Easier approach to $\int_0^{\infty} \frac{\mathrm{e}^{-x} \cosh(2x/5)}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x$? After playing with this integral for a little bit last night, I eventually resorted to complex analysis to solve it.
Can this be solved without complex analysis?  It feels like there ought to be a way.  If not, is there an easier way with complex analysis?  (I'm still pretty beginner tier at this sort of thing.)
My solution is somewhat involved and it is as follows:
First, get rid of the cosh.
$$
\begin{split}
I &= \int_0^{\infty} \frac{\mathrm{e}^{-x} \cosh(2x/5)}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x \\
&= \frac1{2} \int_0^{\infty} \frac{\mathrm{e}^{-3/5 x} + \mathrm{e}^{-7/5 x}}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x \\
 \frac{\mathrm{e}^{-7/5 x}}{1 + \mathrm{e}^{-2x}} &= \frac{\mathrm{e}^{3/5 x}}{1 + \mathrm{e}^{2x}} \\
2I &= \int_0^{\infty} \frac{\mathrm{e}^{-3/5 x}}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x + \int_0^{\infty} \frac{\mathrm{e}^{3/5 x}}{1 + \mathrm{e}^{2x}} \, \mathrm{d}x
\end{split}
$$
Next, do some u-subs to make it nicer.
$$\begin{split}
u = \mathrm{e}^{-x} & \qquad \mathrm{d}u = - \mathrm{e}^{-x} \, \mathrm{d}x \\
\int_0^{\infty} \frac{\mathrm{e}^{-3/5 x}}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x &= \int_{0}^{1} \frac{u^{-2/5}}{1 + u^2} \, \mathrm{d}u \\
\\
u = \mathrm{e}^{x}  & \qquad \mathrm{d}u = \mathrm{e}^{x} \, \mathrm{d}x \\
\int_0^{\infty} \frac{\mathrm{e}^{3/5 x}}{1 + \mathrm{e}^{2x}} \, \mathrm{d}x &= \int_{1}^{\infty} \frac{u^{-2/5}}{1 + u^2} \, \mathrm{d}u \\
\\
2I &= \int_{0}^{\infty} \frac{u^{-2/5}}{1 + u^2} \, \mathrm{d}u \\
\end{split}
$$
Our contour is the counterclockwise semicircular arc of radius $R > 1$ in the upper half of the complex plane.
$$
\begin{split}
\oint_C \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \int_{-R}^0 \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z + \int_0^{R} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z + \int_0^{\pi} \frac{{\left(R \mathrm{e}^{i \phi}\right)}^{-2/5}}{1 + {\left(R \mathrm{e}^{i \phi}\right)}^2} \, iR\mathrm{e}^{i \phi} \, \mathrm{d}\phi \\
\lim_{R \rightarrow \infty} \int_0^{\pi} \frac{{\left(R \mathrm{e}^{i \phi}\right)}^{-2/5}}{1 + {\left(R \mathrm{e}^{i \phi}\right)}^2} \, iR\mathrm{e}^{i \phi} \, \mathrm{d}\phi &= 0 \\
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \int_{-\infty}^0 \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z + \int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z \\
\int_{-\infty}^0 \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= - \int_0^{\infty} \frac{(-z)^{-2/5}}{1 + (-z)^2} \, \mathrm{d}(-z) \\
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \left(1 + \mathrm{e}^{-2\pi i/5}\right) \int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z \\
\end{split}
$$
Finally, take the residue and solve for the original integral.
$$
\begin{split}
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= 2 \pi i \operatorname{Res}_{z = i} \left( \frac{z^{-2/5}}{1+z^2} \right) \\
&= 2 \pi i \left( \frac{i^{-2/5}}{2 i} \right) \\
&= \pi i^{-2/5} \\
\int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \pi \left(\frac{i^{-2/5}}{1 + \mathrm{e}^{-2 \pi i / 5}}\right) \\
&= \frac{\pi}{2} \left(\sqrt{5} - 1 \right) \\
2I &= \frac{\pi}{2} \left(\sqrt{5} - 1 \right) \\
I &= \frac{\pi}{4} \left(\sqrt{5} - 1 \right)
\end{split}
$$
 A: Starting from @Luis Sierra's answer
$$\begin{equation}
I=\frac{1}{2}\int\limits_{0}^{1} \frac{t^{\frac{2}{5}}}{1+t^{2}} \,dt +\frac{1}{2}\int\limits_{0}^{1} \frac{t^{-\frac{2}{5}}}{1+t^{2}}\,dt
\end{equation}$$
Using the quite standard
$$J_a=\int_0^1 \frac {t^a}{1+t^2}\, dt=\frac{1}{4} \left(\psi \left(\frac{a+3}{4}\right)-\psi
   \left(\frac{a+1}{4}\right)\right)\qquad \text{if} \qquad \Re(a)>-1$$ So, rearranging,
$$8I=\Big[\psi\left(\frac{17}{20}\right)-\psi\left(\frac{3}{20}\right)\Big]+\Big[\psi \left(\frac{13}{20}\right)-\psi\left(\frac{7}{20}\right)\Big]=\pi  \cot \left(\frac{3 \pi }{20}\right)+\pi  \tan \left(\frac{3 \pi }{20}\right)$$ that is to say
$$8I=\pi\csc\left(\frac{3 \pi }{20}\right)\,\sec\left(\frac{3 \pi }{20}\right)=2 \left(\sqrt{5}-1\right)\, \pi \implies I=\frac{\sqrt{5}-1}{4} \pi$$
A: Right before "Our contour is", you could substitute $u = v^5; du = 5v^4 dv$, and get yourself an integrand that's
$$
\frac{5v^2}{1 + v^{10}},
$$
after which all you need to do is factor a tenth-degree polynomial and do endless partial-fractions work. I mean...it is elementary, just really unpleasant.
