\\int_{0}^{\\infty}\\frac{x^{2}}{e^{x}-1},dx

Explanation:

Step1: Use the power - series expansion

We know that $\frac{1}{e^{x}-1}=\sum_{n = 1}^{\infty}e^{-nx}$ for $x>0$. Then the integral $\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx=\int_{0}^{\infty}x^{2}\sum_{n = 1}^{\infty}e^{-nx}dx$. By the uniform convergence of the series for $x\in[a,b]$ with $0 < a < b<\infty$, we can interchange the integral and the sum: $\sum_{n = 1}^{\infty}\int_{0}^{\infty}x^{2}e^{-nx}dx$.

Step2: Use integration by parts

Let $u = x^{2}$, $dv=e^{-nx}dx$. Then $du = 2xdx$, $v=-\frac{1}{n}e^{-nx}$. By the integration - by - parts formula $\int_{0}^{\infty}u;dv=uv|{0}^{\infty}-\int{0}^{\infty}v;du$, we have $\int_{0}^{\infty}x^{2}e^{-nx}dx=-\frac{x^{2}}{n}e^{-nx}|{0}^{\infty}+\frac{2}{n}\int{0}^{\infty}xe^{-nx}dx$. The first term $-\frac{x^{2}}{n}e^{-nx}|{0}^{\infty}=0$. For $\int{0}^{\infty}xe^{-nx}dx$, use integration by parts again. Let $u = x$, $dv=e^{-nx}dx$, then $du = dx$, $v = -\frac{1}{n}e^{-nx}$. So $\int_{0}^{\infty}xe^{-nx}dx=-\frac{x}{n}e^{-nx}|{0}^{\infty}+\frac{1}{n}\int{0}^{\infty}e^{-nx}dx$. The first term $-\frac{x}{n}e^{-nx}|{0}^{\infty}=0$, and $\int{0}^{\infty}e^{-nx}dx=-\frac{1}{n}e^{-nx}|{0}^{\infty}=\frac{1}{n}$. Then $\int{0}^{\infty}xe^{-nx}dx=\frac{1}{n^{2}}$, and $\int_{0}^{\infty}x^{2}e^{-nx}dx=\frac{2}{n^{3}}$.

Step3: Evaluate the sum

So $\sum_{n = 1}^{\infty}\int_{0}^{\infty}x^{2}e^{-nx}dx=\sum_{n = 1}^{\infty}\frac{2}{n^{3}}$. We know that the Riemann zeta - function is defined as $\zeta(s)=\sum_{n = 1}^{\infty}\frac{1}{n^{s}}$ for $s>1$. When $s = 3$, $\sum_{n = 1}^{\infty}\frac{2}{n^{3}}=2\zeta(3)$. And $\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx = 2\zeta(3)$. Also, we can use another approach. We know that $\int_{0}^{\infty}\frac{x^{m}}{e^{x}-1}dx=\Gamma(m + 1)\zeta(m+1)$ for $m>0$, where $\Gamma(k)=\int_{0}^{\infty}t^{k - 1}e^{-t}dt$ is the gamma - function and $\Gamma(n)=(n - 1)!$ for positive integers $n$. When $m = 2$, $\Gamma(3)=2! = 2$, so $\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx=\Gamma(3)\zeta(3)=2\zeta(3)$.

Answer:

$2\zeta(3)$