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Differential EquationsSuperchallengeBoundary Values & Fourier Series

Boundary Values & Fourier Series · Superchallenge problem

Show that ln⁡ ⁣(2sin⁡x2)=−∑n≥1cos⁡nxn\ln\!\left(2\sin\dfrac x2\right) = -\displaystyle\sum_{n\ge1}\dfrac{\cos nx}{n} for 0<x<2π0 < x < 2\pi, and use Parseval's identity to evaluate ∫0πln⁡2 ⁣(2sin⁡x2)dx\displaystyle\int_0^{\pi}\ln^2\!\left(2\sin\dfrac x2\right)dx.