The integral is of the form $\int_{-\infty}^\infty \sigma(x)\mu(x)\,\mathrm{d}x$. Where the Fourier transform of the $\sigma$ function is $\tilde \sigma(p)= e^{-iap}\frac{1}{1+e^{-c|p|}}$ and the function $\mu(x)$ is given by $\mu(x)=-2 \tan ^{-1}\left(\frac{2 x-2}{c}\right)$.

The Fourier transform of $\mu(x)$ can be found quite easily $\tilde \mu(p)=\frac{e^{-i p} \left(2 i \pi e^{-\frac{c | p| }{2}}\right)}{p}$.

The question is:

Is it possible to use the the Parseval-Plancherel identity and write the above integral as $\int_{-\infty}^\infty \tilde\sigma(p)\tilde \mu(p)\,\mathrm{d}p$?

If so, the above integral becomes $\int_{-\infty}^\infty dp \frac{i \pi e^{-i (a+1) p} \text{sech}\left(\frac{c p}{2}\right)}{p}$

Which looks like a Fourier Transform of $\frac{sech(\frac{cp}{2})}{p}$ function. How is this Fourier transform computed?

The integral is of the form $\int_{-\infty}^\infty \sigma(x)\mu(x)\,\mathrm{d}x$. Where the Fourier transform of the $\sigma$ function is $\tilde \sigma(p)= e^{-iap}\frac{1}{1+e^{-c|p|}}$ and the function $\mu(x)$ is given by $\mu(x)=-2 \tan ^{-1}\left(\frac{2 x-2}{c}\right)$.

The Fourier transform of $\mu(x)$ can be found quite easily $\tilde \mu(p)=\frac{e^{-i p} \left(2 i \pi e^{-\frac{c | p| }{2}}\right)}{p}$.

The question is:

Is it possible to use the the Parseval-Plancherel identity and write the above integral as $\frac{1}{2 \pi}\int_{-\infty}^\infty \tilde\sigma(p)\tilde \mu(p)\,\mathrm{d}p$?

If so, the above integral becomes $\frac{i}{2}\int_{-\infty}^\infty dp \frac{ e^{-i (a+1) p} \text{sech}\left(\frac{c p}{2}\right)}{p}$

Which looks like a Fourier Transform of $\frac{sech(\frac{cp}{2})}{p}$ function. How is this Fourier transform computed?