\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

Prove that $\ds
\int_0^\infty\frac{x^{a-1}}{e^x-1}dx=\Gamma(a)\zeta(a)$



\vspace{0.25in}

{\bf Answer}

$\ds \int_0^\infty\frac{x^{a-1}}{e^x-1}=
\int_0^\infty\frac{x^{a-1}e^{-x}}{1-e^{-x}}dx=\int_0^\infty
x^{a-1}e^{-x}\sum_{n=0}^\infty e^{-nx}dx$

$\ds =\sum_{n=0}^\infty\int_0^\infty
x^{a-1}e^{-(n+1)x}dx=\sum_{n=0}^\infty\frac{1}{(n+1)^a}\int_0^\infty
t^{a-1}e^{-t}dt=\Gamma(a)\zeta(a)$


\end{document}