How does one get from this $$\zeta(s)=\sum_{k=1}^{\infty}\frac1{k^s}$$ to the integral representation $$\zeta(s)=\frac1{\Gamma(s)}\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx$$ of the Riemann Zeta function?

I can see that it can be rewritten as

$$\Gamma(s)\zeta(s)=\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx$$ and the Gamma function as an integral yields

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

But this approach does not work as the right integral does not converge. So how does one go from the summation to the integral representation?

  • 2
    $\begingroup$ A reference: This is derived on page nine of Riemann's Zeta Function by Edwards. $\endgroup$ – apnorton Jan 13 '15 at 20:23

Recall that for $t>1$, $$\frac{1}{t-1}=\sum_{n=1}^\infty t^{-n}$$

Then substitute $t=e^{x}$ for $x>0$.

Then substituting $x=\frac{v}{n}$ in the $n$th term of the integral, you get:

$$\int_0^{\infty} x^{s-1}e^{-nx}\,dx=\frac{1}{n^s}\int_0^\infty v^{s-1}e^{-v}dv = \frac{1}{n^s}\Gamma(s)$$

  • $\begingroup$ Okay. So. Now you sum over $n$ on both sides and you have a geometric series in the integrand? $\endgroup$ – Eleven-Eleven Jan 14 '15 at 0:17
  • $\begingroup$ Yes. My real analysis memory is somewhat eroded, but you'll need an argument to show why you can switch the sum and integral. $\endgroup$ – Thomas Andrews Jan 14 '15 at 2:46

$$\mathscr{L}\{t^k\}=\int_0^\infty t^ke^{-st} = \dfrac{k!}{s^{k+1}}$$ is the well-known laplace transform of $t^{k}$

then like above in the first answer, sum over $s$ and swap integration and summation


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