How to show $\int_0^{\infty} \frac{\sin^3 x}{\cosh x\>+\>\cos x}\frac{dx}x=\frac{\pi}{8}$ I would like to evaluate the integral below$$\int_0^{\infty} \frac{\sin^3 x}{\cosh x+\cos x}\frac{dx}x $$
which I found to be $\frac \pi8$ numerically. I was able to evaluate a similarly looking, yet simpler, integral
$$\int_0^{\infty} \frac{\sin x}{\cosh x+\cos x}\frac{dx}x =\frac\pi4$$
by writing the integrand in the Frullani format and apply the theorem. However, it does not work in this case and I am not sure if they are related. Any solution is appreciated.
 A: Since
$$
\frac{\sin(x)}{\cosh(x)+\cos(x)}=2\sum_{n\ge 1}(-1)^{n-1}e^{-nx}\sin(nx
)$$
(shown here), we can combine the above with $\sin^2(x)\sin(nx) = \frac14 \left[2 \sin(n x) + \sin((2  - n) x) - \sin((2 + n) x)\right]$ and $\int_0^\infty e^{-ax}\frac{\sin{(bx)}}{x}\mathrm{d}x = \arctan\left(\frac{b}{a}\right)$ to get
\begin{align*}
I &= \frac{1}{2}\sum_{n\ge 1}(-1)^{n-1}\int_{0}^{\infty}\frac{e^{-nx}}{x}\left[2 \sin(n x) + \sin((2  - n) x) - \sin((2 + n) x)\right] \, \mathrm{d}x\\
& = \frac{1}{2}\sum_{n\ge 1}(-1)^{n-1} \left[\frac\pi2 + \arctan\left(\frac{2-n}{n} \right) - \arctan\left(\frac{2+n}{n} \right)\right]\\
& = \frac{1}{2}\sum_{n\ge 1}(-1)^{n-1}  \left[\arctan(n+1) - \arctan(n-1)\right]\\
& = \frac{1}{2}\left[-\arctan(0) + \arctan(1) \right]\\
 & = \frac{\pi}{8}
\end{align*}
as desired.

Note that to re-write the arctangent expressions from equations $2\to 3$ we combine $\arctan(\alpha) - \arctan(\beta) = \arctan\left(\frac{\alpha - \beta}{1 + \alpha\beta} \right)$ and $\arctan(x) + \arctan\left( \frac1x\right)= \frac\pi2,\ x>0$. Thus
\begin{align*}
\frac\pi2 + \arctan\left(\frac{2-n}{n} \right) - \arctan\left(\frac{2+n}{n} \right) &=\frac\pi2 + \arctan\left(\frac{-n^2}{2}\right) \\
&\overset{\color{blue}{n^2/2 >0}}{=}  \arctan\left(\frac{2}{n^2}\right) \\
&=\arctan(n+1) - \arctan(n-1)
\end{align*}
also exploiting the fact that $\arctan(x)$ is odd.
A: It was shown in this answer that if $n$ is a nonnegative integer, then $$\int_{0}^{\infty} \frac{\sin\left((2n+1)x \right)}{\cosh (x) + \cos (x)} \, \, \frac{\mathrm dx}{x} = \frac{\pi}{4}. $$
Therefore, $$\begin{align} \int_{0}^{\infty}\frac{\sin^{3}(x)}{\cosh (x) + \cos (x)} \, \frac{\mathrm dx}{x} &= \frac{1}{4} \int_{0}^{\infty} \frac{3 \sin (x)- \sin(3x)}{\cosh (x) + \cos (x)} \, \frac{\mathrm dx}{x} \\ & = \frac{1}{4} \left(\frac{3 \pi}{4}- \frac{\pi}{4} \right) \\ & = \frac{\pi}{8}. \end{align}$$
In general, if $m$ is a nonnegative integer, then $$ \begin{align} \int_{0}^{\infty}\frac{\sin^{2m+1}(x)}{\cosh (x) + \cos (x)} \, \frac{\mathrm dx}{x} &\overset{(1)}{=} \frac{1}{2^{2m}} \sum_{k=0}^{m} (-1)^{m-k} \binom{2m+1}{k} \int_{0}^{\infty}\frac{\sin \left((2m+1-2k)x \right)}{\cosh(x) + \cos (x)}\frac{\mathrm dx}{x} \\ &= \frac{\pi}{2^{2m+2}}\sum_{k=0}^{m} (-1)^{m-k} \binom{2m+1}{k}. \end{align}$$
$(1)$ https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formulae
A: We know that:
$$2\sin^2(x) = 1-\cos(2x)$$
Therefore, we could re-write your integral as:
$$\int_{0}^{\infty} \frac{\sin^3(x)}{\cosh(x)+\cos(x)} \frac{dx}{x} =\frac{1}{2} \int_{0}^{\infty} \frac{\sin(x)-\sin(x)\cos(2x)}{\cosh(x)+\cos(x)} \frac{dx}{x}$$
But we find that:
$$\sin(3x) = \sin(x+2x) = \sin(x)\cos(2x)+\sin(2x)\cos(x)$$
$$\sin(x) = \sin(2x-x) = \sin(2x)\cos(x)-\sin(x)\cos(2x)$$
from which it follows that $2\sin(x)\cos(2x) = \sin(3x)-\sin(x)$. So:
$$\int_{0}^{\infty} \frac{\sin^3(x)}{\cosh(x)+\cos(x)} \frac{dx}{x} = \frac{\pi}{8} - \frac{1}{4}\int_{0}^{\infty} \frac{\sin(3x)-\sin(x)}{\cosh(x)+\cos(x)} \frac{dx}{x}$$
and we find that the second term vanishes by Frullani's Theorem so the result now follows.
Edit
So, that last term does not vanish by Frullani's Theorem because it is not of Frullani type, I mistakenly thought that it was. However, you could likely show that it vanishes by using the residue theorem.
