# A improper integral with Glaisher-Kinkelin constant

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

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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Define $$f(s) = \int_0^{\infty} x^{s - 1} \frac{e^{-x}}{x^2} \left(\frac1{1 - e^{-x}} - \frac1x - \frac12\right)^2 dx.$$ This defines an analytic function on the domain $\operatorname{Re}(s) > 0$ and the problem is to evaluate $f(1)$.

We have $$f(s) = \int_0^{\infty} x^{s - 3} e^{-x} \left(\frac1{(1 - e^{-x})^2} + \frac1{x^2} + \frac14 - \frac2{x(1 - e^{-x})} - \frac1{1 - e^{-x}} + \frac1x\right) dx$$

$$\quad\ = \int_0^{\infty} \left(\frac{x^{s - 3} e^{x}}{(e^x - 1)^2} + x^{s - 5} e^{-x} + \frac{x^{s - 3} e^{-x}}4 - \frac{2x^{s - 4}}{e^x - 1} - \frac{x^{s - 3}}{e^x - 1} + x^{s - 4} e^{-x}\right) dx.$$

For $\operatorname{Re}(s) > 4$, integrating by parts on the first term gives $$f(s) = \int_0^{\infty} \left(\frac{(s - 3)x^{s - 4}}{e^x - 1} + x^{s - 5} e^{-x} + \frac{x^{s - 3} e^{-x}}4 - \frac{2x^{s - 4}}{e^x - 1} - \frac{x^{s - 3}}{e^x - 1} + x^{s - 4} e^{-x}\right) dx.$$

$$= \int_0^{\infty} \left(\frac{(s - 5)x^{s - 4}}{e^x - 1} + x^{s - 5} e^{-x} + \frac{x^{s - 3} e^{-x}}4 - \frac{x^{s - 3}}{e^x - 1} + x^{s - 4} e^{-x}\right) dx.$$

Again assuming $\operatorname{Re}(s) > 4$, this gives us $$f(s) = (s - 5)\Gamma(s - 3)\zeta(s - 3) + \Gamma(s - 4) + \frac14 \Gamma(s - 2) - \Gamma(s - 2)\zeta(s - 2) + \Gamma(s - 3),$$ but, by analytic continuation, the equation must be valid (where the right side is defined) for $\operatorname{Re}(s) > 0$ and the apparent singularities of the right side at $s = 1$, $2$, $3$, and $4$ must be removable.

We may write $$f(s) = \frac{(s - 4)(s - 5)\zeta(s - 3) + 1 + \frac14 (s - 4)(s - 3) - (s - 4)(s - 3)\zeta(s - 2) + s - 4}{(s - 4)(s - 3)(s - 2)(s - 1)}\Gamma(s)$$ and so $$f(1) = \lim_{s \to 1} \left(\frac{(s - 5)\zeta(s - 3)}{(s - 3)(s - 2)(s - 1)} - \frac{\zeta(s - 2) + \frac1{12}}{(s - 2)(s - 1)} + \frac{\frac13 (s - 4)(s - 3) + s - 3}{(s - 4)(s - 3)(s - 2)(s - 1)}\right)$$

$$= \lim_{s \to 1} \left(\frac{(s - 5)\zeta(s - 3)}{(s - 3)(s - 2)(s - 1)} - \frac{\zeta(s - 2) + \frac1{12}}{(s - 2)(s - 1)} + \frac1{3(s - 4)(s - 2)}\right)$$

$$= -2\zeta'(-2) + \zeta'(-1) + \frac19$$

$$= \frac{\zeta(3)}{2\pi^2} + \frac1{12} - \ln A + \frac19$$

$$= \frac7{36} - \ln A + \frac{\zeta(3)}{2\pi^2}.$$

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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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