# Integral $\int_0^\infty \frac{\sin x}{\cosh ax+\cos x}\frac{x}{x^2-\pi^2}dx=\tan^{-1}\left(\frac{1}{a}\right)-\frac{1}{a}$

Please help me prove the following identity: $$\int_0^\infty \frac{\sin x}{\cosh ax+\cos x}\frac{x}{x^2-\pi^2}dx=\tan^{-1}\left(\frac{1}{a}\right)-\frac{1}{a}\quad a>0$$
This integral is from Gradshteyn and Ryzhik’s tables.

#### Solutions Collecting From Web of "Integral $\int_0^\infty \frac{\sin x}{\cosh ax+\cos x}\frac{x}{x^2-\pi^2}dx=\tan^{-1}\left(\frac{1}{a}\right)-\frac{1}{a}$"

Lemma. For $a>0$ and $x\in\Bbb{R}$,
$$\sum_{n=1}^\infty2(-1)^{n-1}\sin(nx)e^{-anx}=\frac{\sin x}{\cosh(ax)+\cos x}\tag {1}$$

Proof. Indeed,
\eqalign{ \sum_{n=1}^\infty2(-1)^{n-1}\sin(nx)e^{-anx}&= -2\Im\left(\sum_{n=1}^\infty(-1)^n e^{(i-a)nx}\right)\cr &=2\Im\left(\frac{e^{(i-a)x}}{1+e^{(i-a)x}}\right)\cr &=\frac{\sin x}{\cosh(ax)+\cos x}.\qquad\square }

Now, let
$$I(a)=\int_0^\infty\frac{\sin x}{\cosh(ax)+\cos x}\cdot\frac{x}{x^2-\pi^2}\,dx.$$
Then
\eqalign{ I(a)&=\sum_{n=1}^\infty2(-1)^{n-1}\int_0^\infty\frac{x}{x^2-\pi^2}\sin(nx)e^{-anx}dx\cr &=\sum_{n=1}^\infty2(-1)^{n-1}\int_0^\infty\frac{t}{t^2-n^2\pi^2}\sin t e^{-at}dt\cr &=\int_0^\infty\left(\sum_{n=1}^\infty\frac{2t(-1)^{n-1}}{t^2-n^2\pi^2}\right)\sin t e^{-at}dt\cr &\buildrel{(*)}\over{=}\int_0^\infty\left(\frac{1}{t}-\frac{1}{\sin t}\right)\sin t e^{-at}dt\cr &=\int_0^\infty\frac{\sin t}{t} e^{-at}dt-\int_0^\infty e^{-at}dt\cr &=\arctan\frac{1}{a}-\frac{1}{a} }
as desired.

For more information on the partial fraction decomposition $(*)$ of the $\csc$ function, one may consult Alfors’ Book Chapter 5. pp.185–188. $\qquad\square$