If is continuous on the interval with an infinite left on the right at , then

In both cases, if the limit exists and is finite, the improper integral is convergent. Otherwise, it is divergent.

Thus:

The given improper integral is divergent.

Thus:

where the two improper integrals are defined as in Def. 9.4.

We compute each of the terms:

Thus:

By the same way we show that:

Therefore the given improper integral is divergent.

Noah Dana-Picard 2007-12-28