How to prove this identity? $$\pi=\sum_{k=-\infty}^{\infty}\left(\dfrac{\sin(k)}{k}\right)^{2}\;$$
I found the above interesting identity in the book $\bf \pi$ Unleashed.
Does anyone knows how to prove it?
Thanks.
Answer
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
$$c_k = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: f(x) e^{i k x} = \frac{\sin{k}}{k}$$
$$\sum_{k=-\infty}^{\infty} \frac{\sin^2{k}}{k^2} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: |f(x)|^2 = \frac{1}{2 \pi} \int_{-1}^{1} dx \: \pi^2 = \pi $$
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
$$f(x) = \begin{cases} \pi & |x|a \end{cases}$$
No comments:
Post a Comment