I came across this improper integral that I couldn't solve
$$\int_{-\infty}^\infty -\frac{i \pi e^{-i a p} \text{sech}\left(\frac{c p}{2}\right)}{p} dp$$
My guess would be to use Residue theorem but it does not seem help.
My try so far is that it has a pole at $p=0$. With $a>0$, closing the contour upward, I compute the residue
$$\lim_{p\to 0}-\frac{i \pi e^{-i a p} \text{sech}\left(\frac{c p}{2}\right)}{p} p =-I \pi $$
Thus, the value of the integral is $2\pi I Res(f,0)= 2\pi^2$. This is definitely not the correct answer (which I confirmed by numerical integration).
Taking the hint from the comment I proceeded the following way
$\int_{-\infty}^\infty e^{-iap} sech(\frac{cp}{2})=\frac{2 \pi \text{sech}\left(\frac{\pi a}{c}\right)}{c}$ which comes from the fact that Fourier transform of sech function is sech function itself.
Now, to account the p in the denominator, I need to integrate this result and add a delta function which gives
$\int_{-\infty}^\infty \frac{ e^{-i a p} \text{sech}\left(\frac{c p}{2}\right)}{p} dp =\int \frac{2 \pi \text{sech}\left(\frac{\pi a}{c}\right)}{c} da =-\frac{2 \pi ^2 \tanh \left(\frac{\pi a}{c}\right) \text{sech}\left(\frac{\pi a}{c}\right)}{c^2}+ \delta(a)$
Multiplying by the factor $-i\pi$ on both sides, I get
$$\int_{-\infty}^\infty -\frac{i \pi e^{-i a p} \text{sech}\left(\frac{c p}{2}\right)}{p} dp=\frac{2 i \pi ^3 \tanh \left(\frac{\pi a}{c}\right) \text{sech}\left(\frac{\pi a}{c}\right)}{c^2}-i\pi \delta (a)$$
This is supposed to be the correct answer. But that still does not match with the numerical integration.