Log(sinx ) is convergent
and limit is o to π/2

Let , so that and . Then



Integrate by parts, setting




so the integral is equivalent to




The first term vanishes, since



Now for comparison, consider the function . We have that the integral of this function over the same interval is convergent:



so if we could show that at least for , then we're done. Replacing with (which means ), the inequality can be written as



This is certainly true, because as , both sides approach 0, while as , you have the right side approach while the left side approaches . You can compare these two functions' derivatives to see that the left side is monotonic over the interval.

So we've established that



so the first integral must converge, and so must



to some value between -2 and 0.