I encountered this integral in my calculations:
$$\int_0^\infty\frac{\log\left(1+\frac{\pi^2}{4\,x}\right)}{e^{\sqrt{x}}-1}\mathrm dx=2\int_0^\infty\frac{x\log\left(1+\frac{\pi^2}{4\,x^2}\right)}{e^x-1}\mathrm dx=6.041880938342236884944983747836284…,$$
but could not find a closed-form representation for it. I tried to replace a constant factor $\frac{\pi^2}4$ with a parameter and take a derivative, that made the integral look simpler, but I still was not successful in solving it.
I also tried to find possible closed forms using Inverse Symbolic Calculator and WolframAlpha but they did not find anything.
Could you please help me to find a closed form (even using non-elementary special functions), if it exists?
Best Answer
A closed form indeed exists for this integral: $$\int_0^\infty\frac{\log\left(1+\frac{\pi^2}{4\,x}\right)}{e^{\sqrt{x}}-1}\mathrm dx=\pi^2\left(\log\frac{\pi\,A^9\sqrt{2}}{\Gamma(\frac{1}{4})^2}-\frac{9}{8}\right)+2\,\pi\,C,$$ where $A$ is the Glaisher-Kinkelin constant and $C$ is the Catalan constant.
A more general result: $$\int_0^\infty\frac{\log\left(1+\frac{a}{x}\right)}{e^{\sqrt{x}}-1}\mathrm dx=8\,\pi^2\psi^{(-2)}\left(\frac{\sqrt{a}}{2\pi}\right)-\frac{a}2\left(1+\log\frac{4\pi^2}{a}\right)-2\pi\sqrt{a}\left(1+2\log\Gamma\left(\frac{\sqrt{a}}{2\pi}\right)\right),$$ where $\psi^{(-2)}(z)$ is the generalized polygamma function.
The proof is based on Binet's second formula, but I still need to sort out some details.