Editing Basel problem (section)

==Another proof using Parseval's identity==

Given a [[complete orthonormal basis]] in the space <math>L^2_{\operatorname{per}}(0, 1)</math> of [[Lp space#Special cases|L2]] [[periodic function]]s over <math>(0, 1)</math> (i.e., the subspace of [[square-integrable function]]s which are also [[periodic function|periodic]]), denoted by <math>\{e_i\}_{i=-\infty}^{\infty}</math>, [[Parseval's identity]] tells us that
<math display=block>\|x\|^2 = \sum_{i=-\infty}^{\infty} |\langle e_i, x\rangle|^2, </math>

where <math>\|x\| := \sqrt{\langle x,x\rangle}</math> is defined in terms of the [[inner product]] on this [[Hilbert space]] given by
<math display=block>\langle f, g\rangle = \int_0^1 f(x) \overline{g(x)} \, dx,\ f,g \in L^2_{\operatorname{per}}(0, 1).</math>

We can consider the [[orthonormal basis]] on this space defined by <math>e_k \equiv e_k(\vartheta) := \exp(2\pi\imath k \vartheta)</math> such that <math>\langle e_k,e_j\rangle = \int_0^1 e^{2\pi\imath (k-j) \vartheta} \, d\vartheta = \delta_{k,j}</math>. Then if we take <math>f(\vartheta) := \vartheta</math>, we can compute both that
<math display=block>
\begin{align}
\|f\|^2 & = \int_0^1 \vartheta^2 \, d\vartheta = \frac{1}{3} \\
\langle f, e_k\rangle & = \int_0^1 \vartheta e^{-2\pi\imath k\vartheta} \, d\vartheta = \Biggl\{\begin{array}{ll} \frac{1}{2}, & k = 0 \\ -\frac{1}{2\pi\imath k} & k \neq 0, \end{array}
\end{align}
</math>

by elementary [[calculus]] and [[integration by parts]], respectively. Finally, by [[Parseval's identity]] stated in the form above, we obtain that
<math display=block>
\begin{align}
\|f\|^2 = \frac{1}{3} & = \sum_{\stackrel{k=-\infty}{k \neq 0}}^{\infty} \frac{1}{(2\pi k)^2}+ \frac{1}{4}
     = 2 \sum_{k=1}^{\infty} \frac{1}{(2\pi k)^2}+ \frac{1}{4} \\
     & \implies \frac{\pi^2}{6} = \frac{2 \pi^2}{3} - \frac{\pi^2}{2} = \zeta(2).
\end{align}
</math>

===Generalizations and recurrence relations===
Note that by considering higher-order powers of <math>f_j(\vartheta) := \vartheta^j \in L^2_{\operatorname{per}}(0, 1)</math> we can use [[integration by parts]] to extend this method to enumerating formulas for <math>\zeta(2j)</math> when <math>j > 1</math>. In particular, suppose we let
<math display=block>I_{j,k} := \int_0^1 \vartheta^j e^{-2\pi\imath k\vartheta} \, d\vartheta, </math>

so that [[integration by parts]] yields the [[recurrence relation]] that
<math display=block>
\begin{align}
I_{j,k} & = \begin{cases} \frac{1}{j+1}, & k=0; \\[4pt] -\frac{1}{2\pi\imath \cdot k} + \frac{j}{2\pi\imath \cdot k} I_{j-1,k}, & k \neq 0\end{cases} \\[6pt]
     & = \begin{cases} \frac{1}{j+1}, & k=0; \\[4pt] -\sum\limits_{m=1}^j \frac{j!}{(j+1-m)!} \cdot \frac{1}{(2\pi\imath \cdot k)^m}, & k \neq 0. \end{cases}
\end{align}
</math>

Then by applying [[Parseval's identity]] as we did for the first case above along with the linearity of the [[inner product]] yields that
<math display=block>
\begin{align}
\|f_j\|^2 = \frac{1}{2j+1} & = 2 \sum_{k \geq 1} I_{j,k} \bar{I}_{j,k} + \frac{1}{(j+1)^2} \\[6pt]
     & = 2 \sum_{m=1}^j \sum_{r=1}^j \frac{j!^2}{(j+1-m)! (j+1-r)!} \frac{(-1)^r}{\imath^{m+r}} \frac{\zeta(m+r)}{(2\pi)^{m+r}} + \frac{1}{(j+1)^2}.
\end{align}
</math>