Do these integrals have a closed form? $I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$ The following integrals look like they might have a closed form, but Mathematica could not find one. Can they be calculated, perhaps by differentiating under the integral sign?
$$I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$$
$$I_2 = \int_{-\infty }^{\infty } \frac{\sin ^2(x)}{x \sinh (x)} \, dx$$

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*
*$\ds{\large\color{#000}{\tt I_{1}\equiv
            \int_{-\infty }^{\infty }{\sin\pars{x} \over x\cosh\pars{x}} \,\dd x}}$

\begin{align}I_{1}&\equiv\color{#c00000}{
\int_{-\infty }^{\infty }{\sin\pars{x} \over x\cosh\pars{x}} \,\dd x}
=\int_{-\infty }^{\infty }
{2\expo{x} \over \expo{2x} + 1}\int_{0}^{1}\cos\pars{kx}\,\dd k\,\dd x
\\[3mm]&=2\Re\int_{0}^{1}\
\overbrace{\int_{-\infty}^{\infty}
{\pars{\expo{x}}^{\verts{k}\ic} \over \expo{2x} + 1}\,\expo{x}\,\dd x}
^{\ds{\mbox{Set}\ \expo{x} \equiv t}}\ \,\dd k
=2\int_{0}^{1}\color{#00f}{%
\Re\int_{0}^{\infty}{t^{\verts{k}\ic} \over t^{2} + 1}\,\dd x}\,\dd k\tag{1}
\end{align}

\begin{align}&\color{#00f}{%
\Re\int_{0}^{\infty}{t^{\verts{k}\ic} \over t^{2} + 1}\,\dd x}
=\Re\bracks{2\pi\ic\,{\pars{\expo{\pi\ic/2}}^{\verts{k}\ic} \over 2\ic}
+ 2\pi\ic\,{\pars{\expo{3\pi\ic/2}}^{\verts{k}\ic} \over -2\ic}
-\int_{\infty}^{0}
{t^{\verts{k}\ic}\pars{\expo{2\pi\ic}}^{\verts{k}\ic} \over t^{2} + 1}\,\dd t}
\\[3mm]&=\pi\expo{-\pi\verts{k}/2}
-\pi\expo{-3\pi\verts{k}/2}
+\expo{-2\pi\verts{k}}\color{#00f}{%
\Re\int_{0}^{\infty}{t^{\verts{k}\ic} \over t^{2} + 1}\,\dd x}
\\[5mm]&\imp\ \color{#00f}{%
\Re\int_{0}^{\infty}{t^{\verts{k}\ic} \over t^{2} + 1}\,\dd x}
={\pi\expo{-\pi\verts{k}/2}
-\pi\expo{-3\pi\verts{k}/2} \over 1 - \expo{-2\pi\verts{k}}}
=\pi\,{\expo{\pi\verts{k}/2}
-\expo{-\pi\verts{k}/2} \over \expo{\pi\verts{k}} - \expo{-\pi\verts{k}}}
\\[3mm]&=\pi\,{\sinh\pars{\pi\verts{k}/2} \over \sinh\pars{\pi\verts{k}}}
={\pi \over 2}\,\sech\pars{{\pi \over 2}\,k}
\end{align}

Replacing in $\pars{1}$:
\begin{align}
&\color{#66f}{\large\int_{-\infty }^{\infty }{\sin\pars{x} \over x\cosh\pars{x}} \,\dd x}
=2\int_{0}^{1}{\pi \over 2}\,\sech\pars{{\pi \over 2}\,k}\,\dd k
=\left. 4\arctan\pars{\tanh\pars{{\pi \over 4}\,k}}\right\vert_{0}^{1}
\\[3mm]&=\color{#66f}{\large 4\arctan\pars{\tanh\pars{\pi \over 4}}}
\approx {\tt 2.3218}
\end{align}
This result can be expressed in terms of the Gudermannian Function  $\ds{{\rm gd}\pars{z} \equiv 2\arctan\pars{\expo{z}} - {\pi \over 2}}\quad$
as
$\quad\ds{4\arctan\pars{\tanh\pars{\pi \over 4}} = 2\,{\rm gd}\pars{\pi \over 2}}$.



*$\ds{\large\color{#000}{\tt I_{2}\equiv
\int_{-\infty }^{\infty }{\sin^{2}\pars{x} \over x\sinh\pars{x}} \,\dd x}}$

$\tt\mbox{It can be evaluated by following a similar technique}$.

