Proof of my conjecture on closed form of $\int _{0}^{\infty}\frac{x^{a-1}e^{-mbx}}{1-e^{-bx}}$ Let $a$, $b\in \Bbb R^+$ and $m \in \Bbb N$ then My conjectural closed form is $$\int _{0}^{\infty}\frac{x^{a-1}e^{-mbx}}{1-e^{-bx}}\,{\rm d}x
=
\frac{\Gamma(a)}{b^a}\left\lbrack\zeta(a)-\sum^{m-1}_{k=1}\frac{1}{k^a}\right\rbrack$$
Please tell me how to prove it.
 A: If you know what $\zeta(a)$ and $\Gamma(a)$ are, and moreover are able to guess such formula, it is somewhat strange that you are asking for a proof. 
Note that $a$ should be actually greater than $1$ for the integral to converge. Under such assumption, we have
\begin{align}
\zeta(a)-\sum_{k=1}^{m-1}\frac{1}{k^a}&=\sum_{k=m}^{\infty}\frac{1}{k^a}=\\
&=\sum_{k=m}^{\infty}\frac{1}{\Gamma(a)}\int_0^{\infty}x^{a-1}e^{-kx}dx=\\
&=\frac{1}{\Gamma(a)}\int_0^{\infty}\frac{x^{a-1}e^{-mx}}{1-e^{-x}}dx=\\
&=\frac{b^a}{\Gamma(a)}\int_0^{\infty}\frac{x^{a-1}e^{-mbx}}{1-e^{-bx}}dx.
\end{align}
Explanations:


*

*Making the change of variables $kx=y$ in the second line transforms the integral into the definition of $\Gamma(a)$ multiplied by $k^{-a}$;

*To pass from the second to the third line we exchange summation and integration and sum the resulting geometric series;

*The fourth line is obtained from the third by the change of variables $x\rightarrow bx$.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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\begin{align}
\int_{0}^{\infty}{x^{a - 1}\expo{-mbx} \over 1 - \expo{-bx}}\,\dd x
&=
\int _{0}^{\infty}x^{a - 1}\expo{-mbx} \sum_{\ell = 0}^{\infty}
\expo{-\ell bx}\,\dd x
=
\sum_{\ell = 0}^{\infty}\int _{0}^{\infty}x^{a - 1} 
\expo{-\pars{m + \ell}bx}\,\dd x
\\[3mm]&=
\sum_{\ell = 0}^{\infty}{1 \over \bracks{\pars{m + \ell}b}^{a}}
\int _{0}^{\infty}x^{a - 1}\expo{-x}\,\dd x
=
{1 \over b^{a}}\sum_{\ell = m}^{\infty}{1 \over \ell^{a}}\Gamma\pars{a}
\\[3mm]&=
{\Gamma\pars{a} \over b^{a}}\pars{%
\sum_{\ell = 1}^{\infty}{1 \over \ell^{a}}
-
\sum_{\ell = 1}^{m - 1}{1 \over \ell^{a}}}
=
{\Gamma\pars{a} \over b^{a}}\bracks{%
\zeta\pars{a}
-
\sum_{\ell = 1}^{m - 1}{1 \over \ell^{a}}}
\\[1cm]&
\end{align}
$$\large{%
\int_{0}^{\infty}{x^{a - 1}\expo{-mbx} \over 1 - \expo{-bx}}\,\dd x
=
{\Gamma\pars{a} \over b^{a}}\bracks{%
\zeta\pars{a}
-
\sum_{k = 1}^{m - 1}{1 \over k^{a}}}}
$$
