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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?

share|improve this question
    
why do you need a close form so badly, may I ask? why do you think such a thing exists? –  Lost1 May 19 '13 at 22:21
10  
I am not sure it exists, but if it exists I want to know it. Closed forms are easier to manipulate, sometimes closed forms of different integrals or sums contain terms that cancel each other etc. I found this paper very useful and interesting: J. M. Borwein, R. E. Crandall, "Closed Forms: What They Are and Why We Care". –  Oksana Gimmel May 19 '13 at 22:34
7  
Plus, there is always the frivolous answer that they are fun and challenging to arrive at. –  Jon Claus May 19 '13 at 22:55
    
A solution to the same integral was given on another forum, here - integralsandseries.prophpbb.com/post456.html#p456 –  Integrals and Series Jun 19 '13 at 4:35

1 Answer 1

up vote 24 down vote accepted

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.

share|improve this answer
    
waiting for the proof –  Shivam Patel May 25 at 15:18
    
As @ShivamPatel claims: Me too. –  Felix Marin Jul 18 at 23:31

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