Hint on how to prove $\zeta ( 2) =\pi ^{2}/6$ using the complex Fourier series of $f(x)=x$

Question

I know how to prove $\zeta (2)=\pi ^{2}/6$ by using the trigonometric Fourier series expansion of $x^{2}/4$. How can one prove the same result using the complex Fourier series of $f(x)=x$ for $0\leq x\leq 1$? Any suggestion?

Answer 1

Use the definition:

Say $f$ is defined on $[-\pi, \pi]$.

If $f(z) = \sum_{-\infty}^{\infty} {c_{n} e^{inz}}$

then

$c_{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}{f(z)e^{-inz}} dz$

If you put $f(z) = z$, can you work out what $c_{n}$ turns out to be?

To integrate, you can try integration by parts.

Answer 2

This is not related but you would like to see this article: A Short Proof of ζ (2) = π2/6 T.H. Marshall American Math monthly April 2010.

Answer 3

Extending off from Aryabhatta answer:

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For our situation: We have $f(x)=x ~~~{\text{ for }} 0\leq x\leq 1$

$2L=1,\Rightarrow L=\frac{1}{2}$

So restating we have:

$f(x) = \displaystyle\sum_{n=-\infty}^{\infty} {c_{n} e^{inx}}, \text{ where }c_{n} = \displaystyle\frac{1}{2\pi}\int_{-\frac{1}{2}}^{\frac{1}{2}}{f(x)e^{-inx}} ~\mathrm{d}x,~~~~~~n=0,~\pm 1,~\pm 2, \cdots~ $

$ \Rightarrow~~ c_{n} = \displaystyle\frac{1}{2\pi}\int_{-\frac{1}{2}}^{\frac{1}{2}}{xe^{-inx}}~\mathrm{d}x $

After integrating the complex Fourier coefficient we see that we get the following:

$\Rightarrow~~~~\displaystyle c_n=i\left(\frac{\cos(\frac{n}{2})}{2\pi n}-\frac{\sin(\frac{n}{2})}{\pi n^2}\right),~~~\text{for }n \in \mathbb{R}$

Lastly plugging back $c_n$ into $f(x)$ we then get our desired result for $n=0,~\pm 1,~\pm 2, \cdots~$.

Please update if you see any mistakes with any of the work. It has been quite some time since I work with Fourier Series and went off from my head. Feel free to edit mistakes as necessary if willing.

Thanks.