Finding a Definite Integral with Natural Log and Sine

Date: 05/14/2008 at 05:52:27
From: Mohammad
Subject: Integral LN(sin(x)) from 0 to pi

I'm trying to find the following integral:

  -
 |
 |  LN( SIN( x ) ) dx = ?
 |
-

From 0 to pi.

I can evaluate integral of LN(x) but here we have LN(SIN(x)).  I've
tried integration-by-parts but I ended up confusing myself!

Date: 05/19/2008 at 12:38:13
From: Doctor Who
Subject: Re: Integral LN(sin(x)) from 0 to pi

This definite integral can be evaluated by hand as follows:

First we note the symmetric property of sin x in the interval 
[0,pi].  Therefore 

    integral[x=0, x=pi] ln|sin x| dx 

  = 2 integral[x=0, x=pi/2] ln|sin x| dx

Further note that 

    integral[x=0, x=pi/2] ln|sin x| dx 

  = integral[x=0, x=pi/2] ln|sin(pi/2 - x)| dx

You can verify this by using a substitution 

  u = pi/2 - x 

and changing the limits.  Also, 

  sin(pi/2 - x) = cos x

So we have

    integral[x=0, x=pi] ln|sin x| dx 

  = integral[x=0, x=pi/2] ln|sin x| dx 

    + integral[x=0, x=pi/2] ln|cos x| dx

  = integral[x=0, x=pi/2] (ln|sin x| + ln|cos x|) dx

  = integral[x=0, x=pi/2] ln|sin x cos x| dx

  = integral[x=0, x=pi/2] ln|2 sin x cos x| dx - ln(2)x [x=0, x=pi/2]

  = integral[x=0, x=pi/2] ln|sin 2x| dx - pi ln(2)/2

Make the substitution 2x = t and 2 dx = dt; the upper limit becomes 
pi and the lower limit 0.

    integral[x=0, x=pi] ln|sin x| dx

  = integral[t=0, t=pi] (1/2)ln|sin t| dt - pi ln(2)/2

Replace t by x since they are dummy variables and then subtract 

  (1/2) integral[x=0, x=pi] ln|sin x| dx 

from both sides and multiply both sides by 2; we end up with

  integral[x=0, x=pi] ln|sin x| dx = - pi ln(2).

I hope this helps. 

- Doctor Who, The Math Forum
  http://mathforum.org/dr.math/