Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I do not have much experience evaluating improper integrals and I hope someone will please demonstrate how to evaluate this:

$$\int_{0}^{\frac{\pi}{2}} \log \sin x dx$$

Thanks in advance! P.S. :I had accidentally put the word indefinite instead of improper.Sorry for the error!

share|cite|improve this question
3  
This is a definite integral. – Daan Michiels Oct 25 '12 at 11:21
    
WolframAlpha suggests a numerical approximation: bit.ly/ThKVai – Siminore Oct 25 '12 at 11:23
    
Very similar to: goiit.com/posts/list/… – NoChance Oct 25 '12 at 11:33
    
@Emmad ...Where already the first line is wrong since over there the integral is from 0 to pi/4, not to pi/2. :-) – Did Oct 25 '12 at 12:03
    
I am sorry.I meant an improper integral. – user43081 Oct 25 '12 at 12:19
up vote 4 down vote accepted

Hint: (...Since this has been answered several times on the site.) Let $I$ denote the integral to be evaluated, then $$ 2\cdot I=\int_0^{\pi/2}\log\sin x\,\mathrm dx+\int_0^{\pi/2}\log\cos x\,\mathrm dx=\int_0^{\pi/2}\log\left(\tfrac12\sin(2x)\right)\,\mathrm dx=\ldots$$ To check your solution: In the end, you should reach the value $I=-\frac12\pi\log2$.

share|cite|improve this answer
1  
+1, good hint, specially that the limits are accurate :) – NoChance Oct 25 '12 at 12:13
    
@did, thanks. My text had used some intimidating language about this problem calling it an improper integral. So, i thought it better to post it here.(It was referring to "improper integrals treated as definite integrals".Can you please tell me what it was trying to communicate?) – user43081 Oct 27 '12 at 12:29
1  
Try the explanation here. – Did Oct 27 '12 at 12:34
    
@did, thanks a lot.I now get it. – user43081 Oct 27 '12 at 14:03
    
Great. You are welcome. – Did Oct 27 '12 at 14:06

I think first you have to use integration by parts... This is a hint.

share|cite|improve this answer
3  
Not a very useful hint --- integration by parts gets you to $x\cot x$, and how do you propose to integrate that? – Gerry Myerson Oct 25 '12 at 12:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.