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How to show that for a complex number $s$ with $Re (s) > 1$, one has $$\sum_{k=1}^{\infty}\frac1{k^s} = \frac1{\Gamma(s)}\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx$$

where $\Gamma(z) = \int_0^{+\infty} t^{z-1}\mathrm{e}^{-t}\mathrm{d}t$

For info this is the Riemann zeta function

Tian Vlašić
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Julien
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    Expand $$\frac{1}{e^x-1}=\frac{e^{-x}}{1-e^{-x}}=\sum_{n=1}^{\infty}e^{-nx}$$ as a geometric series. Then recall the definition of the $\Gamma$ function and its interval of convergence and similarly for $\zeta$. However you should have $\frac{1}{\Gamma(s)}$ in front of the integral, not $\Gamma(s)$ as you currently do. – KStarGamer Jul 30 '23 at 13:05
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    @KStarGamer not tempted to put your comment as an answer? – peter a g Jul 30 '23 at 13:09
  • https://mathworld.wolfram.com/RiemannZetaFunction.html – Sine of the Time Jul 30 '23 at 13:17
  • I have an answer covering the Mellin transform on RHS. https://math.stackexchange.com/q/4725711/1141581 – bob Jul 30 '23 at 22:00

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Your claim is off. The correct result is:

\begin{align*} \int_{0}^{\infty} \frac{x^{s-1}}{e^x-1}\, \mathrm{d}x &= \int_{0}^{\infty} x^{s-1} \sum_{n=1}^{\infty} e^{-nx} \, \mathrm{d}x \\ &= \sum_{n=1}^{\infty} \int_{0}^{\infty} e^{-nx} x^{s-1}\, \mathrm{d}x \\ &= \sum_{n=1}^{\infty} \frac{\Gamma(s)}{n^s}\\ &= \Gamma(s) \sum_{n=1}^{\infty} \frac{1}{n^s}\\ &= \Gamma(s) \zeta(s) \end{align*}

As a side note,

$$\int_{0}^{\infty} \frac{x^{s-1}}{e^x+1}\, \mathrm{d}x = \Gamma(s) \eta(s)$$

where $\eta$ is the Dirichlet's eta function.

Tolaso
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