Understanding $\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$

Question

In the answer, of this question Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$?, given by Christian Blatter,

• Why $|c_{\pm k}|^2={1\over4}a_k^2={4\over k^4}$ $\ (k\geq1)$ ?

According to my calculations $|c_{\pm k}|^2=\vert1/2\pi\int_{-\pi}^\pi t^2e^{-i(\pm k)x}dx\vert^2=\vert t^2\sin(\pi(\pm k))/\pi(\pm k)\vert^2=0$

• And how did he get $\frac{\pi^4}{90}?$

Consider the function $f(t):=t^2\ \ (-\pi\leq t\leq \pi)$, extended to all of ${\mathbb R}$ periodically with period $2\pi$. Developping $f$ into a Fourier series we get $$t^2 ={\pi^2\over3}+\sum_{k=1}^\infty {4(-1)^k\over k^2}\cos(kt)\qquad(-\pi\leq t\leq \pi).$$ If we put $t:=\pi$ here we easily find $\zeta(2)={\pi^2\over6}$. For $\zeta(4)$ we use Parseval's formula $$\|f\|^2=\sum_{k=-\infty}^\infty |c_k|^2\ .$$ Here $$\|f\|^2={1\over2\pi}\int_{-\pi}^\pi t^4\>dt={\pi^4\over5}$$ and the $c_k$ are the complex Fourier coefficients of $f$. Therefore $c_0={\pi^2\over3}$ and $|c_{\pm k}|^2={1\over4}a_k^2={4\over k^4}$ $\ (k\geq1)$. Putting it all together gives $\zeta(4)={\pi^4\over 90}$.

Answer 1

With the inner product, $\langle a , b \rangle = {1 \over 2 \pi} \int_{-\pi}^\pi \overline{a(t)} b(t) dt $, the complex Fourier series for $f$ is given by $\hat{f_k} = {1 \over 2 \pi} \int_{-\pi}^\pi e^{-ikt} f(t) dt $.

The function is reconstituted (in a $L^2$ sense) by $f(t) = \sum_{k \in \mathbb{Z}}\hat{f_k} e^{ikt}$.

Parseval's theorem gives $\|f\|^2_2 = { 1\over 2 \pi}\int_{-\pi}^\pi |f(t)|^2 dt = \sum_{k \in \mathbb{Z}}|\hat{f_k}|^2$.

In this case, $f(t) = t^2$, hence $\|f\|^2_2 = {\pi^4 \over 5}$, and the complex Fourier series given by $\hat{f_0} = {\pi ^2 \over 3}$ and for $k \neq 0$, we have $\hat{f_k}={1 \over 2 \pi} \int_{-\pi}^\pi t^2 \cos (kt) dt = (-1)^k {2 \over k^2}$.

Then ${\pi^4 \over 5} = \sum_{k <0} {4 \over k^4} + {\pi^4 \over 9} + \sum_{k >0} {4 \over k^4} = {\pi^4 \over 9} + 2 \sum_{k >0} {4 \over k^4}$, which simplifies to the desired result.