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math.stackexchange.com/questions/331404/how-t … ac-sin/582932#582932
How to prove this identity?
$$\pi=\sum_{k=-\infty}^{\infty}\left(\dfrac{\sin(k)}{k}\right)^{2}$$
Find a function whose Fourier coefficients are $\sin{k}/k$. Then evaluate the integral of the square of that function.
To wit, let
$$f(x) = \begin{cases} \pi & |x|<1\\0&|x|>1 \end{cases}$$
Then, if
$$f(x) = \sum_{k=-\infty}^{\infty} c_k e^{-i k x}$$
then
$$c_k = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: f(x) e^{i k x} = \frac{\sin{k}}{k}$$
By Parseval's Theorem:
$$\sum_{k=-\infty}^{\infty} \frac{\sin^2{k}}{k^2} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: |f(x)|^2 = \frac{1}{2 \pi} \int_{-1}^{1} dx \: \pi^2 = \pi $$
ADDENDUM
This result is easily generalizable to
$$\sum_{k=-\infty}^{\infty} \frac{\sin^2{a k}}{k^2} = \pi a$$
where $a \in[0,\pi)$, using the function
$$f(x) = \begin{cases} \pi & |x|< a\\0&|x|>a \end{cases}$$
Then, if
$$f(x) = \sum_{k=-\infty}^{\infty} c_k e^{-i k x}$$
then
$$c_k = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: f(x) e^{i k x} = \frac{\sin{ak}}{k}$$
By Parseval's Theorem:
$$\sum_{k=-\infty}^{\infty} \frac{\sin^2{ak}}{k^2} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: |f(x)|^2 = \frac{1}{2 \pi} \int_{-a}^a dx \: \pi^2 = \pi a$$
Corollary
$$\sum_{n = 1}^{\infty}{\sin n \over n} = \frac12\left(\sum_{n = -\infty}^{\infty}{\sin n \over n} - 1\right)$$
yields
$$\sum_{n = 1}^{\infty}{\sin n \over n} = \frac12(\pi - 1)$$ |
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