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Show that : $$\int_0^\infty \frac{\text{e}^{-x}}{x^2} \left( \frac{1}{1-\text{e}^{-x}} - \frac{1}{x} - \frac{1}{2} \right)^2 \, \text{d}x = \frac{7}{36}-\ln A+\frac{\zeta \left( 3 \right)}{2\pi ^2}$$ Where $\displaystyle A$ is Glaisher-Kinkelin constant

I see Chris's question is a bit related with this Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm{dx}$

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The TeX code here was done by one of those hideous programs that are intended to make it hard for humans to understand how to code this stuff. Stuff like {{{x}^{{2}}}} and putting {} or {{}} around every use of \frac{}{} or {.........} after every instance of \int or \sum, etc. I've cleaned it up. –  Michael Hardy Feb 16 '13 at 19:32
... where $\ln A = 1/12 - \zeta'(-1)$ for example –  GEdgar Feb 16 '13 at 20:41
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2 Answers

up vote 5 down vote accepted

Here is an identity for log(A) that may assist.

$\displaystyle \ln(A)-\frac{1}{4}=\int_{0}^{\infty}\frac{e^{-t}}{t^{2}}\left(\frac{1}{e^{t}-1}-\frac{1}{t}+\frac{1}{2}-\frac{t}{12}\right)dt$.

I think Coffey has done work in this area. Try searching for his papers on the Stieltjes constant, log integrals, Barnes G, log Gamma, etc.

Another interesting identity is $\displaystyle 2\int_{0}^{1}\left(x^{2}-x+\frac{1}{6}\right)\log\Gamma(x)dx=\frac{\zeta(3)}{2{\pi}^{2}}$.

Just some thoughts that may help put it together.

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You may start with the decomposition $$ \int_0^\infty {\frac{e^{-x}}{x^2} \left( \frac{1}{\left(e^x - 1\right)^2} - \frac{1}{x^2} + \frac{1}{x} - \frac{5}{12} + \frac{x}{12} \right)dx} - 2\int_0^\infty {\frac{e^{-x}}{x^3} \left(\frac{1}{e^x - 1} - \frac{1}{x} + \frac{1}{2}-\frac{x}{12} \right)dx} + \int_0^\infty {\frac{e^{-x}}{x^2} \left(\frac{1}{e^x-1}- \frac{1}{x} + \frac{1}{2}-\frac{x}{12} \right)dx} $$ and use the theory of the Multiple Gamma functions.

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