$ Now all that remains is to find a suitable contour that includes $[0,infty)$, and apply residue theorem.
answered Jan 13 at 16:17
aledenaleden
2,352511
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add a comment |
$begingroup$
Not an answer but I believe that the form below is better for contour integration $$int_0^infty frac{sin(2arctan(x))}{(1+x^2)(e^{2pi x} -1)}dx=2int_0^infty frac{sin(arctan(x))cos(arctan(x))}{(1+x^2)(e^{2 pi x} -1)}dx=2int_0^infty frac{x}{(1+x^2)^2 (e^{2 pi x} -1)}dx$$ Now all that remains is to find a suitable contour that includes $[0,infty)$, and apply residue theorem.
answered Jan 13 at 16:17
aledenaleden
2,352511
$endgroup$
add a comment |
$begingroup$
Not an answer but I believe that the form below is better for contour integration $$int_0^infty frac{sin(2arctan(x))}{(1+x^2)(e^{2pi x} -1)}dx=2int_0^infty frac{sin(arctan(x))cos(arctan(x))}{(1+x^2)(e^{2 pi x} -1)}dx=2int_0^infty frac{x}{(1+x^2)^2 (e^{2 pi x} -1)}dx$$ Now all that remains is to find a suitable contour that includes $[0,infty)$, and apply residue theorem.
answered Jan 13 at 16:17
aledenaleden
2,352511
$endgroup$
Not an answer but I believe that the form below is better for contour integration $$int_0^infty frac{sin(2arctan(x))}{(1+x^2)(e^{2pi x} -1)}dx=2int_0^infty frac{sin(arctan(x))cos(arctan(x))}{(1+x^2)(e^{2 pi x} -1)}dx=2int_0^infty frac{x}{(1+x^2)^2 (e^{2 pi x} -1)}dx$$ Now all that remains is to find a suitable contour that includes $[0,infty)$, and apply residue theorem.
answered Jan 13 at 16:17
aledenaleden
2,352511
answered Jan 13 at 16:17
aledenaleden
2,352511
answered Jan 13 at 16:17
aledenaleden
2,352511
answered Jan 13 at 16:17
aledenaleden
2,352511
2,352511
add a comment |
add a comment |
$begingroup$
Here is an approach that first converts the integral to a double integral.
Let
$$I = int_0^infty frac{sin (2 tan^{-1} x)}{(1 + x^2) (e^{2 pi x} - 1)} , dx = 2 int_0^infty frac{x}{(1 + x^2)^2 (e^{2 pi x} - 1)} , dx.$$
Observe that
$$frac{1}{2} int_0^infty y e^{-y} sin (xy) , dy = frac{x}{(1 + x^2)^2}.$$
Thus the integral can be rewritten as
$$I = int_0^infty y e^{-y} int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx , dy, tag1$$
after a change of order has been made.
For the inner $x$-integral
begin{align}
int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx &= int_0^infty frac{e^{-2 pi x} sin (xy)}{1 - e^{-2 pi x}} , dx\
&= int_0^infty sum_{n = 0}^infty e^{-2 pi x} sin (xy) cdot e^{-2pi n x} , dx\
&= sum_{n = 1}^infty int_0^infty e^{-2pi n x} sin (xy) , dx,
end{align}
where a shift in the index of $n mapsto n - 1$ has been made. Now integrating by parts twice leads to
$$int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx = sum_{n = 1}^infty frac{y}{4 pi^2 n^2 + y^2} = frac{1}{2 pi} sum_{n = 1}^infty frac{frac{y}{2pi}}{left (frac{y}{2 pi} right )^2 + n^2}. tag2$$
Using the well-known result of
$$pi coth (pi z) = frac{1}{z} + 2 sum_{n = 1}^infty frac{z}{z^2 + n^2} , quad z neq 0,$$
the sum in (2) can be expressed as
$$int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx = frac{1}{4} coth left (frac{y}{2} right ) - frac{1}{2y}.$$
On returning to our integral in (1), we have
begin{align}
I &= -frac{1}{2} int_0^infty e^{-y} + frac{1}{4} int_0^infty y e^{-y} coth left (frac{y}{2} right ) , dy = -frac{1}{2} + frac{1}{4} J,
end{align}
where
$$J = int_0^infty y e^{-y} coth left (frac{y}{2} right ) , dy.$$
Finding $J$ we have
begin{align}
J &= int_0^infty y e^{-y} frac{e^{y/2} + e^{-y/2}}{e^{y/2} - e^{-y/2}} , dy\
&= int_0^infty y e^{-y} frac{1 + e^{-y}}{1 - e^{-y}} , dy\
&= int_0^infty y e^{-y} left (1 + frac{2 e^{-y}}{1 - e^{-y}} right ) , dy\
&= int_0^infty y e^{-y} , dy + 2 int_0^infty frac{y e^{-2y}}{1 - e^{-y}} , dy\
&= 1 + 2 sum_{n = 0}^infty int_0^infty y e^{-y(n + 2)} , dy\
&= 1 + 2 sum_{n = 0}^infty frac{1}{(n + 2)^2} qquad text{(by parts)}\
&= 1 + 2 sum_{n = 2}^infty frac{1}{n^2} \
&= 1 + 2 sum_{n = 1}^infty frac{1}{n^2} - 2\
&= -1 + 2 cdot frac{pi^2}{6}\
&= -1 + frac{pi^2}{3}.
