Comparison Test for Improper Integrals

The comparison test for improper integrals decides convergence without integrating. If 0 ≤ g(x) ≤ f(x) and the integral of the bigger function f converges, then the integral of g converges too. If the integral of the smaller function g diverges, then the integral of f diverges too.

Before this lesson: Improper Integrals and Improper Integrals with Infinite Discontinuities

Trapping an area

Many improper integrals, like $\int_1^{\infty} e^{-x^2}\,dx$ , have no antiderivative you can write down. You can still tell whether the area is finite by comparing the function with one you know.

If a region sits inside a region of finite area, it has finite area too. If a region contains a region of infinite area, its area is infinite. The comparison test is just this idea written with integrals.

Graphs of f(x) = 1/x² and g(x) = 1/(x² + 3) for x from 1 to 6. The curve g lies below f everywhere. The region under g is shaded and sits inside the region under f, whose area from 1 to infinity is 1. — Since g(x) ≤ f(x) and the area under f is 1, the area under g is finite, and less than 1.

Comparison test for improper integrals

Suppose f and g are continuous and 0 ≤ g(x) ≤ f(x) for all x ≥ a.

If you know…	Then…
$\int_a^{\infty} f(x)\,dx$ converges (bigger one)	$\int_a^{\infty} g(x)\,dx$ converges
$\int_a^{\infty} g(x)\,dx$ diverges (smaller one)	$\int_a^{\infty} f(x)\,dx$ diverges

The same rule works for integrals with an infinite discontinuity. The other two cases tell you nothing: being smaller than something divergent, or bigger than something convergent, proves nothing.

Good functions to compare with: $\dfrac{1}{x^p}$ on [1, ∞), which converges only when p is greater than 1; $\dfrac{1}{x^p}$ on (0, 1], which converges only when p is less than 1; and $e^{-ax}$ with a positive, which converges on [0, ∞).

Comparison test examples

Example 1: Smaller than a convergent integral

Does $\int_1^{\infty} \dfrac{1}{x^2 + 3}\,dx$ converge?

x^2 + 3 \gt x^2

Adding 3 makes the bottom bigger.

0 \lt \dfrac{1}{x^2 + 3} \lt \dfrac{1}{x^2}

A bigger bottom makes a smaller fraction.

\int_1^{\infty} \dfrac{1}{x^2}\,dx = 1

Converges by the p-rule, with p = 2.

Answer: the integral converges, and its value is less than 1.

Example 2: Bigger than a divergent integral

Does $\int_2^{\infty} \dfrac{1}{\sqrt{x} - 1}\,dx$ converge?

0 \lt \sqrt{x} - 1 \lt \sqrt{x} \text{ for } x \ge 2

Subtracting 1 makes the bottom smaller, and it stays positive.

\dfrac{1}{\sqrt{x} - 1} \gt \dfrac{1}{\sqrt{x}}

A smaller positive bottom makes a bigger fraction.

\int_2^{\infty} \dfrac{1}{\sqrt{x}}\,dx = \lim_{t \to \infty} (2\sqrt{t} - 2\sqrt{2}) = \infty

Diverges, since p = 1/2.

Answer: the integral diverges, because it is bigger than a divergent integral.

Example 3: A bounded top

Does $\int_1^{\infty} \dfrac{\sin^2 x}{x^2}\,dx$ converge?

0 \le \sin^2 x \le 1

A square of sine is between 0 and 1 (see graphs of sine and cosine).

0 \le \dfrac{\sin^2 x}{x^2} \le \dfrac{1}{x^2}

Divide by x², which is positive.

\int_1^{\infty} \dfrac{1}{x^2}\,dx = 1

Converges.

Answer: the integral converges.

Example 4: The bell curve integral

Does $\int_1^{\infty} e^{-x^2}\,dx$ converge?

x^2 \ge x \text{ for } x \ge 1

Multiply both sides of x ≥ 1 by x.

-x^2 \le -x

Multiply by −1 and flip the sign.

e^{-x^2} \le e^{-x}

eᵘ grows as u grows, so the order stays the same.

\int_1^{\infty} e^{-x}\,dx = \lim_{t \to \infty} (e^{-1} - e^{-t}) = e^{-1}

The bigger integral converges.

Answer: the integral converges, and its value is less than 1/e ≈ 0.368. Its actual value is about 0.1394.

