Improper Integrals

By the end of this lesson, you will be able to evaluate improper integrals with infinite limits and discontinuous integrands, and determine convergence or divergence.

Two Types of Improper Integrals

Type I — Infinite limits: $$\int_a^\infty f(x), dx = \lim_{b\to\infty}\int_a^b f(x), dx$$

$$\int_{-\infty}^b f(x), dx = \lim_{a\to-\infty}\int_a^b f(x), dx$$

$$\int_{-\infty}^\infty f(x), dx = \int_{-\infty}^c f(x), dx + \int_c^\infty f(x), dx$$

Type II — Discontinuous integrand: If $f$ has a discontinuity at $b$: $\int_a^b f(x), dx = \lim_{t\to b^-}\int_a^t f(x), dx$

If discontinuity at interior point $c$: split there.

Convergence and Divergence

The integral converges if the limit exists (is a finite number). It diverges if the limit is $\pm\infty$ or does not exist.

The p-Integral: A Critical Reference

$$\int_1^\infty \frac{1}{x^p}, dx = \begin{cases} \frac{1}{p-1} & \text{if } p > 1 \text{ (converges)}\ +\infty & \text{if } p \leq 1 \text{ (diverges)}\end{cases}$$

Proof for $p \neq 1$: $\int_1^b x^{-p}, dx = \left[\frac{x^{1-p}}{1-p}\right]_1^b = \frac{b^{1-p}-1}{1-p}$

If $p > 1$: $1-p < 0$, so $b^{1-p}\to 0$ as $b\to\infty$ → converges to $\frac{1}{p-1}$.

If $p < 1$: $1-p > 0$, so $b^{1-p}\to\infty$ → diverges.

If $p = 1$: $\int_1^\infty \frac{dx}{x} = [\ln x]_1^\infty = \infty$ → diverges.

Examples

Example 1: $\int_1^\infty e^{-x}, dx = \lim_{b\to\infty}[-e^{-x}]1^b = \lim{b\to\infty}(-e^{-b}+e^{-1}) = e^{-1}$ Converges to $1/e$.

Example 2: $\int_0^1 \frac{1}{\sqrt{x}}, dx = \lim_{t\to 0^+}[2\sqrt{x}]t^1 = \lim{t\to 0^+}(2-2\sqrt{t}) = 2$ Converges to 2.

Example 3: $\int_1^\infty \frac{1}{x}, dx = \lim_{b\to\infty}[\ln x]_1^b = \infty$ Diverges.

Example 4: $\int_{-\infty}^\infty e^{-x^2}, dx = \sqrt{\pi}$ (famous Gaussian integral — converges).

Comparison Test for Improper Integrals

If $0 \leq f(x) \leq g(x)$ for $x \geq a$:

• If $\int_a^\infty g, dx$ converges, so does $\int_a^\infty f, dx$