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The full problem asked: Let $\sigma$ be real, and not an integer. Find the complex form of the Fourier series for the $2\pi$ periodic function $F(x)=e^{-i\sigma x}$ on $[-\pi,\pi]$. Use this and Parseval's theorem to show that $$\csc^2(\pi\sigma)=\frac{1}{\pi^2}\sum_{n=-\infty}^{\infty}\frac{1}{(n+\sigma)^2}.$$

I found the Fourier series for $F$ to be $$\sum_{n=-\infty}^\infty\frac{\sin[\pi(n+\sigma)]}{\pi(n+\sigma)}e^{inx},$$ and using Parseval's theorem, I have that $$\frac{1}{2\pi}\int_{-\pi}^\pi|e^{-i\sigma x}|^2dx= \sum_{n=-\infty}^\infty\bigg|\frac{\sin[\pi(n+\sigma)]}{\pi(n+\sigma)}\bigg|^2,$$ but I don't see how to get from here to $\csc^2(\pi\sigma)\frac{1}{\pi^2}\sum_{n=-\infty}^{\infty}\frac{1}{(n+\sigma)^2}.$

Matt G
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1 Answers1

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Using the angle addition formula for the sine we have, for all integers $n$,

$$\sin[\pi(n+\sigma)] = \sin(\pi n + \pi \sigma) = \sin(\pi n) \cos(\pi \sigma) + \cos(\pi n) \sin(\pi \sigma) = (-1)^n \sin(\pi \sigma),$$

where the last step follows from the identities $\sin(\pi n) = 0$ and $\cos(\pi n) = (-1)^n$. Thus

$$\sum_{n = -\infty}^\infty \left\lvert \frac{\sin[\pi(n+\sigma)]}{\pi(n+\sigma)}\right\rvert^2 = \sum_{n = -\infty}^\infty \frac{\sin^2(\pi \sigma)}{\pi^2(n+\sigma)^2} = \frac{\sin^2(\pi\sigma)}{\pi^2}\sum_{n = -\infty}^\infty \frac{1}{(n+\sigma)^2}.$$

Since $\frac{1}{2\pi}\int_{-\pi}^\pi |e^{-i\sigma x}|^2\, dx = 1$, your equation reduces to

$$1 = \frac{\sin^2(\pi \sigma)}{\pi^2} \sum_{n = -\infty}^\infty \frac{1}{(n+\sigma)^2}.$$

Multiply both sides of this equation by $\csc^2(\pi \sigma)$ to get the result.

kobe
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