- #1
polygamma
- 229
- 0
For a few of you, this probably isn't very challenging. But I'm going to post it anyways since I find it interesting.
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}}$.
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}}$.
- I'd like to propose the following one for you... Something to get you teeth into.For \(\displaystyle a,\, b,\, c > 0 \in \mathbb{R}\), \(\displaystyle m\in \mathbb{Z}^+\ge 1\), and \(\displaystyle 0 < \theta \le \pi/2\), find the additional conditions on the parameters as well as the closed form solution for:\(\displaystyle \int_0^{\theta}\log^m(a+b\cos x +c\sin x)\,dx\)