Improper Integrals

Convergence and Divergence

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Improper integrals extend the idea of definite integration to cases where the interval is unbounded or the function becomes infinite at some point. They matter because many important quantities in physics, probability, and engineering are modeled by areas that stretch forever or blow up near a boundary. A normal definite integral cannot be applied directly in these cases, so calculus replaces it with a limit process. The key question is whether that limit exists and gives a finite value.

There are two main types of improper integrals. One type occurs on an infinite interval, such as $\int_1^{\infty} \frac{1}{x^2}\,dx$ , and the other occurs when the integrand has a vertical asymptote or discontinuity. In both situations, the integral is defined by taking a limit of ordinary definite integrals over safer intervals. If the limit exists, the improper integral converges, and if not, it diverges.

Key Facts

For an infinite upper limit, $\int_a^{\infty} f(x)\,dx = \lim_{b\to\infty} \int_a^b f(x)\,dx$ .
For an infinite lower limit, $\int_{-\infty}^b f(x)\,dx = \lim_{a\to -\infty} \int_a^b f(x)\,dx$ .
If $c$ is a discontinuity in $[a,b]$ , split the integral and take one-sided limits from each side of $c$ .
An improper integral converges only if the defining limit exists and is finite.
$p$ -test on $[1, \infty)$ : $\int_1^{\infty} \frac{1}{x^p}\,dx$ converges if $p > 1$ and diverges if $p \leq 1$ .
$p$ -test near $0$ : $\int_0^1 \frac{1}{x^p}\,dx$ converges if $p < 1$ and diverges if $p \geq 1$ .

Vocabulary

Improper integral: An integral defined by a limit because the interval is $\infty$ or the function is unbounded at some point.
Converges: Describes an improper integral whose defining limit exists and equals a finite number.
Diverges: Describes an improper integral whose defining limit does not exist or is $\infty$ .
Vertical asymptote: A vertical line where a function grows without bound and is not defined in the usual way.
Infinite interval: An interval that extends to infinity, such as [2, infinity) or (-infinity, 5].

Common Mistakes to Avoid

Treating $\infty$ like a regular endpoint, which is wrong because $\infty$ is not a number and the integral must be rewritten as a limit first.
Ignoring a discontinuity inside the interval, which is wrong because the integral must be split at the point where the function is undefined or unbounded.
Combining one-sided limits automatically, which is wrong because each side of a vertical asymptote must converge separately before adding them.
Assuming a small-looking graph means a finite area, which is wrong because a region can extend forever and still diverge depending on the rate of decay.

Practice Questions

1 Evaluate $\int_1^{\infty} \frac{1}{x^3} \,dx$ and state whether it converges or diverges.
2 Evaluate $\int_0^1 \frac{1}{\sqrt{x}} \,dx$ and state whether it converges or diverges.
3 Explain why $\int_{-1}^{1} \frac{1}{x}\,dx$ is not a convergent improper integral even though the areas on the left and right may seem to cancel.