Definite integral $\int_{0}^{\infty} \frac{dx}{(1+e^{ax})(1+e^{x/a})}$

Question

I'm trying to solve the following definite integral; no success so far with standard approaches. Any help welcome!

$$\int_{0}^{\infty} \frac{\mathrm{d}x}{(1+e^{ax})(1+e^{x/a})}$$

Answer 1

Assuming $a>0$ $$\frac{1}{\left(1+e^{a x}\right)\left(1+e^{\frac{x}{a}}\right) }=\sum_{n=0}^\infty(-1)^n\,\frac {e^{-\frac{(n+1) }{a}x}}{1+e^{a x} }$$

$$I_n=\int_0^\infty \frac {e^{-\frac{(n+1) }{a}x}}{1+e^{a x}} \,dx=\frac 1{2a}\left(H_{\frac{n+1}{2 a^2}}-H_{\frac{n+1}{2 a^2}-\frac 12}\right)$$ $$\frac {I_{n+1}}{I_n}=1-\frac{1}{n}+\frac{a^2+4}{2 n^2}+O\left(\frac{1}{n^3}\right)$$

$$\color{blue}{\int_0^\infty \frac{dx}{\left(1+e^{a x}\right)\left(1+e^{\frac{x}{a}}\right) }=\frac 1{2a}\sum_{n=0}^\infty(-1)^n\,\left(H_{\frac{n+1}{2 a^2}}-H_{\frac{n+1}{2 a^2}-\frac 12}\right)}$$

When $a$ increases, the convergence is slower and slower since, when $n$ is large $$\frac 1{2a}\left(H_{\frac{n+1}{2 a^2}}-H_{\frac{n+1}{2 a^2}-\frac 12}\right)=\frac{a}{2 n}-\frac{a \left(a^2+2\right)}{4 n^2}+O\left(\frac{1}{n^3}\right)$$

Computing the partial sums for $a=2$ from $n=0$ to $n=10^k$ $$\left( \begin{array}{cc} \large{ k} & \large{\sum_{n=0}^{10^k} (-1)^n \,I_n }\\ 0 & 0.051452188718 \\ 1 & 0.184663765025 \\ 2 & 0.153241327722 \\ 3 & 0.148910576860 \\ 4 & 0.148462305363 \\ 5 & 0.148417322684 \\ \cdots & \cdots \\ \infty & 0.148412322859 \\ \end{array} \right)$$

Answer 2

I tried with Wolfram.
If $a\in\mathbb{N}$ is has an analytic solution (very complex but is anaytic)
If $a\not\in\mathbb{N}$ the solution is not analytic.
For $a=1$ you have $I_1=x+\dfrac{1}{e^x+1}-\ln(e^x+1)+C$
For $a=2$ you have $\displaystyle -\frac{1}{8} \sum_{ω: ω^4 + 1 = 0}(x( ω^3 - ω^2+ ω) -(ω^3-ω^2+ω-1) \ln((e^{x/2} - ω)^2) + x - \ln(e^{x/2} + 1) + C$
Where, $$\sum_{ω: ω^4 + 1 = 0}\text{ means that the sum is over the root of }z^4+1=0, z\in\mathbb{C}$$ $$\text{So you consider }\omega\in\{e^{\pi i/4},e^{-\pi i/4},e^{3\pi i/4},e^{-3\pi i/4}\}$$ For $a=3$ you have $$\displaystyle I_3=\\\frac{1}{18} \left(-2 \sum_{ω: ω^6 - ω^3 + 1 = 0} \frac{ω^5- ω^4+ ω^3- ω^2+ ω-1}{ω^3 - 2}(\ln((e^{x/3} - ω)^3)-x)+ 12 x + \frac{6}{e^{x/3} + 1} - 30 \ln(e^{x/3} + 1) - 3 \ln(e^{2x/3}-e^{x/3} + 1) - 2 \sqrt{3} \arctan\left(\frac{2 e^{x/3} - 1}{\sqrt{3}}\right)\right) + C$$ For $a=4$ is longer and the sum is over the root of $z^{16}+1=0$
For $a=5$ the sum is over the root of $z^4-z^3+z^2-z+1=0$
For $a=6$ the sum is over the root of $z^4+1=0$
For $a=7$ the sum is over the root of $z^6-z^5+z^4-z^3+z^2-z+1=0$
For $a=8$ the sum is over the root of $z^{64}+1=0$
For $a=9$ the sum is over the root of $z^{6}-z^3+1=0$
For $a=10$ the sum is over the root of $z^{4}+1=0$
Sorry but I can't see a pattern in this integral, even if the polynomials remember me the cyclotomic polynomials.
In general since the value of the integral is a real value, I suggest you to take the real values of these big integral and work on them.
Remember that $$\Re[\ln(x+iy)]=\frac{1}{2}\ln(x^2+y^2)$$