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

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
$$ I_2(1)=\int_0^1\pi\tan(a\pi)da=\ln(\cosh(a\pi)). $$
