Improper Definite integral $\int_{-\infty}^\infty -\frac{i \pi e^{-i a p} \text{sech}\left(\frac{c p}{2}\right)}{p}dp$ 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.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[10px,#ffd]{\left. -\ic\pi\,\mrm{P.V.}\int_{-\infty}^{\infty} {\expo{-\ic ap}
\mrm{sech}\pars{cp/2} \over p}\,\dd p
\,\right\vert_{\ a, c\ \in\ \mathbb{R}}}:\ {\Large ?}}$.

I assumed you are dealing with a principal value of the integral.

\begin{align}
&\bbox[5px,#ffd]{\left. -\ic\pi\,\mrm{P.V.}\int_{-\infty}^{\infty} {\expo{-\ic ap}
\mrm{sech}\pars{cp/2} \over p}\,\dd p
\,\right\vert_{\ a, c\ \in\ \mathbb{R}}}
\\[5mm] = &\
-2\pi\,\mrm{sgn}\pars{a}\int_{0}^{\infty} {\sin\pars{\verts{a}p}
 \over p\cosh\pars{\verts{c}p/2}}\,\dd p
\\[5mm] \,\,\,\stackrel{2\verts{c}p\ \mapsto\ p}{=}\,\,\,&
\left. -\,{a \over \verts{c}}\,\pi\int_{0}^{\infty}
\!\!\!\!\!{1
 \over \cosh\pars{p/4}}{\sin\pars{bp} \over bp}\,\dd p
\,\right\vert_{\ds{\ 
b = \color{red}{\verts{a}/\pars{2\verts{c}}}}}
\label{1}\tag{1}
\\[5mm] = &\
-\,{a \over \verts{c}}\,\pi\int_{0}^{\infty} {1
 \over \cosh\pars{p/4}}
\pars{{1 \over 2}\int_{-1}^{1}{\expo{\ic kbp} \,\dd k}}\,\dd p
\\[5mm] = &\
-\,{a \over 2\verts{c}}\,\pi\int_{-1}^{1}\
\underbrace{\int_{0}^{\infty}{\expo{\ic kbp}
 \over \cosh\pars{p/4}}\,\dd p}
_{\ds{\equiv\ \mathcal{I}\pars{b}}}\ \dd k
\label{2}\tag{2}
\end{align}

$\ds{\large\mathcal{I}\pars{b}\ \mbox{Evaluation:}}$
\begin{align}
\mathcal{I}\pars{b} & \equiv
\bbox[5px,#ffd]{\int_{0}^{\infty}{\expo{\ic kbp}
 \over \cosh\pars{p/4}}\,\dd p}
\\[5mm] = &\
\int_{0}^{\infty}{\expo{-\pars{-1/4 - \ic kb}p} -
\expo{-\pars{1/4 - \ic kb}p}
 \over \sinh\pars{p/2}}\,\dd p
\\[5mm] & =
2\int_{0}^{\infty}{\expo{-\pars{1/4 - \ic kb}p} -
\expo{-\pars{3/4 - \ic kb}p}
 \over 1 - \expo{-p}}\,\dd p
\\[5mm] & =
2\bracks{\Psi\pars{{3 \over 4} - \ic kb} -
\Psi\pars{{1 \over 4} - \ic kb}}
\label{3}\tag{3}
\end{align}
With (\ref{2}) and (\ref{3}):
\begin{align}
&\bbox[5px,#ffd]{\left. -\ic\pi\,\mrm{P.V.}\int_{-\infty}^{\infty} {\expo{-\ic ap}
\mrm{sech}\pars{cp/2} \over p}\,\dd p
\,\right\vert_{\ a, c\ \in\ \mathbb{R}}}
\\[5mm] = &\
-\,{a\,\pi \over \verts{c}}\bracks{%
\ln\pars{\Gamma\pars{3/4 - \ic kb}} -
\ln\pars{\Gamma\pars{1/4 - \ic kb}} \over -\ic b}
_{\ k\ =\ -1}^{\ k\ =\ 1}
\\[5mm] = &
\left.
-2\pi\ic\,\mrm{sgn}\pars{a}\ln\pars{\Gamma\pars{3/4 - \ic kb} \over
\Gamma\pars{1/4 - \ic kb}}\right\vert_{\ k\ =\ -1}^{\ k\ =\ 1}
\\[5mm] = &\
-2\pi\ic\,\mrm{sgn}\pars{a}\ln\pars{\Gamma\pars{3/4 - \ic b}\Gamma\pars{1/4 +\ic b} \over
\Gamma\pars{1/4 - \ic b}\Gamma\pars{3/4 + \ic b}}
\\[5mm] = &\
-2\pi\ic\,\mrm{sgn}\pars{a}\,\ln\pars{\sin\pars{\pi\bracks{1/4 - \ic b}} \over
\sin\pars{\pi\bracks{1/4 + \ic b}}}
\\[5mm] = &\
4\pi\,\mrm{sgn}\pars{a}\,\Im\ln\pars{\sin\pars{\pi\bracks{{1 \over 4} - \ic b}}}
\\[5mm] = &\
4\pi\,\mrm{sgn}\pars{a}\,\Im\ln\pars{{\root{2} \over 2}\cosh\pars{\pi b} -
{\root{2} \over 2}\sinh\pars{\pi b}\ic}
\\[5mm] = &\
-4\pi\,\mrm{sgn}\pars{a}\,\arctan\pars{\tanh\pars{\pi b}}
\\[5mm] = &\
\bbx{\large%
-4\pi\arctan\pars{\tanh\pars{\pi a \over 2\verts{c}}}} \\ &\
\end{align}

(\ref{3}): See $\ds{\color{black}{\bf 6.3.22}}$ Digamma Identity in A & S Table.
