How do you use the Riemann Zeta Function?

Question

I know that the Riemann Zeta Function is defined as: $$\zeta (s)=\sum_{n=1}^\infty \frac {1}{n^s}=\frac {1}{\Gamma (s)} \int _0^{\infty}\frac { x^{s-1}}{e^x-1} dx$$ Which I think would prove useful for solving the Basel Problem without having to use Euler's tricks and just evaluating the integral. However, when I try to evaluate the integral, I end up with a polylogarithm function, which is defined precisely as an infinite sum of inverse powers. So how can the Riemann Zeta function be used to solve the problem without having to prove every single result?

Answer 1

let me propose a sketch of proof :

for solving the Basel problem from $$\Gamma(s)\zeta(s) = \int_0^\infty \frac{x^{s-1}}{e^x-1} dx\qquad\qquad\qquad(Re(s) > 1)$$

you can't directly compute $$\zeta(2) = \frac{1}{\Gamma(2)}\int_0^\infty \frac{x}{e^x-1} dx$$

but you'll have to prove first that $$\frac{x}{e^x-1} = \sum_{k=0}^\infty \frac{B_k}{k!} x^k$$

for $|x| < 2\pi$, with $B_k$ the Bernouilli numbers

hence that

$$\int_0^1 x^{s-2}\left(\frac{x}{e^x-1}-\sum_{k=0}^{K} \frac{B_k}{k!} x^k\right) dx$$

is holomorphic for $Re(s) > -K$, hence that when $s \to -K$ :

$$\Gamma(s) \zeta(s) \sim \frac{B_{k+1}}{s+K}$$

which from $\Gamma(s+K) \sim \frac{(-1)^K}{(s+K)K!}$ tells that $$\zeta(-K) = (-1)^K \frac{B_{K+1}}{K+1}$$

then you'll have to prove the functional equation

$$\zeta(s) = 2^{s-1} \pi^s \sin(\pi s/2) \Gamma(1-s) \zeta(1-s)$$

for example by showing that for every $Re(a) > 0$ :

$$F(s,a) = \int_0^\infty \frac{(a x)^{s-1}}{e^{a x}-1} d(ax)= \Gamma(s) \zeta(s)$$

(a change of contour, using the Cauchy integral formula)

an show that $$\lim_{a \to i} \frac{F(s,a)+F(s,-a)}{2} = 2^{s-1} \pi^{s+1} \frac{\sin(\pi s/2)}{\sin(\pi s)} \zeta(1-s)$$

showing that $$\zeta(2) = \frac{\pi^2}{6}$$

Answer 2

One of the quickest ways to evaluate $\zeta(2k)$ at even natural numbers involves integrating the function $$\frac{z^{-2k}}{e^z - 1}$$ over a large rectangular contour (not the integral $(0,\infty)$, though).

The residue of $\frac{z^{-2k}}{e^z - 1}$ at $2\pi i n$, $n \in \mathbb{Z} \backslash \{0\}$, is $$\lim_{z \rightarrow 2\pi i n} \frac{(z - 2\pi i n) z^{-2k}}{e^z - 1} = (2\pi i n)^{-2k}.$$

The residue at $0$ follows from the Taylor expansion $$\frac{z}{e^z - 1} = \sum_{n=0}^{\infty} \frac{B_n}{n!} z^n:$$ $$\mathrm{Res}_0 \Big( \frac{z^{-2k}}{e^z - 1} \Big) = \frac{B_{2k}}{(2k)!}.$$

Integrating over the rectangular contour with vertices at $\pm (2R+1)/2 \pm (2R+1)i/2$, $R \in \mathbb{N}$ will give you zero in the limit as $R \rightarrow \infty$, so the residue theorem implies $$2\pi i \Big( \sum_{n \ne 0} (2\pi i n)^{-2k} + \frac{B_{2k}}{(2k)!}\Big) = 0.$$ This implies $$\zeta(2k) = - \frac{(2\pi i)^{2k} B_{2k}}{2 * (2k)!}.$$