Solve $$I=\int_{0}^{\infty }\frac{x^{6}e^{x}}{\left ( e^{x}-1 \right )^{2}}dx$$ I tried integration by part to get $$I=6\int_{0}^{\infty }\frac{x^{5}}{e^{x}-1}dx$$ But I stuck here.How can I go further after this.
Solve $\int_{0}^{\infty }\frac{x^{6}e^{x}}{\left ( e^{x}-1 \right )^{2}}dx$
1
$\begingroup$
calculus
integration
definite-integrals
improper-integrals
1 Answers
4
For $n>1$ we have $$\int_{0}^{\infty }\frac{x^{n-1}}{e^{x}-1}\, \mathrm{d}x=\sum_{k=1}^{\infty }\int_{0}^{\infty }x^{n-1}e^{-kx}\, \mathrm{d}x\overset{[1]}{=}\sum_{k=1}^{\infty }\frac{1}{k^{n}}\int_{0}^{\infty }x^{n-1}e^{-x}\, \mathrm{d}x=\zeta \left ( n \right )\Gamma \left ( n \right )$$ where $\zeta(\cdot)$ is Riemann zeta function and $\Gamma(\cdot)$ is Gamma function.
$[1]:$ Let $kx\to x$
Hence let $n=6$ we obtain the answer $$I=\frac{16}{21}\pi ^{6}$$