Convergence theorems.

First we give a very important improper integral, which will be used together with the tests for convergence.

Proposition 8.9.9   The improper integral

$\displaystyle \int_1^{+\infty } \frac {dx}{x^p}$    

is convergent for $ p>1$ and is divergent for $ p \leq 1$ .

Proof. For $ p \neq -1$ , we have:

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

Then:

$\displaystyle \int_1^{+\infty } \frac {dx}{x^p} = \underset{ \lambda \rightarro...
...y}{\text{lim}} \left( \frac {1}{-p+1} \lambda^{-p+1} - \frac {1}{-p+1} \right).$    

This limit is infinite for $ p \geq 1$ , and is equal to 1 for $ p <1$ .

For $ p=1$ , we have:

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

This integral has infinite limit for $ \lambda$ in a neighborhood of $ +\infty$ , i.e. the given improper integral diverges. $ \qedsymbol$

Proposition 8.9.10 (Direct Comparison)   Let $ f$ and $ g$ be two functions, continuous on the interval $ [a,+\infty)$ . Suppose that for every $ x \in [a, + \infty )$ , they verify $ 0 \leq f(x) \leq g(x)$ . Then:
  1. If $ \int_a^{+ \infty } g(x) \; dx$ is convergent, then $ \int_a^{+ \infty } f(x) \; dx$ is convergent.
  2. If $ \int_a^{+ \infty } f(x) \; dx$ is divergent, then $ \int_a^{+ \infty } g(x) \; dx$ is divergent.

Example 8.9.11   Let $ k(x)= \frac {cos^2 (3x+1)}{x^3}$ , for $ x \geq 1$ . The improper integral

$\displaystyle \int_1^{+ \infty } \frac {cos^2 (3x+1)}{x^3} \; dx$    

is convergent because:

Proposition 8.9.12 (Limit Comparison)   Let $ f$ and $ g$ be two functions, continuous and positive on the interval $ [a,+\infty)$ . Suppose that there exists a real positive number $ l$ such that

$\displaystyle \underset{x \rightarrow + \infty }{\text{lim}} \frac {f(x)}{g(x)} =l.$    

Then the improper integrals $ \int_a^{+ \infty } f(x) \; dx$ and $ \int_a^{+ \infty } g(x) \; dx$ are either both convergent or both divergent.

Example 8.9.13   The improper integral

$\displaystyle \int_1^{+ \infty } \frac {2}{e^x+3} \; dx$    

is convergent because:

Noah Dana-Picard 2007-12-28