Improper Integrals
Evaluate improper integrals with infinite bounds or discontinuous integrands using limits. Determine convergence or divergence and apply the comparison test. Connect to the p-series and geometric series from sequences and series.
A lesson of Calculus, in Advanced Integration Techniques, about 10 minutes.
By the end of this lesson, you will be able to evaluate improper integrals (infinite limits and unbounded integrands) using limits, determine convergence or divergence, and apply the Comparison Test.
Types of Improper Integrals
Type 1: Infinite limits of integration
$$\int_a^{\infty}f(x),dx = \lim_{t\to\infty}\int_a^t f(x),dx$$
$$\int_{-\infty}^b f(x),dx = \lim_{t\to -\infty}\int_t^b f(x),dx$$
$$\int_{-\infty}^{\infty}f(x),dx = \int_{-\infty}^c f(x),dx + \int_c^{\infty}f(x),dx$$
Type 2: Unbounded integrand (vertical asymptote within the interval)
$$\int_a^b f(x),dx = \lim_{t\to b^-}\int_a^t f(x),dx \quad \text{(if } f \to \pm\infty \text{ as } x\to b\text{)}$$
Convergent: the limit is a finite number.
Divergent: the limit is $\pm\infty$ or doesn't exist.
Example 1 — Type 1 Convergent
$$\int_1^{\infty}\frac{1}{x^2},dx = \lim_{t\to\infty}\left[-\frac{1}{x}\right]1^t = \lim{t\to\infty}\left(-\frac{1}{t}+1\right) = 1$$
Converges to 1.
Example 2 — Type 1 Divergent
$$\int_1^{\infty}\frac{1}{x},dx = \lim_{t\to\infty}[\ln x]1^t = \lim{t\to\infty}\ln t = +\infty$$
Diverges. The harmonic integral diverges.
The $p$-Test
$$\int_1^{\infty}\frac{1}{x^p},dx = \begin{cases} \frac{1}{p-1} & p > 1 \text{ (converges)} \ \text{diverges} & p \leq 1 \end{cases}$$
This simple test classifies all $1/x^p$ integrals.
Example 3 — Type 2 (Unbounded Integrand)
$$\int_0^1\frac{1}{\sqrt{x}},dx = \lim_{t\to 0^+}\int_t^1 x^{-1/2},dx = \lim_{t\to 0^+}[2\sqrt{x}]t^1 = \lim{t\to 0^+}(2-2\sqrt{t}) = 2$$
Converges to 2. Despite the singularity at $x = 0$, the area is finite.
Example 4 — Divergent Type 2
$$\int_0^1\frac{1}{x},dx = \lim_{t\to 0^+}[\ln x]t^1 = \lim{t\to 0^+}(0-\ln t) = +\infty$$
Diverges.
Example 5 — Double-Sided
$$\int_{-\infty}^{\infty}e^{-x^2},dx = 2\int_0^{\infty}e^{-x^2},dx = \sqrt{\pi}$$
(This famous result requires advanced techniques; we state it without proof.)