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.