I need to prove this, today my Instructor solved an integral using this formula but didn't gave a proof $$\displaystyle \int_{0}^{\infty}\dfrac{x^{s-1}}{e^{x}-1}\,dx=\zeta(s).\Gamma(s) $$ i tried to solve it using a series of $e^{x}$ but ended up nowhere.


We have $\int_{0}^{+\infty}z^{s-1}e^{-z}\,dz = \Gamma(s)$ for any $s>0$ by the very definition of the $\Gamma$ function.
Moreover $$ \frac{1}{e^x-1} = e^{-x}+e^{-2x}+e^{-3x}+\ldots $$ with uniform convergence over any compact subset of $\mathbb{R}^+$.
By the dominated convergence theorem it follows that $$\begin{eqnarray*} \int_{0}^{+\infty}\frac{x^{s-1}}{e^x-1}\,dx &=& \sum_{n\geq 1}\int_{0}^{+\infty}x^{s-1}e^{-nx}\,dx\\ &\stackrel{x\mapsto z/n}{=}& \sum_{n\geq 1}\frac{1}{n^s}\int_{0}^{+\infty}z^{s-1}e^{-z}\,dz\\&=&\Gamma(s)\sum_{n\geq 1}\frac{1}{n^s}\\&=&\Gamma(s)\,\zeta(s)\end{eqnarray*}$$ as wanted.

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    $\begingroup$ better: monotone convergence $\endgroup$ – Rüdiger Apr 12 '17 at 20:59

From Ramanujan's Master Theorem we know.... $\int_{0}^{\infty} x^{s-1} \sum_{0}^{\infty} \phi({k}){(-1)^k}\frac{x^k}{k!} dx=\Gamma({s})\phi({-s})$

If we put $\zeta(-k)$ in place of $\phi(k)$ then we get...

$\int_{0}^{\infty} x^{s-1} \sum_{0}^{\infty} \zeta({-k}){(-1)^k}\frac{x^k}{k!} dx=\Gamma({s})\zeta({s})$

or, $\int_{0}^{\infty} x^{s-1} \sum_{0}^{\infty} (\sum_{1}^\infty n^{k}){(-1)^k}\frac{x^k}{k!} dx=\Gamma({s})\zeta({s})$

$\int_{0}^{\infty} x^{s-1} \sum_{1}^{\infty} (\sum_{0}^\infty \frac{(-nx)^k}{k!}) dx=\Gamma({s})\zeta({s})$

$\int_{0}^{\infty} x^{s-1} \sum_{1}^{\infty} (e^{-x})^n dx=\Gamma({s})\zeta({s})$

$\int_{0}^{\infty} \frac{x^{s-1}}{e^{x}-1} dx=\Gamma(s)\zeta(s)$

  • $\begingroup$ Did you ever look at a proof of Ramanujan master theorem ? If you do you'll see how it is ridiculous to use it here. $\endgroup$ – reuns Aug 8 at 5:08
  • $\begingroup$ Also how do you know that $ \int_{0}^{\infty} e^{-x} \sum_{0}^{\infty} \zeta({-k}){(-1)^k}\frac{x^k}{k!}$ converges. $\endgroup$ – reuns Aug 8 at 5:11

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