Test the following improper integral for convergence:

The integral converges.

*Proof.* We can compute the value of the improper integral directly. First, we can evaluate the indefinite integral using integration by parts with

Therefore,

So, to evaluate the improper integral we take the limit,

The substitution for will not give the limit 0, since logarithm is not defined there – it is an indeterminate form

I believe you can rewrite

can be rewritten as

then we can just spam l’Hôpital and find that the limit is, in fact, equal to 0.