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

The initial statement is opposite the conclusion of the proof.