Help with integral using compelx contour : x/(e^x-1) from 0 to infinity

[c0nfig]

Hi,

i need to solve this integral :
$$\int _{0}^{\infty }\!{\frac {x}{{{\rm e}^{x}}-1}}{dx}$$
i solved it using series and i got the right answer of Pi^2 / 6
but i need a solution using complex analysis
i need help with finding the right contour for this problem.

i tried change of variables and got the integral :
$$\int _{0}^{\infty }\!{\frac {\ln \left( x+1 \right) }{x \left( x+1\right) }}{dx}$$

but i couldn't find the right contour for this one as well

thanks,

 

[hunt_mat]

Try a semi circular contour, with an indent around x=1 of radius epsilon.

 

[c0nfig]

Yeah alright, well you can do it like Riemann did it using what we now call the Hankel integral:

$$I(s)=\frac{1}{2\pi i}\int_{C^-} \frac{z^{s-1}}{e^{-z}-1}dz$$

where $C^{-}$ is the mirror-image Hankel contour. Then:

$$\pi I(s)=\sin(s\pi)\int_0^{\infty} \frac{r^{s-1}}{e^r-1}dr,\quad Re(s)>1$$

Bingo-bango ain't it? I'll leave the details for you to research and fill-in.

thanks for the information,
i found some related info and learned how to get another relation between this integral I and zeta function using a contour that contains all the residues.
but i want to understand also your answer,
in the presented case s=2 , so sin(2*pi) = 0 , so how can i find the answer for the integral: $$\int_0^{\infty}\frac{x^{2-1}}{e^x-1}dx$$ ?

 

[jackmell]

thanks for the information,
i found some related info and learned how to get another relation between this integral I and zeta function using a contour that contains all the residues.
but i want to understand also your answer,
in the presented case s=2 , so sin(2*pi) = 0 , so how can i find the answer for the integral: $$\int_0^{\infty}\frac{x^{2-1}}{e^x-1}dx$$ ?

Ok. I made a mistake saying that so that's why I deleted it above but you got it in the interim period. Sorry for that. The integral I(s) is zero for s=2, 3, ... so that expression would not help in evaluating your integral. Your integral is usually expressed as:

$$\int_0^{\infty} \frac{x^{s-1}}{e^x-1}dx=\zeta(s)\Gamma(s),\quad Re(s)>1$$

but that does not take complex analysis to derive. I do not know how to use a contour integral directly to evaluate your integral and also I don't follow what Hunt is suggesting above.

 

[c0nfig]

someone told me that it might be solved with mellin transformation.
but i couldn't figure it out how it may help.
could anyone throw me an hint ?
thanks.