end{align}
So finally we have
$$I = -frac{1}{2} + frac{1}{4} left (-1 + frac{pi^2}{3} right ),$$
or
$$int_0^infty frac{sin (2 tan^{-1} x)}{(1 + x^2) (e^{2 pi x} - 1)} , dx = frac{pi^2 - 9}{12},$$
as expected.
answered Jan 16 at 5:53
omegadotomegadot
5,9122728
$endgroup$
add a comment |
$begingroup$
Here is an approach that first converts the integral to a double integral.
Let
$$I = int_0^infty frac{sin (2 tan^{-1} x)}{(1 + x^2) (e^{2 pi x} - 1)} , dx = 2 int_0^infty frac{x}{(1 + x^2)^2 (e^{2 pi x} - 1)} , dx.$$
Observe that
$$frac{1}{2} int_0^infty y e^{-y} sin (xy) , dy = frac{x}{(1 + x^2)^2}.$$
Thus the integral can be rewritten as
$$I = int_0^infty y e^{-y} int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx , dy, tag1$$
after a change of order has been made.
For the inner $x$-integral
begin{align}
int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx &= int_0^infty frac{e^{-2 pi x} sin (xy)}{1 - e^{-2 pi x}} , dx\
&= int_0^infty sum_{n = 0}^infty e^{-2 pi x} sin (xy) cdot e^{-2pi n x} , dx\
&= sum_{n = 1}^infty int_0^infty e^{-2pi n x} sin (xy) , dx,
end{align}
where a shift in the index of $n mapsto n - 1$ has been made. Now integrating by parts twice leads to
$$int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx = sum_{n = 1}^infty frac{y}{4 pi^2 n^2 + y^2} = frac{1}{2 pi} sum_{n = 1}^infty frac{frac{y}{2pi}}{left (frac{y}{2 pi} right )^2 + n^2}. tag2$$
Using the well-known result of
$$pi coth (pi z) = frac{1}{z} + 2 sum_{n = 1}^infty frac{z}{z^2 + n^2} , quad z neq 0,$$
the sum in (2) can be expressed as
$$int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx = frac{1}{4} coth left (frac{y}{2} right ) - frac{1}{2y}.$$
On returning to our integral in (1), we have
begin{align}
I &= -frac{1}{2} int_0^infty e^{-y} + frac{1}{4} int_0^infty y e^{-y} coth left (frac{y}{2} right ) , dy = -frac{1}{2} + frac{1}{4} J,
end{align}
where
$$J = int_0^infty y e^{-y} coth left (frac{y}{2} right ) , dy.$$
Finding $J$ we have
begin{align}
J &= int_0^infty y e^{-y} frac{e^{y/2} + e^{-y/2}}{e^{y/2} - e^{-y/2}} , dy\
&= int_0^infty y e^{-y} frac{1 + e^{-y}}{1 - e^{-y}} , dy\
&= int_0^infty y e^{-y} left (1 + frac{2 e^{-y}}{1 - e^{-y}} right ) , dy\
&= int_0^infty y e^{-y} , dy + 2 int_0^infty frac{y e^{-2y}}{1 - e^{-y}} , dy\
&= 1 + 2 sum_{n = 0}^infty int_0^infty y e^{-y(n + 2)} , dy\
&= 1 + 2 sum_{n = 0}^infty frac{1}{(n + 2)^2} qquad text{(by parts)}\
&= 1 + 2 sum_{n = 2}^infty frac{1}{n^2} \
&= 1 + 2 sum_{n = 1}^infty frac{1}{n^2} - 2\
&= -1 + 2 cdot frac{pi^2}{6}\
&= -1 + frac{pi^2}{3}.
