Can you please show me how to solve the following integral, I'll use
words where I can't use math symbols.
The integral is the sum over the zero to pi/2 interval of the follwing
expression: Log(sin x) dx.
I know the answer: - (pi/2) * Log 2.
But what I don't know is:
1.  How to establish the integral is indeed definite.  For x = 0, sin
x = 0 and Log 0 isn't definite.  That's the hardest part, if I
remember correctly.
2.  The juggling of variable substitutions, that's only allowed after
determining the integral is definite.

Thanks in advance and feel free to contact me at gassee@gmail.com if
my limited command of math vocabulary in English makes the above
unclear.

JLG

This is an improper integral because log(sin x) is only integrable on
(0, pi/2] not [0, pi/2]. To compute an improper integral, replace the
nonintegrable point with a constant, then take the limit as that
constant approaches the nonintegrable point.

In this case, take the limit as c approaches 0, of the integral from c
to pi/2 of log(sin x).

To compute the integral, use integration by parts. Take the derivative
of log(sin x) and the integral of 1dx. What you get then should be
manageable.

Some links:
Improper Integrals
http://www.sosmath.com/calculus/improper/intro/intro.html

Integration by parts
http://www.sosmath.com/calculus/integration/byparts/byparts.html

This is a difficult integral to evaluate and needs complex analysis techniques. 
The integral is dicussed in "Complex Analysis - Ahlfors" where you
will find the method as well as the answer.

If you don't feel like opening Ahlfors and want to get a sense of why
this integral is finite note that

Integral 0 to delta ln(sin x) dx goes as delta*ln(sin (delta)) which
goes to zero as delta goes to zero.

If you buy into my "delta*ln(sin delta)" argument in previous comment
then the integral can be evaluated by formal manipulations as follows
without invoking complex analysis

Using x=pi/2-t we can show for integration over (0, pi/2)
I = Int( ln (sin x) dx) = Int (ln (cos x) dx) Now adding these

2I = Int{0 to pi/2}( ln sin 2x dx)  - ln 2 * pi/2

put 2x =t to get

2I = Int{0 to pi} (ln (sin t)dt/2) - ln 2 *pi/2

Split the first integral from 0 to pi/2 and pi/2 to pi
2I = 1/2 I + 1/2 Int{pi/2 to pi} (ln(sin t) dt) - ln 2* pi/2

Choosing t = pi/2+h you will get

2I = 1/2I + 1/2 I - ln 2 * pi/2
I = - pi/2*ln 2

It seems you don't understand the basics of integration very well...
you asked, for example, " How to establish the integral is indeed
definite. "   It is a definite integral because it is an integral of a
function over a specific integral.  The actual function being
integrated has nothing to do with whether or not it is a definite
integral.

This definite integral is _improper_ because Lim x->0 Log(Sin(x)) does
not exist (it has a vertical asymptote at x = 0).

Another poster suggested integration by parts - this will only get you
another integral which is no easier to evaluate.

Write log(sin(x)) = log(sqrt(1-cos(x)^2)) = 1/2 log(1-cos(x)^2) and
use the series expansion for the logarithm, log(1-u) = -u -u^2/2 -
u^3/3 ..., valid for -1 < u < 1. Then determine the integrals
int_0^(pi/2) cos(x)^(2n) dx and sum the resulting infinite series.