Another integral

[ZaidAlyafey]

\(\displaystyle \int^{\infty}_0 \frac{\sin(ax)}{e^{2\pi x}-1} \, dx \)

 

[ZaidAlyafey]

The integral can be solved by contour integration but I made a mistake somewhere , anyways the following approach is somehow a transformation to the contour to famous one .

\(\displaystyle \int^{\infty}_0 \frac{\sin(ax)}{e^{2\pi x}-1} \, dx \)

Let \(\displaystyle t = 2 \pi x\)

\(\displaystyle
\begin{align*}

\frac{1}{2\pi }\int^{\infty}_0 \frac{\sin \left(\frac{at}{2\pi } \right) }{e^{t}-1} \, dt


&=\frac{1}{2\pi }\sum^{\infty}_{n =0}\frac{(-1)^n}{\Gamma(2n+2)}\int^{\infty}_0 \frac{ \left(\frac{at}{2\pi } \right)^{2n+1} }{e^{t}-1} \, dt \\
\\

&=\frac{1}{2\pi }\sum^{\infty}_{n =0}\frac{(-1)^n \left(\frac{a}{2\pi } \right)^{2n+1}}{\Gamma(2n+2)}\int^{\infty}_0 \frac{t^{2n+1} }{e^{t}-1} \, dt \\

&=\frac{1}{2\pi }\sum^{\infty}_{n =0}\frac{(-1)^n \left(\frac{a}{2\pi } \right)^{2n+1} \Gamma (2n+2)\zeta(2n+2)}{\Gamma(2n+2)} \\

&=\frac{1}{2\pi }\sum^{\infty}_{n =0}(-1)^n \left(\frac{a}{2\pi } \right)^{2n+1}\zeta(2n+2)\\

&=\frac{1}{2\pi }\sum^{\infty}_{n =0}(-1)^n \left(\frac{a}{2\pi } \right)^{2n+1}\sum_{k=1}^{\infty} \frac{1}{k^{2n+2}}\\

&=\frac{1}{2\pi }\sum_{k=1}^{\infty}\frac{1}{k^2}\sum^{\infty}_{n =0}(-1)^n \frac{\left(\frac{a}{2\pi } \right)^{2n+1} }{k^{2n}}\\

&=\frac{a}{4\pi^2 } \sum_{k=1}^{\infty} \frac{1}{k^2} \sum^{\infty}_{n =0}\left( -\frac{a^2}{4\pi^2 k^2 } \right)^n \\

&=\frac{a}{4\pi^2 }\sum_{k=1}^{\infty}\frac{1}{k^2(1+\frac{a^2}{4\pi^2 k^2})} \\

&= \sum_{k=1}^{\infty}\frac{a}{k^2+a^2}\\

\end{align*}


\)

\(\displaystyle \int^{\infty}_0 \frac{\sin(ax)}{e^{2\pi x}-1} \, dx = \sum_{k=1}^{\infty}\frac{a}{k^2+a^2} \)


If anyone wants to try this sum , otherwise I will solve it in the next thread .

 

[ZaidAlyafey]

My solution is based on that \(\displaystyle |a|<2 \pi \) . I will try to find a general solution.

 

[chisigma]

The series...


$$\sum_{n=1}^{\infty} \frac{a}{n^{2} + a^{2}}\ (1)$$


... can be computed finding the Fourier series expansion of the function $\cosh ax$ in $[-\pi,\pi]$ obtaining...


$$\cosh ax = \frac{\sinh \pi a}{\pi\ a} + 2\ a\ \frac{\sinh \pi a}{\pi}\ \sum_{n=1}^{\infty} \frac{\cos n \pi}{n^{2} + a^{2}}\ \cos n x\ (2)$$

... and setting in (2) $x=\pi$ we obtain...

$$ \sum_{n=1}^{\infty} \frac{a}{n^{2} + a^{2}} = \frac{\pi}{2} (\coth \pi a - \frac{1}{\pi\ a})\ (3)$$

Kind regards

$\chi$ $\sigma$

 

[ZaidAlyafey]

The other way is using complex analysis

\(\displaystyle \sum_{k=-\infty}^{\infty }\frac{a}{k^2+a^2} =- \text{Res}\left(\frac{a \pi \cot ( \pi z) }{z^2+a^2} ;\pm ai \right)\)


\(\displaystyle \sum_{k \leq -1}\frac{a}{k^2+a^2}+ \frac{1}{a}+ \sum_{k \geq 1}\frac{a}{k^2+a^2}=- \frac{\pi \cot(a \pi i)}{2ai}-\frac{\pi \cot(-a \pi i)}{-2ai}\)


\(\displaystyle \frac{1}{a}+ 2 \sum_{k \geq 1}\frac{a}{k^2+a^2}=- \frac{\pi \coth(a \pi i)}{2ai}-\frac{\pi \cot(a\pi i )}{2ai}=-2\frac{\pi \cot(a\pi i )}{2ai} = \frac{\pi \coth( \pi a) }{a} \)

\(\displaystyle \sum_{k \geq 1}\frac{a}{k^2+a^2}=\frac{\pi \coth( \pi a) }{2a} -\frac{1}{2a}\)