Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm{dx}$

share|improve this question
1  
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
1  
... where $\ln A = 1/12 - \zeta'(-1)$ for example –  GEdgar Feb 16 '13 at 20:41

3 Answers 3

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.

share|improve this answer

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 $$f(1) = \lim_{s \to 1} \left(\frac{(s - 5)\zeta(s - 3)}{(s - 3)(s - 2)(s - 1)} - \frac{\zeta(s - 2) + \frac1{12}}{(s - 2)(s - 1)} + \frac{\frac13 (s - 4)(s - 3) + s - 3}{(s - 4)(s - 3)(s - 2)(s - 1)}\right)$$

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

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

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

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.