Editing Basel problem (section)

===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>