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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!

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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/… –  Emmad Kareem 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

2 Answers 2

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 $I=-\frac12\pi\log2$.

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+1, good hint, specially that the limits are accurate :) –  Emmad Kareem 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
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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.

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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

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