Proof of my conjecture on closed form of $\int _{0}^{\infty}\frac{x^{a-1}e^{-mbx}}{1-e^{-bx}}$

Question

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.

Answer 1

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 $x\rightarrow bx$.

Answer 2

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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}

$$\large{% \int_{0}^{\infty}{x^{a - 1}\expo{-mbx} \over 1 - \expo{-bx}}\,\dd x = {\Gamma\pars{a} \over b^{a}}\bracks{% \zeta\pars{a} - \sum_{k = 1}^{m - 1}{1 \over k^{a}}}} $$