It is known that (see for example)
\begin{align}
&\int_0^{\infty} \frac{\sin x}{\cosh x+\cos x}\frac{dx}x =\frac\pi4\\
&\int_0^{\infty} \frac{\sin^3 x}{\cosh x+\cos x}\frac{dx}x =\frac\pi8
\end{align}
I am wondering if the similar integral below
$$\int_0^{\infty} \frac{\sin^2 x}{\cosh x+\cos x}\frac{dx}x $$
can be evaluated to a simple close-form as well. I have tried the same approaches for the known integrals above and they do not lead to any results. Alternatively, I have also manipulated the integral and expressed it in the equivalent form
$$\int_0^\infty e^{-x}\cos x\tanh x \>\frac{dx}x
$$
which, though appearing simpler, is not any easier.
Best Answer
I'm not sure how you showed the two integrals are equivalent, but the following is an evaluation of $$\int_{0}^{\infty} e^{-x} \cos (x) \tanh(x) \, \frac{\mathrm dx}{x}.$$
For $\Re(s) >0$, we have
$$ \begin{align} \int_{0}^{\infty} \tanh(t) e^{-st} \, \mathrm dt &= \int_{0}^{\infty} \frac{1-e^{-2t}}{1+e^{-2t}} \, e^{-st} \, \mathrm dt \\ &= \int_{0}^{\infty} (1-e^{-2t})e^{-st} \sum_{n=0}^{\infty} (-1)^{n}e^{-2tn} \, \mathrm dt \\ &= \sum_{n=0}^{\infty} (-1)^{n}\int_{0}^{\infty}\left(e^{-(2n+s)t} -e^{-(2n+s+2)t} \right) \, \mathrm dt \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+s} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+s+2} \\ &= \frac{1}{2} \left(\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+\frac{s}{2}} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+\frac{s}{2}+1} \right) \\ &\overset{(1)}{=} \frac{1}{4} \left(\psi \left(\frac{s}{4} + \frac{1}{2}\right)- \psi \left( \frac{s}{4}\right) - \psi \left(\frac{s}{4}+1 \right) + \psi \left(\frac{s}{4}+\frac{1}{2} \right)\right) \\ &\overset{(2)}{=} \frac{1}{2} \left( \psi \left(\frac{s}{4}+ \frac{1}{2} \right) - \psi \left(\frac{s}{4} \right) - \frac{2}{s} \right). \end{align}$$
Therefore, $ \begin{align} \int_{0}^{\infty} e^{-x} \cos (x) \tanh(x) \, \frac{\mathrm dx}{x} &= \Re\int_{0}^{\infty} e^{-(1+i)x} \frac{\tanh (x)}{x} \, \mathrm dx \\ &= \Re \int_{0}^{\infty}e^{-(1+i)x} \tanh(x) \int_{0}^{\infty} e^{-xt} \, \mathrm dt \, \mathrm dx \\ &= \Re \int_{0}^{\infty} \int_{0}^{\infty}\tanh(x) e^{-(1+i+t)x} \, \mathrm dx \, \mathrm dt \\ &= \Re \int_{0}^{\infty} \frac{1}{2}\left(\psi \left(\frac{1+i+t}{4}+ \frac{1}{2} \right) - \psi \left(\frac{1+i+t}{4} \right) - \frac{2}{1+i+t} \right) \, \mathrm dt \\ &= \Re \left(2 \ln\Gamma \left(\frac{3+i+t}{4} \right) -2 \ln \Gamma \left(\frac{1+i+t}{4} \right) -\ln (1+i+t)\right) \Bigg|_{0}^{\infty} \\ &\overset{(3)}= \Re \left(-2 \ln(2) - 2 \ln \Gamma \left(\frac{3+i}{4} \right)+2 \ln \Gamma \left(\frac{1+i}{4} \right) + \ln(1+i)\right)\\ &= - \frac{3 \ln (2)}{2} + 2\Re \left(\ln \Gamma \left(\frac{1+i}{4} \right)- \ln \Gamma \left(\frac{3+i}{4} \right) \right). \end{align}$
$(1)$ For $\Re(z) > 0$, $\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k+z} = \frac{1}{2} \left(\psi \left(\frac{z+1}{2} \right) - \psi \left(\frac{z}{2} \right) \right).$
$(2)$ Recurrence relation of the digamma function
$(3)$ For $a >0$, $\ln \Gamma(z_{1}+ax)- \ln \Gamma(z_{2}+ax) \sim (z_{1}-z_{2}) \ln(ax) + \mathcal{O} \left(\frac{1}{x}\right)$ as $x \to \infty$.