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I have the function $$f(x)=\begin{cases}1, \space |x|\leq a \\ 0, \space a<|x|\leq 1/2\end{cases}$$ I have calculated the Fourier series of this even function: $$Sf(x)=2a+2\sum_{n=0}^\infty \frac{\sin(2\pi na)}{\pi n}\cos(2\pi nx)$$ Now I need to calculate $$\sum_{n=0}^\infty\frac{\sin^2(2\pi na)}{n^2}$$ How do I use Parserval's Identity to calculate this last sum? I have tried $$2a=\int_{-a}^a1^2=2a^2+\sum_{n=0}^\infty\frac{\sin^2(2\pi na)}{\pi^2 n^2}$$ And therefore $$(2a-2a^2)\pi^2=\sum_{n=0}^\infty\frac{\sin^2(2\pi na)}{n^2}$$ Is this correct? If not, what is the correct way to do it?

Rócherz
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codingnight
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    Those sums should probably run from $1$ to $\infty$; you don't want a literal $\frac00$ term in there, and the zero term $2a$ has a different formula anyway. – jmerry Jan 30 '19 at 00:17
  • The leading constant should be part of the first fourier coefficient to have the form $Sf = \sum_n c_n cos(2\pi nx)$ for which we can apply Parserval's identity. – Nicholas Parris Jan 30 '19 at 00:25
  • @NicholasParris How would you plug in the leading constant into the sum? I'm also interested in this problem – John Keeper Feb 08 '19 at 01:55

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