First type.

Definition 8.9.1   If $ f$ is continuous on the interval $ [a,+\infty)$ , then

$\displaystyle \int_a^{+ \infty} f(x) \; dx = \underset{ \lambda \rightarrow + \infty}{\text{lim}} \int_a^{\lambda } f(x) \; dx.$    

If $ f$ is continuous on the interval $ ( -\infty , b]$ , then

$\displaystyle \int_{ \infty}^b f(x) \; dx = \underset{\lambda \rightarrow \infty}{\text{lim}} \int_{\lambda }^b f(x) \; dx.$    

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

Example 8.9.2   Let $ I=\int_1^{+\infty} \frac 1x \; dx$ .

We have:

$\displaystyle I( \lambda ) = \int_1^{\lambda } \frac 1x \; dx = \left[ \ln x \right]_1^{\lambda } = \ln \lambda - \ln 1 = \ln \lambda.$    

Thus:

$\displaystyle I = \underset{ \lambda \rightarrow + \infty}{\text{lim}} \ln \lambda = + \infty.$    

The given integral is divergent.

Example 8.9.3   Let $ I=\int_1^{+\infty} \frac {1}{x^2} \; dx$ .

We have:

$\displaystyle I( \lambda ) = \int_1^{\lambda }\frac {1}{x^2} \; dx = \left[ \frac {-1}{x} \right]_1^{\lambda } =\frac {-1}{\lambda } + 1.$    

Thus:

$\displaystyle I=\underset{ \lambda \rightarrow + \infty}{\text{lim}} \left( 1-\frac {-1}{\lambda } \right) =1.$    

The given improper integral is convergent.

Noah Dana-Picard 2007-12-28