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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?

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    $\begingroup$ If you want to use Riemann zeta function, then you need to prove every single result about Riemann zeta. Sorry, that's how things work in maths. $\endgroup$ – Wojowu Apr 1 '16 at 20:25
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    $\begingroup$ For solving the Basel problem Euler compared the taylor series for the sine function with the product representation. You can also have a look at this resource: youtube.com/…. $\endgroup$ – MrYouMath Apr 1 '16 at 20:25
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    $\begingroup$ Yes solving the Basel Problem is simply evaluating $\zeta (2)$, but evaluating a well defined function isn't always easy, for instance $\zeta (3)$ is still sort of a mystery (Apéry's constant). One of course need more tools beyond the definition to do explicit computations. Also, technically the Riemann Zeta function is the analytic continuation of this Dirichlet series, not the series itself. $\endgroup$ – Qidi Apr 1 '16 at 20:30
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    $\begingroup$ You don't use the Riemann Zeta function to solve the Basil problem, instead the Basil problem solves $\zeta(2)$. The Riemann Zeta function is a bit too unwieldy. $\endgroup$ – Kaynex Apr 1 '16 at 20:41
  • $\begingroup$ So if you need to solve something like the Basel Problem for every value of $\zeta (s)$, what is its use? $\endgroup$ – Guacho Perez Apr 1 '16 at 20:48
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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 $$\zeta(2k) = - \frac{(2\pi i)^{2k} B_{2k}}{2 * (2k)!}.$$

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    $\begingroup$ so you are saying I'm wrong we can directly evaluate $\Gamma(2) \zeta(2)$ without the functional relation ? please give some details I'm not convinced $\endgroup$ – reuns Apr 1 '16 at 21:02
  • $\begingroup$ Of course you can. The values of $\zeta(2k)$ have been known much longer than the functional equation. Section 6 in the lecture notes people.reed.edu/~jerry/311/lec08.pdf follows essentially the same idea - although you can apply complex methods directly to $\frac{z^{-2k+1}}{e^z - 1}$ rather than going through $\cot(z).$ $\endgroup$ – user327837 Apr 1 '16 at 21:13
  • $\begingroup$ I know that you can use $\cot(z)$ (or the product formula for $\sin(z)$, that's the same and that's the original Euler's proof of $\zeta(2) = \pi^2/6$), that's not what I asked. if you use contour integration on $f(z) = \frac{z^{-2k+1}}{e^z-1} $ you'll have to use next the functional equation... $\endgroup$ – reuns Apr 1 '16 at 21:21
  • $\begingroup$ and when you integrate $z^{s-1}/(e^z-1)$ on some contour including all the poles at $2 i \pi k$, with proper regularization and limits, you get in fact the functional relation, hence we always go back to the functional relation $\endgroup$ – reuns Apr 1 '16 at 21:28
  • $\begingroup$ @user1952009 OK, I worked out the details. $\endgroup$ – user327837 Apr 1 '16 at 21:42
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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 $$\zeta(2) = \frac{\pi^2}{6}$$

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