A: Let
\begin{equation}
I=\int\limits_{0}^{+\infty} \frac{e^{-x}\cosh\left(\frac{2x}{5}\right)}{1+e^{-2x}} \,dx
\end{equation}
Now let $t=e^{-x}$, which implies that $-dt=e^{-x}\,dx$. Plugging everything in yields:
\begin{equation}
I=\int\limits_{0}^{1} \frac{\cosh\left(\frac{2}{5}\ln(t)\right)}{1+t^{2}} \,dt
\end{equation}
Using the exponential definition of $\cosh(x)$ and the fact that $\ln(x^{a})=a\ln(x)$, we derive that:
\begin{equation}
\cosh\left(\frac{2}{5}\ln(t)\right)=\frac{t^{\frac{2}{5}}+t^{-\frac{2}{5}}}{2}
\end{equation}
Thus:
\begin{equation}
I=\frac{1}{2}\int\limits_{0}^{1} \frac{t^{\frac{2}{5}}}{1+t^{2}} \,dt +\frac{1}{2}\int\limits_{0}^{1} \frac{t^{-\frac{2}{5}}}{1+t^{2}}\,dt
\end{equation}
Now, consider the following integral:
\begin{equation}
I(a,b)=\int\limits_{0}^{1}\frac{t^{a}}{1+t^{2}}e^{-bt}\,dt
\end{equation}
We can compute the two integrals above with this generalized integral, note that:
\begin{equation} 
I=\frac{1}{2}I\left(\frac{2}{5},0\right)+\frac{1}{2}I\left(-\frac{2}{5},0\right)
\end{equation}
We want to obtain a differential equation with respect to $I(a,b)$, so that when we solve it, we can compute $I$. Let's take the first derivative with respect to $b$:
\begin{equation}
I'(a,b)=\int\limits_{0}^{1}\frac{\partial}{\partial b}\left[\frac{t^{a}}{1+t^{2}}e^{-bt}\right]\,dt=\int\limits_{0}^{1}\frac{(-t)t^{a}e^{-bt}}{1+t^{2}}\,dt
\end{equation}
Let's differentiate once again:
\begin{equation}
I''(a,b)=\int\limits_{0}^{1}\frac{\partial}{\partial b}\left[\frac{(-t)t^{a}e^{-bt}}{1+t^{2}}\right]\,dt=\int\limits_{0}^{1}\frac{t^{2}t^{a}e^{-bt}}{1+t^{2}}\,dt
\end{equation}
If we add and substract $1$ in the $t^{2}$, we can simplify things:
\begin{equation}
I''(a,b)=\int\limits_{0}^{1}\frac{(t^{2}+1-1)t^{a}e^{-bt}}{1+t^{2}}\,dt=\int\limits_{0}^{1}\frac{(t^{2}+1)t^{a}e^{-bt}}{1+t^{2}}\,dt-\underbrace{\int\limits_{0}^{1}\frac{t^{a}e^{-bt}}{1+t^{2}}\,dt}_{I(a,b)}
\end{equation}
Note that  the second integral is just our original $I(a,b)$. The first integral can be expressed in terms of the lower incomplete gamma function:
\begin{equation}
\int\limits_{0}^{1}\frac{(t^{2}+1)t^{a}e^{-bt}}{1+t^{2}}\,dt=\int\limits_{0}^{1}t^{a}e^{-bt}\,dt
\end{equation}
If we let $z=bt$, calculate and plug everything in, one gets the following integral:
\begin{equation}
\frac{1}{b}\int\limits_{0}^{b}\left(\frac{z}{b}\right)^{a}e^{-z}\,dz=\frac{1}{b^{1+a}}\underbrace{\int\limits_{0}^{b}z^{a}e^{-z}\,dz}_{\gamma\left(1+a,b\right)}=\frac{1}{b^{1+a}}\gamma\left(1+a,b\right)
\end{equation}
Finally, we obtain the following differential equation:
\begin{equation}
I''(a,b)+I(a,b)-\frac{1}{b^{1+a}}\gamma\left(1+a,b\right)=0
\end{equation}
Given that we have differentiated $I$ exclusively with respect to $b$, then we only need to consider the dependence with respect to $b$. Also, it is known that $\gamma\left(a+1,b\right)=a\gamma(a,b)-b^{a}e^{-b}$, then:
\begin{equation}
I''(b)+I(b)-\frac{1}{b^{1+a}}\left[a\gamma(a,b)-b^{a}e^{-b}\right]=0
\end{equation}
Solving the differential equation will allow us to compute $I(a,b)$, and once this is calculated, we can just plug the necessary values to determine $I$.
A: Yet another special-function solution, this time using beta-integrals: $$\mathrm{B}(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx=\int_0^\infty\frac{y^{a-1}\,dy}{(1+y)^{a+b}}\color{blue}{=\int_0^1\frac{z^{a-1}+z^{b-1}}{(1+z)^{a+b}}\,dz},$$ and your integral reduces to a particular case after $z=\mathrm{e}^{-2x}$: $$I=\frac14\int_0^1\frac{z^{-3/10}+z^{-7/10}}{1+z}\,dz=\frac14\mathrm{B}\left(\frac{7}{10},\frac{3}{10}\right)=\frac{\pi}{4\sin(3\pi/10)}=\frac{\pi}{4\cos(\pi/5)}=\ldots$$
