Second type.

Definition 8.9.4   If $ f$ is continuous on the interval $ [a,b)$ with an infinite limit on the left at $ b$ , then

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

If $ f$ is continuous on the interval $ (a, b]$ with an infinite left on the right at $ a$ , then

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

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

Example 8.9.5   Let $ I= \int_0^1 \frac {1}{x^2} \; dx$ . We have:

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

Thus:

$\displaystyle I= \underset{ \lambda \underset{>}{\rightarrow} 0}{\text{lim}} \left( -1 + \frac {1}{\lambda } \right) = + \infty.$    

The given improper integral is divergent.

Example 8.9.6   Let $ I= \int_0^1 \frac {1}{\sqrt{x}} \; dx$ . We have:

$\displaystyle I( \lambda ) = \int_{ \lambda}^1 \frac {1}{\sqrt{x}} \; dx = \left[ 2 \sqrt{x} \right]_{\lambda }^1 =2 ( 1 - \sqrt{\lambda } ).$    

Thus:

$\displaystyle I=\underset{ \lambda \underset{>}{\rightarrow} 0}{\text{lim}} \; 2 ( 1 - \sqrt{\lambda } ) =2.$    

Definition 8.9.7   Suppose that the function $ f$ is continuous at all the points of $ [a,b]$ , but not at the interior point $ c$ . Moreover suppose that $ f$ has an infinite (at least one-sided) limit at $ c$ . Then:

$\displaystyle \int_a^b f(x) \; dx = \int_a^c f(x) \; dx + \int_c^b f(x) \; dx$    

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

Example 8.9.8   Let $ I= \int_{-1}^{2} \frac {1}{x^2} \; dx$ . By definition, we have:

$\displaystyle I=\int_{-1}^{0} \frac {1}{x^2} \; dx + \int_{0}^{2} \frac {1}{x^2} \; dx$    

We compute each of the terms:

$\displaystyle I_1(\lambda ) = \int_{-1}{0} \frac {1}{x^2} \; dx = int_{-1}^{\la...
... \; dx = \left[ \frac {-1}{x} \right]_{-1}^{\lambda } =\frac {-1}{\lambda } -1.$    

Thus:

$\displaystyle \int_{-1}{0} \frac {1}{x^2} \; dx =\underset{ \lambda \underset{<}{\rightarrow} 0}{\text{lim}} \left( \frac {-1}{\lambda } -1 \right)= + \infty.$    

By the same way we show that:

$\displaystyle \int_{0}{2} \frac {1}{x^2} \; dx =\underset{ \lambda \underset{>}...
...tarrow} 0}{\text{lim}} \left( \frac 12 +\frac {1}{\lambda } \right) = + \infty.$    

Therefore the given improper integral is divergent.

Noah Dana-Picard 2007-12-28