Test the following improper integral for convergence:
The integral converges if and diverges for .
Proof. We can compute this directly. We evaluate the indefinite integral (using the substitution ) for :
But then the limit
is finite for and diverges for (since as and so the limit diverges when and converges for ). Hence, the integral
converges for and diverges for . In the case that the indefinite integral is
and as , so the integral divers for as well. Therefore, we have the integral converges for and diverges for