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- Jan 31, 2012

- 253

Show that for $0 \le \theta \le \pi$, $ \displaystyle \int_{0}^{\theta} \ln(\sin x) \ dx = - \theta \ln 2 - \frac{1}{2} \sum_{n=1}^{\infty} \frac{\sin (2n \theta)}{n^{2}}$.

Also show that for $0 \le \theta \le \frac{\pi}{2}$, $ \displaystyle \int_{0}^{\theta} \ln(\cos x) \ dx = - \theta \ln 2 + \frac{1}{2} \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sin (2n \theta)}{n^{2}}$.