Close form solution for $\int_{0}^{\infty}\frac{x^\alpha}{e^x-1}$

Question

Is there the close form solution for the integral is given by $$\int_{0}^{\infty}\frac{x^\alpha}{e^x-1}\,dx$$ where $\alpha>0$

Answer 1

\begin{align} & I=\int_{0}^{+\infty }{\frac{{{x}^{\alpha }}}{{{e}^{x}}-1}}\,\,dx=\int_{0}^{+\infty }{{{x}^{\alpha }}\frac{{{e}^{-x}}}{1-{{e}^{-x}}}}\,\,dx=\int_{0}^{+\infty }{\sum\limits_{n=0}^{+\infty }{{{x}^{\alpha }}{{e}^{-x}}{{e}^{-nx}}}}\,\,dx \\ & I=\sum\limits_{n=1}^{\infty }{\,\int_{0}^{\infty }{{{x}^{\alpha }}{{e}^{-nx}}\,dx}}=\Gamma (\alpha +1)\sum\limits_{n=1}^{\infty }{\frac{1}{{{n}^{\alpha +1}}}}=\Gamma (\alpha +1)\zeta (\alpha +1) \\ \end{align} Note $$\int_{0}^{\infty }{{{x}^{\alpha }}}{{e}^{-a{{x}^{m}}}}dx=\frac{1}{m{{a}^{\frac{\alpha +1}{m}}}}\Gamma \left( \frac{\alpha +1}{m} \right)$$