Improper integral of $\dfrac{x^2}{ e^x−1}$

Question

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.

Answer 1

Depending on whether a series might do, it looks easier to integrate $x^2(e^{-x}+e^{-2x}+e^{-3x}+\cdots)$ by parts instead.

Answer 2

-$1^{st}$ Approach:

Using integration by parts you get

$$\begin{align}\int \frac{x^2}{e^x-1}dx &= x^2 \ (\ln (1-e^x) - x) - \int 2x\ln (1-e^x) - 2x^2 dx \\&= x^2\ln (1-e^x) - \frac{x^3}{3} +2xLi_{2}(e^x) - 2Li_{3}(e^x) +C \end{align}$$

Evaluating separetly $\int x\ln(1-e^x) dx$ we have

$$\int x\ln(1-e^x) dx = -xLi_{2}(e^x) + Li_{3}(e^x)$$

Where $$\int \ln (1-e^x) = -Li_2(e^x) + C_1 (*) \ \ \text{and}\ \ \int Li_2(e^x) = Li_3(e^x) + C_2$$

$(*)$ Write $\ln(1 - e^x)$ in series and use integration term by term.

- $2^{nd}$ Approach:

$$\zeta (3) = \frac{1}{\Gamma(3)}\int_{0}^{\infty} \frac{x^{3-1}}{e^x-1} dx \Rightarrow \int_{0}^{\infty} \frac{x^{2}}{e^x-1} dx = 2\zeta(3) $$

See more about Riemann Zeta Function.

Answer 3

Note that $$\eqalign{\dfrac{d}{dx} \text{Li}_n(e^x) &= \text{Li}_{n-1}(e^x),\ n \ge 3\cr \dfrac{d}{dx} \text{Li}_2(e^x) &= -\ln(1-e^x)\cr \dfrac{d}{dx} \left(-\ln(1 - e^x)\right) &= \dfrac{e^x}{1-e^x} = \dfrac{1}{1-e^x} - 1\cr}$$ So we can do integration by parts $$ \eqalign{\int \dfrac{x^n \; dx}{1-e^x} &= \dfrac{x^{n+1}}{n+1} + \int \dfrac{x^n e^x\; dx}{1 - e^x} \cr &= \dfrac{x^{n+1}}{n+1} - x^n\ln(1 - e^x) + n \int x^{n-1} \ln(1 - e^x)\; dx\cr &= \dfrac{x^{n+1}}{n+1} - x^n\ln(1 - e^x) - n x^{n-1} \text{Li}_2(e^x) + n(n-1) \int x^{n-2} \text{Li}_2(e^x)\; dx}$$ etc.

Answer 4

$\int_0^t\dfrac{x^2}{e^x-1}dx$

$=\int_0^t\sum\limits_{n=0}^\infty\dfrac{B_nx^{n+1}}{n!}dx$ (with the formula in http://en.wikipedia.org/wiki/Bernoulli_number#Generating_function)

$=\left[\sum\limits_{n=0}^\infty\dfrac{B_nx^{n+2}}{n!(n+2)}\right]_0^t$

$=\sum\limits_{n=0}^\infty\dfrac{B_nt^{n+2}}{n!(n+2)}$