end{align}
So finally we have
$$I = -frac{1}{2} + frac{1}{4} left (-1 + frac{pi^2}{3} right ),$$
or
$$int_0^infty frac{sin (2 tan^{-1} x)}{(1 + x^2) (e^{2 pi x} - 1)} , dx = frac{pi^2 - 9}{12},$$
as expected.
answered Jan 16 at 5:53
omegadotomegadot
5,9122728
$endgroup$
add a comment |
$begingroup$
Here is an approach that first converts the integral to a double integral.
Let
$$I = int_0^infty frac{sin (2 tan^{-1} x)}{(1 + x^2) (e^{2 pi x} - 1)} , dx = 2 int_0^infty frac{x}{(1 + x^2)^2 (e^{2 pi x} - 1)} , dx.$$
Observe that
$$frac{1}{2} int_0^infty y e^{-y} sin (xy) , dy = frac{x}{(1 + x^2)^2}.$$
Thus the integral can be rewritten as
$$I = int_0^infty y e^{-y} int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx , dy, tag1$$
after a change of order has been made.
For the inner $x$-integral
begin{align}
int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx &= int_0^infty frac{e^{-2 pi x} sin (xy)}{1 - e^{-2 pi x}} , dx\
&= int_0^infty sum_{n = 0}^infty e^{-2 pi x} sin (xy) cdot e^{-2pi n x} , dx\
&= sum_{n = 1}^infty int_0^infty e^{-2pi n x} sin (xy) , dx,
end{align}
where a shift in the index of $n mapsto n - 1$ has been made. Now integrating by parts twice leads to
$$int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx = sum_{n = 1}^infty frac{y}{4 pi^2 n^2 + y^2} = frac{1}{2 pi} sum_{n = 1}^infty frac{frac{y}{2pi}}{left (frac{y}{2 pi} right )^2 + n^2}. tag2$$
Using the well-known result of
$$pi coth (pi z) = frac{1}{z} + 2 sum_{n = 1}^infty frac{z}{z^2 + n^2} , quad z neq 0,$$
the sum in (2) can be expressed as
$$int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx = frac{1}{4} coth left (frac{y}{2} right ) - frac{1}{2y}.$$
On returning to our integral in (1), we have
begin{align}
I &= -frac{1}{2} int_0^infty e^{-y} + frac{1}{4} int_0^infty y e^{-y} coth left (frac{y}{2} right ) , dy = -frac{1}{2} + frac{1}{4} J,
end{align}
where
$$J = int_0^infty y e^{-y} coth left (frac{y}{2} right ) , dy.$$
Finding $J$ we have
begin{align}
J &= int_0^infty y e^{-y} frac{e^{y/2} + e^{-y/2}}{e^{y/2} - e^{-y/2}} , dy\
&= int_0^infty y e^{-y} frac{1 + e^{-y}}{1 - e^{-y}} , dy\
&= int_0^infty y e^{-y} left (1 + frac{2 e^{-y}}{1 - e^{-y}} right ) , dy\
&= int_0^infty y e^{-y} , dy + 2 int_0^infty frac{y e^{-2y}}{1 - e^{-y}} , dy\
&= 1 + 2 sum_{n = 0}^infty int_0^infty y e^{-y(n + 2)} , dy\
&= 1 + 2 sum_{n = 0}^infty frac{1}{(n + 2)^2} qquad text{(by parts)}\
&= 1 + 2 sum_{n = 2}^infty frac{1}{n^2} \
&= 1 + 2 sum_{n = 1}^infty frac{1}{n^2} - 2\
&= -1 + 2 cdot frac{pi^2}{6}\
&= -1 + frac{pi^2}{3}.
end{align}
So finally we have
$$I = -frac{1}{2} + frac{1}{4} left (-1 + frac{pi^2}{3} right ),$$
or
$$int_0^infty frac{sin (2 tan^{-1} x)}{(1 + x^2) (e^{2 pi x} - 1)} , dx = frac{pi^2 - 9}{12},$$
as expected.
answered Jan 16 at 5:53
omegadotomegadot
5,9122728
$endgroup$
Here is an approach that first converts the integral to a double integral.
