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?
$$I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$$
$$I_2 = \int_{-\infty }^{\infty } \frac{\sin ^2(x)}{x \sinh (x)} \, dx$$

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,
$$ \int_{-\infty}^{\infty} \frac{\sin x}{x \cosh x} \ dx = 4 \arctan (e^{\pi /2}) - \pi  \approx 2.3217507819 . $$
