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?
$$I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$$
$$I_2 = \int_{-\infty }^{\infty } \frac{\sin ^2(x)}{x \sinh (x)} \, dx$$
 A: For $I_2$, we can use a well-known result:
$$ \int_{-\infty }^{\infty } \frac{\sinh (ax)}{\sinh(bx)}dx=\frac{\pi}{b}\tan\frac{a\pi}{2b}. $$
Note $\sinh(ix)=\sin(x), \tanh(ix)=\tan(x)$. Thus
$$ \int_{-\infty }^{\infty } \frac{\sin (ax)}{\sinh(bx)}dx=\int_{-\infty }^{\infty } \frac{\sinh (iax)}{\sinh(bx)}dx=\frac{\pi}{b}\tanh\frac{a\pi}{2b}. $$
For $I_2$, define
$$ I_2(a)=\int_{-\infty }^{\infty } \frac{\sin^2(ax)}{x\sinh (x)}dx. $$
Then $I_2(0)=0$ and $I_2(1)=I_2$. Now
\begin{eqnarray}
I_2'(a)&=&\int_{-\infty }^{\infty } \frac{2\sin(ax)\cos(ax)}{\sinh (x)}dx\\
&=&\int_{-\infty }^{\infty } \frac{\sin(2ax)}{\sinh (x)}dx\\
&=&\pi\tanh(a\pi).
\end{eqnarray}
So
$$ I_2(1)=\int_0^1\pi\tan(a\pi)da=\ln(\cosh(a\pi)). $$
A: $\newcommand{\+}{^{\dagger}}
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*
*$\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
$\quad\ds{4\arctan\pars{\tanh\pars{\pi \over 4}} = 2\,{\rm gd}\pars{\pi \over 2}}$.



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

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

A: First I'm going to evaluate $$\int_{-\infty}^{\infty} \frac{\cos ax}{\cosh x} \ dx .$$
Integrate the function $ \displaystyle f(z) = \frac{e^{iaz}}{\cosh z}$ around a rectangle on the complex plane with vertices at $z= R$, $ z= R + i \pi$, $z= -R + i \pi$, and $z= - R$.
As $R \to \infty$, $ \displaystyle \int f(z) \ dz$ vanishes on the left and right sides of the rectangle.
So going around the rectangle counterclockwise, we get
$$ \int_{-\infty}^{\infty} f(x) \ dx + \int_{\infty}^{-\infty} f(t + i \pi) \ dt = 2 \pi i \ \text{Res} [f(z),i \pi] ,$$
which implies
$$ (1+ e^{- a \pi}) \int_{-\infty}^{\infty} \frac{e^{iax}}{\cosh x} \ dx = 2 \pi i \lim_{z \to i \pi /2} \frac{e^{iaz}}{\sinh z} = 2 \pi \ e^{- a \pi /2} .$$
And equating the real parts on both sides of the equation, we get
$$  \int_{-\infty}^{\infty} \frac{\cos ax}{\cosh x} \ dx = \frac{2 \pi}{e^{a \pi /2} + e^{- a \pi/2}} = \pi \ \text{sech} \left( \frac{a \pi}{2}\right) .$$
Then
$$ \begin{align} \int_{0}^{a} \int_{-\infty}^{\infty} \frac{\cos ax}{\cosh x} \ dx \ da &= \int_{-\infty}^{\infty} \int_{0}^{a} \frac{\cos ax}{\cosh x} \ da \ dx \\ &= \int_{-\infty}^{\infty} \frac{\sin ax}{x \cosh x} \ dx \\ &= \pi \int_{0}^{a} \text{sech} \left(\frac{a \pi}{2} \right) \ da \\ &= 2 \int_{0}^{a \pi /2} \text{sech}(u) \ du  \\ &= 4 \int_{0}^{a \pi /2} \frac{e^{u}}{1+e^{2u}} \ du \\ &= 4 \int_{1}^{e^{a \pi /2}} \frac{1}{1+w^{2}} \ dw \\ &= 4 \left(\arctan (e^{a \pi /2}) - \frac{\pi}{4} \right) . \end{align}$$
Therefore,
$$ \int_{-\infty}^{\infty} \frac{\sin x}{x \cosh x} \ dx = 4 \arctan (e^{\pi /2}) - \pi  \approx 2.3217507819 . $$
A: 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:
$$
\begin{aligned}\int _{0}^{\infty }\!{\frac { \sin^2 \left( x \right)  }{x\sinh \left( x \right) }}{dx}&=\frac{1}{2}\,\ln\!\left( \frac{1}{2}\,{\frac {\sinh
 \left( 2\,\pi  \right) }{\sinh \left( \pi  \right) }} \right)\\
&=\frac{1}{2}\,\ln  \left( \cosh \left( \pi 
 \right)  \right) 
\end{aligned}$$ 
A: I do not know of a closed form but it seems that the integral can be
converted into an infinite series. Let
$$
I_{1}(k)=\int_{-\infty }^{+\infty }dx\frac{\sin kx}{x\cosh x}
$$
so $I_{1}=I_{1}(1).$ Now (note that $I_{1}(0)$ vanishes)
\begin{eqnarray*}
\partial _{k}I_{1}(k) &=&\int_{-\infty }^{+\infty }dx\frac{\cos kx}{\cosh x}=%
{Re}\int_{-\infty }^{+\infty }dx\frac{\exp [ikx]}{\cosh x}\\&=&2{Re}%
\int_{-\infty }^{+\infty }dx\exp [ikx]\frac{1}{\exp [x]+\exp [-x]} \\
I_{1} &=&\int_{0}^{1}dk\partial _{k}I_{1}(k)
\end{eqnarray*}
If this Fourier transform is known we can try to integrate over $k$ and
obtain $I_{1}(k)$ and hence $I_{1}$. If not we note that $(\exp [z]+\exp
[-z])^{-1}$ has poles $z_{n}=i\pi (1+2n)$ so closing the contour in the
upper half plane results in
$$
\int_{-\infty }^{+\infty }dx\exp [ikx]\frac{1}{\exp [x]+\exp [-x]}=2\pi
i\sum_{n}r_{n}\exp [ikz_{n}]
$$
where the $r_{n}$ are the residues and the sum over $n$ is that with $z_{n}$
in the upper half plane. Now the $k$-integral can be done and we end up with
a sum over $n$.
