01 / 20
\(\int\)
\(\infty\)

Evaluating
Improper Integrals

Master integrals with infinite limits and discontinuous integrands: convergence, divergence, and the limit techniques that connect them.

AP Calculus BC

What Is an Improper Integral?

A definite integral is called improper when it violates one of the standard assumptions of the Riemann integral: either the interval of integration is infinite, or the integrand has an infinite discontinuity on the interval.

Type I
At least one limit of integration is \(\pm\infty\). Example: \(\displaystyle\int_1^{\infty}\frac{1}{x^2}\,dx\)
Type II
The integrand has an infinite discontinuity on \([a,b]\). Example: \(\displaystyle\int_0^{1}\frac{1}{\sqrt{x}}\,dx\)

Type I: Infinite Limits

x y 1
\(y = \frac{1}{x^2}\)
→ \(\infty\)

Type II: Infinite Discontinuities

x y 0 1 asymptote
\(y = \frac{1}{\sqrt{x}}\)

Convergence vs Divergence

After replacing the problematic bound with a limit, we evaluate the resulting definite integral. The outcome falls into exactly one of two cases:

Converges
The limit exists and is finite. We say the improper integral converges to that value.
Diverges
The limit is \(\pm\infty\) or does not exist. The integral has no finite value.

The Formal Definitions

1
Type I (upper limit \(\infty\)): \(\displaystyle\int_a^{\infty} f(x)\,dx = \lim_{t\to\infty}\int_a^{t} f(x)\,dx\)
2
Type I (lower limit \(-\infty\)): \(\displaystyle\int_{-\infty}^{b} f(x)\,dx = \lim_{t\to -\infty}\int_t^{b} f(x)\,dx\)
3
Type I (both limits): Split at any \(c\): \(\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^{c} f(x)\,dx + \int_c^{\infty} f(x)\,dx\). Both must converge.

The \(p\)-Integral Test

One of the most important reference results: the behavior of \(\displaystyle\int_1^{\infty}\frac{1}{x^p}\,dx\) depends entirely on \(p\).

Converges
When \(p > 1\).
Value: \(\displaystyle\frac{1}{p-1}\)
Diverges
When \(p \le 1\).
Includes the critical case \(p = 1\) (harmonic: \(\ln x \to \infty\)).

Type I vs Type II: Side by Side

Type I (Infinite Bounds) Type II (Discontinuity)
Problem Limit of integration is \(\pm\infty\) Integrand undefined at a point in \([a,b]\)
Fix Replace \(\infty\) with \(t\), take \(\lim_{t\to\infty}\) Replace bad point \(c\) with \(t\), take \(\lim_{t\to c}\)
Classic example \(\displaystyle\int_1^{\infty}\frac{1}{x^2}\,dx = 1\) \(\displaystyle\int_0^1\frac{1}{\sqrt{x}}\,dx = 2\)
Both limits bad? Split at any finite \(c\) Split at the discontinuity

Step-by-Step Evaluation

1
Identify the problem: Is the interval infinite (Type I), or does the integrand blow up somewhere (Type II)?
2
Replace with a limit: Swap each problematic bound with a variable (\(t\)) and write the corresponding limit.
3
Evaluate the definite integral: Find the antiderivative and apply the Fundamental Theorem on the resulting finite interval.
4
Take the limit: Let \(t\) approach its target (\(\infty\), \(-\infty\), or the point of discontinuity).
5
Conclude: Finite limit means converges. Infinite or DNE means diverges.

Example 1: \(\displaystyle\int_1^{\infty}\frac{1}{x^2}\,dx\)

1
Identify: Upper limit is \(\infty\): Type I.
2
Rewrite: \(\displaystyle\lim_{t\to\infty}\int_1^{t}\frac{1}{x^2}\,dx\)
3
Antiderivative: \(\displaystyle\int \frac{1}{x^2}\,dx = -\frac{1}{x} + C\)
4
Evaluate: \(\displaystyle\lim_{t\to\infty}\left[-\frac{1}{x}\right]_1^t = \lim_{t\to\infty}\left(-\frac{1}{t}+1\right) = 0 + 1 = 1\)

Example 2: \(\displaystyle\int_0^{1}\frac{1}{\sqrt{x}}\,dx\)

1
Identify: The integrand \(\frac{1}{\sqrt{x}}\) is undefined at \(x=0\): Type II.
2
Rewrite: \(\displaystyle\lim_{t\to 0^+}\int_t^{1}\frac{1}{\sqrt{x}}\,dx\)
3
Antiderivative: \(\displaystyle\int x^{-1/2}\,dx = 2\sqrt{x} + C\)
4
Evaluate: \(\displaystyle\lim_{t\to 0^+}\left[2\sqrt{x}\right]_t^1 = \lim_{t\to 0^+}\left(2 - 2\sqrt{t}\right) = 2 - 0 = 2\)

Example 3: \(\displaystyle\int_1^{\infty}\frac{1}{x}\,dx\)

1
Identify: Upper limit is \(\infty\): Type I.
2
Rewrite: \(\displaystyle\lim_{t\to\infty}\int_1^{t}\frac{1}{x}\,dx\)
3
Antiderivative: \(\displaystyle\int\frac{1}{x}\,dx = \ln|x| + C\)
4
Evaluate: \(\displaystyle\lim_{t\to\infty}\left[\ln x\right]_1^t = \lim_{t\to\infty}(\ln t - 0) = \infty\)

Example 4: \(\displaystyle\int_0^{\infty} e^{-x}\,dx\)

x y 0 1
\(y = e^{-x}\)
1
\(\displaystyle\lim_{t\to\infty}\int_0^t e^{-x}\,dx = \lim_{t\to\infty}\left[-e^{-x}\right]_0^t = \lim_{t\to\infty}(-e^{-t}+1) = 1\)

The Comparison Test for Integrals

Sometimes evaluating the antiderivative is hard or impossible. The comparison test lets you determine convergence by bounding the integrand against a simpler function.

