1. ## Tricky Integral

Hi I'm doing a question to find the pressure of a gas made of massless bosons. I've done everything else and now the question has resulted in me needing to evaluate this integral
$\int_{0}^{\infty}\frac{z^2}{e^{z}+1}$

This I'm pretty stuck on I've tried so far to multiply the integrand by

$\frac{e^z-1}{e^z-1}$

which gives me a second term which is easy to evaluate in terms of the zeta function but again a first term I can't seem to be able to do much with.

Any hints in the right direction would be much appreciated!

Thank you =)

2. With the substitution $e^{-x}= t$ the integral becomes...

$\displaystyle \int_{0}^{\infty} \frac{x^{2}}{e^{x}+1}\ dx = \int_{0}^{1} \frac{\ln^{2} t}{1+t}\ dt$ (1)

With a little of patience You can arrive to the 'general formula'...

$\displaystyle \int t^{m}\ \ln^{n} t \ dt = t^{m+1}\ \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(m+1)^{k+1}}\ \frac{n!}{(n-k)!}\ \ln^{n-k} t + c$ (2)

... and from (2) to the 'particular formula'...

$\displaystyle \int_{0}^{1} t^{m}\ \ln^{n} t \ dt = (-1)^{n}\ \frac{n!}{(m+1)^{n+1}}$ (3)

Very well!... now using (3) You can 'meet the goal'...

$\displaystyle \int_{0}^{1} \frac{\ln^{2} t}{1+t}\ dt = \sum_{k=0}^{\infty} (-1)^{k}\ \int_{0}^{1} t^{k}\ \ln^{2} t\ dt = 2\ \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(k+1)^{3}} =$

$\displaystyle = 2\ \{\sum_{k=1}^{\infty} \frac{1}{k^{3}} - 2\ \sum_{k=1}^{\infty} \frac{1}{(2k)^{3}} \} = 2\ (1-\frac{1}{4}) \ \zeta(3) = \frac{3}{2}\ \zeta (3)$ (4)

... where $\zeta(s)$ is the so called 'Riemann Zeta Function'...

Kind regards

$\chi$ $\sigma$

3. Thank you very much - nice to see how you would deal with a problem like this in general as well