Let
$$I = int_0^infty frac{sin (2 tan^{-1} x)}{(1 + x^2) (e^{2 pi x} - 1)} , dx = 2 int_0^infty frac{x}{(1 + x^2)^2 (e^{2 pi x} - 1)} , dx.$$
Observe that
$$frac{1}{2} int_0^infty y e^{-y} sin (xy) , dy = frac{x}{(1 + x^2)^2}.$$
Thus the integral can be rewritten as
$$I = int_0^infty y e^{-y} int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx , dy, tag1$$
after a change of order has been made.
For the inner $x$-integral
begin{align}
int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx &= int_0^infty frac{e^{-2 pi x} sin (xy)}{1 - e^{-2 pi x}} , dx\
&= int_0^infty sum_{n = 0}^infty e^{-2 pi x} sin (xy) cdot e^{-2pi n x} , dx\
&= sum_{n = 1}^infty int_0^infty e^{-2pi n x} sin (xy) , dx,
end{align}
where a shift in the index of $n mapsto n - 1$ has been made. Now integrating by parts twice leads to
$$int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx = sum_{n = 1}^infty frac{y}{4 pi^2 n^2 + y^2} = frac{1}{2 pi} sum_{n = 1}^infty frac{frac{y}{2pi}}{left (frac{y}{2 pi} right )^2 + n^2}. tag2$$
Using the well-known result of
$$pi coth (pi z) = frac{1}{z} + 2 sum_{n = 1}^infty frac{z}{z^2 + n^2} , quad z neq 0,$$
the sum in (2) can be expressed as
$$int_0^infty frac{sin (xy)}{e^{2 pi x} - 1} , dx = frac{1}{4} coth left (frac{y}{2} right ) - frac{1}{2y}.$$
On returning to our integral in (1), we have
begin{align}
I &= -frac{1}{2} int_0^infty e^{-y} + frac{1}{4} int_0^infty y e^{-y} coth left (frac{y}{2} right ) , dy = -frac{1}{2} + frac{1}{4} J,
end{align}
where
$$J = int_0^infty y e^{-y} coth left (frac{y}{2} right ) , dy.$$
Finding $J$ we have
begin{align}
J &= int_0^infty y e^{-y} frac{e^{y/2} + e^{-y/2}}{e^{y/2} - e^{-y/2}} , dy\
&= int_0^infty y e^{-y} frac{1 + e^{-y}}{1 - e^{-y}} , dy\
&= int_0^infty y e^{-y} left (1 + frac{2 e^{-y}}{1 - e^{-y}} right ) , dy\
&= int_0^infty y e^{-y} , dy + 2 int_0^infty frac{y e^{-2y}}{1 - e^{-y}} , dy\
&= 1 + 2 sum_{n = 0}^infty int_0^infty y e^{-y(n + 2)} , dy\
&= 1 + 2 sum_{n = 0}^infty frac{1}{(n + 2)^2} qquad text{(by parts)}\
&= 1 + 2 sum_{n = 2}^infty frac{1}{n^2} \
&= 1 + 2 sum_{n = 1}^infty frac{1}{n^2} - 2\
&= -1 + 2 cdot frac{pi^2}{6}\
&= -1 + frac{pi^2}{3}.
end{align}
So finally we have
$$I = -frac{1}{2} + frac{1}{4} left (-1 + frac{pi^2}{3} right ),$$
or
$$int_0^infty frac{sin (2 tan^{-1} x)}{(1 + x^2) (e^{2 pi x} - 1)} , dx = frac{pi^2 - 9}{12},$$
as expected.
answered Jan 16 at 5:53
omegadotomegadot
5,9122728
answered Jan 16 at 5:53
omegadotomegadot
5,9122728
answered Jan 16 at 5:53
omegadotomegadot
5,9122728
answered Jan 16 at 5:53
omegadotomegadot
5,9122728
5,9122728
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$begingroup$
You have the value of the integral. What you're seeking is to confirm this answer.
$endgroup$
– David G. Stork
Jan 13 at 16:05
$begingroup$
Hint: $sin2theta=2sinthetacostheta=2tanthetacos^2theta=2tantheta/sec^2theta=2tantheta/(1+tan^2theta)$
$endgroup$
– Barry Cipra
Jan 13 at 16:13