3
$\begingroup$

I was going through a question on finding the Fourier sine transform of the following function: $$\frac{e^{ax}+e^{-ax}}{e^{\pi x}-e^{-\pi x}}.$$


Attempt:

So I got stuck with the following integral: $$ \int_{0}^\infty \frac{e^{(a+ip)x}-e^{-(a+ip)x}dx}{e^{\pi x}-e^{-\pi x}} = \frac{1}{2} \tan {\frac{a+ip}{2}}$$

The second one which i guess must be quite similar to the former that I encountered in another similar question is:

$$\int_{0}^\infty \frac{e^{(a+ip)x}+e^{-(a+ip)x}dx}{e^{\pi x}-e^{-\pi x}} = \frac{1}{2} \sec {\frac{a+ip}{2}}$$

I am not able to understand how to proceed with these two. Any help would be appreciated.

$\endgroup$
  • $\begingroup$ Regarding the Fourier transform, see here. $\endgroup$ – Maxim Apr 1 at 17:37
3
$\begingroup$

Note that we can write for $|a|<\pi$

$$\begin{align} \int_0^\infty \frac{e^{(a+ip)x}-e^{-(a+ip)x}}{\left(e^{\pi x}-e^{-\pi x}\right)}\,dx&=\int_0^\infty \frac{e^{-\pi x}\left(e^{(a+ip)x}-e^{-(a+ip)x}\right)}{1-e^{-2\pi x}}\,dx\\\\ &=\sum_{n=0}^\infty \int_0^\infty e^{-(2n+1)\pi x}\left(e^{(a+ip)x}-e^{-(a+ip)x}\right)\\\\ &=\sum_{n=0}^\infty\left(\frac{1}{(2n+1)\pi -(a+ip)}-\frac{1}{(2n+1)\pi +(a+ip)}\right)\\\\ &=-\frac12\sum_{n=-\infty}^\infty \frac{1}{\frac{a+ip}{2}+\frac\pi2+n\pi}\tag1\\\\ &=\frac12\tan\left(\frac{a+ip}{2}\right)\tag2 \end{align}$$

where in going from $(1)$ to $(2)$ we noted that right-hand side of $(1)$ was the partial fraction representation of $-\cot\left(\frac{a+ip}{2}+\frac\pi2\right)=\tan\left(\frac{a+ip}{2}\right)$ (See THIS ANSWER and the Appendix of THIS ONE).

$\endgroup$
  • $\begingroup$ Partial fraction expansion representation of the tangent function. Could you please elaborate on this. That would be really helpful. $\endgroup$ – Shatabdi Sinha Mar 28 at 16:43
  • 2
    $\begingroup$ Sure. See THIS and the Appendix of THIS answer. $\endgroup$ – Mark Viola Mar 28 at 17:26

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.