Evaluate $\int_{0}^{\pi} x \ln(\sin x)dx.$

Asked May 22 '17 at 10:13

Active May 22 '17 at 10:42

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Evaluate $$\int_{0}^{\pi} x \ln(\sin x)dx.$$

My attempt,

My first glance on this question I know that I'm going to use Integration by Parts.

So, it equals to $$-\frac{1}{2}\int_{0}^{\pi} x^2 \cot x dx$$

I've tried every methods that I've learnt for integration for this and I ended up in nowhere. Hope someone could provide a solution and explain for it. Thanks in advance.

Updated

Obviously, my attempt is wrong as the integral does not converge.

edited May 22 '17 at 10:42

asked May 22 '17 at 10:13

Mathxx

7,570

1

I am expecting a problem since close to $x=\pi$, $$x^2\cot(x)=\frac{\pi ^2}{x-\pi }+2 \pi +\left(1-\frac{\pi ^2}{3}\right) (x-\pi )+O\left((x-\pi )^2\right)$$ – Claude Leibovici May 22 '17 at 10:21
I checked with WA for the original integral, it equals to -3.42054. – Mathxx May 22 '17 at 10:23
Yes, and the definite integral you have after the integration by parts does not converge. – projectilemotion May 22 '17 at 10:25
@Mathxx. Which is $-\frac{1}{2} \pi ^2 \log (2)$ but the result is not important. How to compute this integral ? The antiderivative of the initial function is just a nightmare. – Claude Leibovici May 22 '17 at 10:29
Hoping there's someone who could solve this. @ClaudeLeibovici – Mathxx May 22 '17 at 10:30
3

Parts isn't the way to go here. Not only does the second integral diverge, but the "uv" part also diverges. – B. Goddard May 22 '17 at 10:31
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I totally support projectilemotion's suggestion. – Claude Leibovici May 22 '17 at 10:35
Edited with your suggestion. – Mathxx May 22 '17 at 10:37
Hint : $\log\sin(x) = \log\sin(\pi-x)$ and $\int_0^{\pi} \log\sin(x) dx = -\pi\log 2$ (see this answer) – achille hui May 22 '17 at 10:41

Evaluate $\int_{0}^{\pi} x \ln(\sin x)dx.$

0 Answers0