When is it possible to use the Parseval-Plancherel identity to solve an integral?

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?

Recall the identity that Fourier transform of $$K(x)=\text{sech}(x)$$ is $$\tilde K(p)=\pi \text{sech}\left(\frac{\pi p}{2}\right)$$.
Using this identity the Fourier transform of $$\frac{\text{sech} {x}}{x}$$ can be easily computed
$$$$\int_{-\infty}^{-\infty} e^{-i x p} \frac{\text{sech}{x}}{x} \, \mathrm{d}x= -i \int \pi \text{sech}\left(\frac{\pi p}{2}\right) \mathrm{d}p= -2 i \tan ^{-1}\left(\sinh \left(\frac{\pi p}{2}\right)\right) \label{ident}$$$$
$$$$\frac{i}{2}\int_{-\infty}^\infty dp \frac{ e^{-i (a+1) p} \text{sech}\left(\frac{c p}{2}\right)}{p}= \tan ^{-1}\left(\sinh \left(\frac{\pi (\Lambda_h+1)}{|c|}\right)\right) \label{rest}$$$$