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.

  • 1
    $\begingroup$ I managed to reduce it down to $$2\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\frac{\partial}{\partial a}\left(\int_0^{\infty} \frac{\sin(nx)e^{-anx}}{x^2-\pi^2}\,dx\right)$$ but I can't evaluate the integral in terms of $n$ and $a$ and I am not sure if what I have reached is correct. $\endgroup$ – Pranav Arora Jun 6 '14 at 9:49
  • $\begingroup$ Maybe Plancherels Theorem can help? $\endgroup$ – TheOscillator Jun 6 '14 at 10:46

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. $\qquad\square$

  • $\begingroup$ Thanks! It's very nice and easy :) $\endgroup$ – Shobhit Bhatnagar Jun 16 '14 at 7:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.