How was the zeta function integrated from

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


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

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.

  • $\begingroup$ Residue theorem maybe? $\endgroup$
    – Wouter
    Jul 4, 2016 at 17:56
  • 3
    $\begingroup$ 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


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 $Re(s) > 1$ :

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.