<s>[INST] 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?
$$I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$$
$$I_2 = \int_{-\infty }^{\infty } \frac{\sin ^2(x)}{x \sinh (x)} \, dx$$
 [/INST] For the first one we need:
$$\int _{-1/2}^{1/2}\!{{\rm e}^{2\,iax}}{da}={\frac {\sin \left( x
 \right) }{x}}\tag{1}$$
$$ \frac{1}{\cosh \left( x \right)}=-2\,\sum _{n=1}^{\infty 
} \left( -1 \right) ^{n}{{\rm e}^{- \left| x \right|  \left( 2\,n-1
 \right) }}\tag{2}$$
$$\int _{-\infty }^{\infty }\!{{\rm e}^{2\,iax}}{{\rm e}^{- \left| x
 \right|  \left( 2\,n-1 \right) }}{dx}=- \frac{1}{\left( 2\,ia-2\,n+1 \right) }-  \frac{1}{\left( -2\,ia-2\,n+1 \right)}\tag{3}$$
$$-2\,\sum _{n=1}^{\infty } \left( -1 \right) ^{n} \left(- \frac{1}{\left( 2\,ia-2\,n+1 \right) }-  \frac{1}{\left( -2\,ia-2\,n+1 \right)} \right) ={
\frac {\pi }{\cosh \left( \pi \,a \right) }}\tag{4}$$
we get:
$$
\begin{aligned}
\int _{-\infty }^{\infty }\!{\frac {\sin \left( x \right) }{x\cosh
 \left( x \right) }}{dx}&=\int _{-1/2}^{1/2}\!{\frac {\pi }{\cosh
 \left( \pi \,a \right) }}{da}\\
&=2\,\arctan \left( {
{\rm e}^{1/2\,\pi }} \right)-2\,\arctan \left( {{\rm e}^{-1/2\,\pi }} \right)\\
&=2\,\arctan \left( \sinh \left( \frac{1}{2}\,\pi  \right)  \right) 
\end{aligned}\tag{5}$$
where the last part follows from $(2)$ and the Taylor series for arctan:
$$\arctan \left( x \right) =\sum _{n=0}^{\infty }{\frac { \left( -1
 \right) ^{n}{x}^{2\,n+1}}{2\,n+1}}\tag{6}$$
For the second one we need:
$$ \frac{1}{\sinh \left( x \right) }=2\,\sum _{n=1}^{\infty }
{{\rm e}^{-x \left( 2\,n-1 \right) }}\tag{7}$$
$${\frac { \sin^2 \left( x \right)}{x}}=-\frac{1}{2}\,\sum _{
m=1}^{\infty }{\frac { \left( -1 \right) ^{m}{2}^{2\,m}{x}^{2\,m-1}}{
 \left( 2\,m \right) !}}\tag{8}$$
$$\int _{0}^{\infty }\!{x}^{2\,m-1}{{\rm e}^{-x \left( 2\,n-1 \right) }}
{dx}={\frac { \left( 2\,m-1 \right) !}{ \left( 2\,n-1 \right) ^{2\,m}}
}\tag{9}$$
$$\cot \left( z \right) -\frac{1}{z}=-\frac{2}{\pi}\,\sum _{m=1}^{\infty }\zeta 
 \left( 2\,m \right)  \left( {\frac {z}{\pi }} \right) ^{2\,m-1}\tag{10}$$ 
From $(6,7,8)$:
$$
\begin{aligned}
\int _{0}^{\infty }\!{\frac { \sin^2 \left( x \right)}{x\sinh \left( x \right) }}{dx}&=-\frac{1}{2}\,\sum _{m=1}^{\infty } \left( 
\frac{\left( -4 \right) ^{m}}{m}\sum _{n=1}^{\infty }{\frac {1}{\left( 2\,n-1
 \right) ^{2\,m}}} \right)\\
&=\frac{1}{4}\sum _{m=1}^{\infty }\,{\frac 
{\zeta  \left( 2\,m \right)  \left( {4}^{m}-1 \right)  \left( -1
 \right) ^{m}}{m}}
\end{aligned}\tag{11}$$
and after integrating $(10)$ once we know that:
$$\ln  \left( {\frac {\sin \left( z \right) }{z}} \right) =-\sum _{m=1}^
{\infty }\frac{\zeta  \left( 2\,m \right)}{m}  \left( {\frac {z}{\pi }} \right) 
^{2\,m}\tag{12}
$$
so by comparing $(11)$ with $(12)$ we know that:
$$
\begin{aligned}\int _{0}^{\infty }\!{\frac { \sin^2 \left( x \right)  }{x\sinh \left( x \right) }}{dx}&=\frac{1}{2}\,\ln\!\left( \frac{1}{2}\,{\frac {\sinh
 \left( 2\,\pi  \right) }{\sinh \left( \pi  \right) }} \right)\\
&=\frac{1}{2}\,\ln  \left( \cosh \left( \pi 
 \right)  \right) 
\end{aligned}$$ 
</s>