Question: "How to prove this identityπ=∑k=−∞∞(sin⁡(k)k)2 ?"

leciazyg

Answered question

2022-02-27

How to prove this identity

π = \sum_{k = - \infty}^{\infty} {(\frac{\sin (k)}{k})}^{2}

Ulgelmorgs

Beginner2022-02-28Added 8 answers

Assume

a \in [0, \frac{π}{2}]

.
An integral

\int_{0}^{a} \frac{\sin (2 k x)}{k} d x = \int_{0}^{a} \frac{2 \sin (k x)}{k^{2}} d \sin (k x)

= \frac{\sin^{2} (k x)}{k^{2}}]_{0}^{a}

= \frac{\sin^{2} (k a)}{k^{2}}

(1)
and a sum

\sum_{k = 1}^{\infty} \frac{\sin (2 k x)}{k} = \sum_{k = 1}^{\infty} \frac{e^{i 2 k x} - e^{- i 2 k x}}{2 i k}

= \frac{1}{2 i} (- \log (1 - e^{i 2 x}) + \log (1 - e^{- i 2 x}))

= \frac{1}{2 i} \log (- e^{- 2 x})

= \frac{π}{2} - x for x \in (0, π)

(2)
Putting (1) and (2) together

\sum_{k = 1}^{\infty} \frac{\sin^{2} (k a)}{k^{2}} = \int_{0}^{a} (\frac{π}{2} - x) d x

= \frac{π}{2} a - \frac{a^{2}}{2}

If we take

\frac{\sin (k a)}{k a} = 1

when

k = 0

, we get the answer to the question using

a = 1

\sum_{k \in Z} {(\frac{\sin (k a)}{k a})}^{2} = \frac{π}{a}

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