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How was the zeta function integrated from

$$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^{s}}$$

To

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


I've tried googling this and surprisingly can't find much on it, even the wikipedia article for zeta function doesn't explain how this integral is derived or cite a source anywhere. I have no doubt it's true; I am just curious how it was obtained. I know very little about converting infinite sums to integrals.

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  • $\begingroup$ Residue theorem maybe? $\endgroup$
    – Wouter
    Jul 4, 2016 at 17:56
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    $\begingroup$ all you need is that for $Re(s) > 0$ : $ \ \Gamma(s) n^{-s} =n^{-s}\int_0^\infty x^{s-1} e^{-x} dx = \int_0^\infty x^{s-1} e^{-nx} dx$ so that for $Re(s) > 1$ : $\Gamma(s) \zeta(s) = \sum_{n=1}^\infty \int_0^\infty x^{s-1} e^{-nx} dx =\int_0^\infty x^{s-1} \sum_{n=1}^\infty e^{-nx} dx = \int_0^\infty \frac{x^{s-1} }{e^x-1} dx$ where the inversion of $\int$ and $\sum$ has to be justified by absolute/monotone/dominated convergence @Wouter $\endgroup$
    – reuns
    Jul 4, 2016 at 18:03

1 Answer 1

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Instead of using the dominated or monotone convergence theorem, I like to prove it by elementary means.

For $Re(s) > 0$ and $n > 0$ (change of variable $y = nx$) : $$\Gamma(s)n^{-s} = \int_0^\infty x^{s-1} e^{-nx}dx$$

So that for $Re(s) > 0$ (using the geometric series) : $$\Gamma(s) \sum_{n=1}^N n^{-s} = \int_0^\infty x^{s-1} \sum_{n=1}^N e^{-nx}dx = \int_0^\infty x^{s-1}\frac{1-e^{-Nx}}{e^{x}-1} dx$$

For $Re(s) > 1$ it is known that

$$\zeta(s) = \lim_{N \to \infty} \sum_{n=1}^N n^{-s}$$

Finally, we need to prove that (again for $Re(s) > 1$) : $$\lim_{N \to \infty}\int_0^\infty x^{s-1}\frac{1-e^{-Nx}}{e^{x}-1} dx = \int_0^\infty \frac{x^{s-1}}{e^{x}-1} dx$$ which is obvious once we showed that for $x >0$ : $\displaystyle\left|\frac{x}{e^{x}-1}\right| < 1$ whence $\displaystyle\int_0^\infty \frac{x^{s-1}}{e^x-1}e^{-Nx}dx$ converges absolutely and $\to 0$ as $N \to \infty$

Overall, for $Re(s) > 1$ :

$$\Gamma(s) \zeta(s) = \lim_{N \to \infty} \Gamma(s) \sum_{n=1}^N n^{-s} = \lim_{N \to \infty} \int_0^\infty x^{s-1} \frac{1-e^{-Nx}}{e^x-1}dx = \int_0^\infty \frac{x^{s-1}}{e^{x}-1} dx$$

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  • $\begingroup$ Thank you! This is good stuff $\endgroup$ Jul 4, 2016 at 20:37

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