TITLE: Improper integral of $\dfrac{x^2}{ e^x−1}$
QUESTION [9 upvotes]: I am trying to evaluate 
$$\int_0^t \frac{x^2}{ e^x−1}dx$$
for $t>0$.
I tried integrating by parts, like follows: 
$$\int_0^t x\cdot\frac{xe^{-x}}{1-e^{-x}}dx=t\cdot Li_2(1-e^{-t})-\int_0^t Li_2(1-e^{-x}) dx,$$ where $Li$ is the dilogarithm function. But then, how can I proceed? 
Any suggestion? Thank you.

REPLY [3 votes]: $\int_0^t\dfrac{x^2}{e^x-1}dx$
$=\int_0^t\sum\limits_{n=0}^\infty\dfrac{B_nx^{n+1}}{n!}dx$ (with the formula in http://en.wikipedia.org/wiki/Bernoulli_number#Generating_function)
$=\left[\sum\limits_{n=0}^\infty\dfrac{B_nx^{n+2}}{n!(n+2)}\right]_0^t$
$=\sum\limits_{n=0}^\infty\dfrac{B_nt^{n+2}}{n!(n+2)}$