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What are the complex fourier coefficients of the function $f:\mathbb{R}\rightarrow \mathbb{R}$ defined by the $2\pi-periodic$ continuation of $f(x)=\pi-x$ , for x $0\le x < 2\pi$ ?

And how can one use that fact together with the Bessel function to show that $\sum \frac{1}{k^2} = \frac{\pi^2}{6}$?

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    @ cassandrao It says to conclude from the bessel equality that : $\sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi ^2}{6}$2012-11-28

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The fourier coefficients for nonzero $n$ are

$\begin{eqnarray} \hat f(n) &=& \frac{1}{2 \pi} \int_0^{2 \pi} (\pi-t) e^{-i n t} dt \\ &=& \frac{1}{2} \int_0^{2 \pi} e^{-i n t} dt - \frac{1}{2 \pi} \int_0^{2 \pi} t e^{-i n t} dt \\ &=& - \frac{1}{2 \pi} \left(\left[t \frac{e^{- i n t}}{- i n}\right]_0^{2 \pi} - \int_0^{2 \pi} \frac{e^{- i n t}}{- i n} dt \right) \\ &=& \frac{1}{-in} \\ \end{eqnarray}$

and $\hat f(0) = 0$.

Thus applying Parseval's identity we have

$\sum_{n=-\infty,n\neq 0}^\infty \frac{1}{n^2} = \frac{1}{2\pi}\int_{-\pi}^\pi (\pi - t)^2 \, dt = \frac{1}{2\pi}\int_{0}^{2\pi} t^2 \, dt = \frac{\pi^{2}}{3}$

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    Just did it for you, @cassandrao2012-11-29