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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?

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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

\begin{equation} \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} \end{equation}

Using equation this relation, the given integral can be easily integrated

\begin{equation} \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} \end{equation}

Checking the answer numerically. Plot: Constant a Plot Constant c

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