1
Direct Comparison: If \(0 \le f(x) \le g(x)\) for all \(x \ge a\): if \(\int_a^{\infty}g(x)\,dx\) converges, so does \(\int_a^{\infty}f(x)\,dx\). If \(\int_a^{\infty}f(x)\,dx\) diverges, so does \(\int_a^{\infty}g(x)\,dx\).
2
Limit Comparison: If \(\displaystyle\lim_{x\to\infty}\frac{f(x)}{g(x)} = L\) where \(0 < L < \infty\), then both integrals converge or both diverge.

Common Mistakes

✕ Forgetting the Limit
Plugging \(\infty\) directly into the antiderivative without writing the limit. Always use \(\lim_{t\to\infty}\).
✕ Missing a Discontinuity
Not checking whether the integrand blows up inside \([a,b]\). For example, \(\int_0^2 \frac{1}{x-1}\,dx\) has a discontinuity at \(x=1\).
✕ One Limit for Both
Writing \(\int_{-\infty}^{\infty}f\,dx = \lim_{t\to\infty}\int_{-t}^{t}f\,dx\). This is the Cauchy Principal Value, not the improper integral. You must split and take independent limits.

Quick Reference Table

Integral Result Why
\(\displaystyle\int_1^{\infty}\frac{1}{x^p}\,dx, p > 1\) Converges to \(\frac{1}{p-1}\) \(x^{1-p}\to 0\)
\(\displaystyle\int_1^{\infty}\frac{1}{x^p}\,dx, p \le 1\) Diverges \(x^{1-p}\to\infty\) or \(\ln x\to\infty\)
\(\displaystyle\int_0^{\infty}e^{-x}\,dx\) Converges to \(1\) Exponential decay
\(\displaystyle\int_0^{1}\frac{1}{x^p}\,dx, p < 1\) Converges to \(\frac{1}{1-p}\) \(x^{1-p}\to 0\) at \(0^+\)
\(\displaystyle\int_0^{1}\frac{1}{x^p}\,dx, p \ge 1\) Diverges Singularity too strong

Worked practice: Practice Problems, Problem 4 — applied

Work through this original lesson problem, then reveal the solution and illustrations.

In a physics model, the gravitational potential energy of a satellite launched radially outward from Earth's surface (radius R) is modeled by the integral W = ∫ᵣ∞ (GMm/r²) dr, where G, M, and m are positive constants. (a) Classify this integral and explain why it is improper. (1 point) (b) Use proper limit notation to set up the evaluation. (1 point) (c) Find the antiderivative and evaluate the limit. (2 points) (d) Interpret your result: what does the finite value of this integral mean physically? (1 point)

Show answer and explanation

Point-by-point rubric: (a) [1 point] This is a Type I improper integral because the upper limit of integration is ∞. The integrand GMm/r² is continuous for r ≥ R > 0, so the only issue is the infinite bound. (b) [1 point] W = lim (b→∞) ∫ᵣᵇ (GMm/r²) dr. The student must explicitly write the limit notation with a finite upper bound b replacing ∞. (c) [2 points — 1 for antiderivative, 1 for limit evaluation] The antiderivative of GMm/r² = GMm · r⁻² is GMm · (−r⁻¹) = −GMm/r. Evaluating: lim (b→∞) [−GMm/b − (−GMm/R)] = lim (b→∞) [−GMm/b + GMm/R] = 0 + GMm/R = GMm/R. (d) [1 point] The finite value W = GMm/R represents the total work needed to move the satellite from the surface to infinitely far away — the escape energy. Because this integral converges, only a finite amount of energy is required to escape Earth's gravity, which is why escape velocity is a well-defined, finite quantity.

Inverse-square force versus distance, with work from R to 4R and the remaining tail toward infinity. Total work is GMm/R.
Inverse-square gravity requires finite positive work GMm/R to reach arbitrarily distant space, with potential energy referenced separately.
The finite work from R to 4R is three quarters of GMm/R; the remaining work to infinity is one quarter. Potential is −GMm/R at R and zero at infinity.
The finite work from R to 4R is three quarters of GMm/R; the remaining work to infinity is one quarter. Potential is −GMm/R at R and zero at infinity.

Question 1 of 5

Evaluate \(\displaystyle\int_1^{\infty}\frac{1}{x^3}\,dx\).

Question 2 of 5

Does \(\displaystyle\int_1^{\infty}\frac{1}{\sqrt{x}}\,dx\) converge or diverge?

Question 3 of 5

Evaluate \(\displaystyle\int_0^{4}\frac{1}{\sqrt{4-x}}\,dx\).

Question 4 of 5

Which statement about \(\displaystyle\int_{-\infty}^{\infty}\frac{1}{1+x^2}\,dx\) is correct?

Question 5 of 5

Using the Comparison Test, determine whether \(\displaystyle\int_1^{\infty}\frac{1}{x^2 + x}\,dx\) converges or diverges.