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

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

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

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

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

Then:

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

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

For $p=1$ , we have:

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

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

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

is convergent because:

• For any $x \geq 1$ , we have $0 \leq k(x) \leq \frac {1}{x^3}$ .
• The improper integral $\int_1^{+ \infty } \frac {dx}{x^3}$ is convergent (by Prop. 9.9).

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

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

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

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

is convergent because:

• the improper integral $\int_1^{+ \infty } \frac {1}{e^x} \; dx$ is convergent (please check it directly!)
• $\underset{x \rightarrow + \infty }{\text{lim}} \frac {\frac {1}{e^x}}{\quad \frac {2}{e^x+3} \quad } = \frac 12$ .