Example 5: Bounding the energy from a machine

After its first hour, a cooling machine gives off heat at a rate of $E(t) = \dfrac{40}{t^2 + 2t}$ kilojoules per hour. Is the total heat given off after t = 1 finite? Bound it, then find it exactly.

\dfrac{40}{t^2 + 2t} \le \dfrac{40}{t^2} \text{ for } t \ge 1

Dropping 2t makes the bottom smaller and the fraction bigger.

\int_1^{\infty} \dfrac{40}{t^2}\,dt = 40

40 times the p-integral with p = 2.

\dfrac{40}{t(t + 2)} = \dfrac{20}{t} - \dfrac{20}{t + 2}

Now find it exactly with partial fractions (see integration by partial fractions).

\int_1^{b} \left(\dfrac{20}{t} - \dfrac{20}{t + 2}\right)dt = 20\ln\dfrac{b}{b + 2} - 20\ln\dfrac{1}{3}

Integrate and combine the logs (see properties of logarithms).

\lim_{b \to \infty} 20\ln\dfrac{b}{b + 2} = 20\ln 1 = 0

b/(b + 2) → 1 (see limits at infinity).

0 - 20\ln\dfrac{1}{3} = 20\ln 3 \approx 21.97

−ln(1/3) = ln 3.

Answer: the total is finite. The comparison shows it is at most 40 kJ, and the exact total is 20 ln 3 ≈ 22.0 kJ.

Common mistakes

Mistake	Example of the error	Fix
Comparing in the wrong direction	$\dfrac{1}{x^2 + 1} \le \dfrac{1}{x}$ , and ∫1/x diverges, so it diverges	Smaller than divergent proves nothing. Compare with 1/x² instead: it converges.
An inequality that fails somewhere	$\dfrac{1}{x - 1} \le \dfrac{1}{x}$ on [2, ∞)	It’s backward: 1/(x − 1) is bigger. Check the inequality before you use it.
Using it on negative functions	Comparing $\dfrac{\sin x}{x^2}$ with 1/x² directly	The test needs g(x) ≥ 0. Compare $\left\|\dfrac{\sin x}{x^2}\right\|$ with 1/x² instead.
Picking a divergent benchmark	Using $\dfrac{1}{x}$ as the bigger function	Bigger functions must have convergent integrals. Use 1/xᵖ with p greater than 1, or e⁻ˣ.
Limit comparison with L = 0	$\lim \dfrac{f}{g} = 0$ , so both act the same	Limit comparison needs a finite, positive limit L.

Related methods

Limit comparison test

If f and g are positive and $\lim_{x \to \infty} \dfrac{f(x)}{g(x)} = L$ with L finite and positive, then the two integrals both converge or both diverge. For example, $\dfrac{2x + 1}{x^3 + 5}$ behaves like $\dfrac{2}{x^2}$ , since the ratio of the two approaches 1. So $\int_1^{\infty} \dfrac{2x + 1}{x^3 + 5}\,dx$ converges.

Absolute convergence

If $\int_a^{\infty} |f(x)|\,dx$ converges, then $\int_a^{\infty} f(x)\,dx$ converges too. This handles integrands that change sign, like sin x / x².

The same idea for series

Infinite series have their own comparison tests, which work the same way. You’ll see them later in this course.

Practice problems

Decide whether each integral converges or diverges, and say what you compared it with.

$\int_1^{\infty} \dfrac{1}{x^3 + 1}\,dx$
$\int_1^{\infty} \dfrac{1 + e^{-x}}{x}\,dx$
$\int_1^{\infty} \dfrac{\cos^2 x}{x^2 + 1}\,dx$
$\int_2^{\infty} \dfrac{1}{\ln x}\,dx$
$\int_1^{\infty} \dfrac{x}{x^3 + 2}\,dx$ (use limit comparison)
$\int_1^{\infty} \dfrac{\sqrt{x}}{x + 1}\,dx$ (use limit comparison)
$\int_1^{\infty} \dfrac{e^{-x}}{x}\,dx$
$\int_0^{1} \dfrac{e^{x}}{\sqrt{x}}\,dx$
$\int_0^{1} \dfrac{1}{x^2 + x}\,dx$
Show that $\int_1^{\infty} \dfrac{1}{x^4 + x}\,dx$ converges to a number at most 1/3.

Show answers

x^3 + 1 \gt x^3, \text{ so } \dfrac{1}{x^3 + 1} \lt \dfrac{1}{x^3}

A bigger bottom gives a smaller fraction.

\int_1^{\infty} \dfrac{1}{x^3}\,dx = \dfrac{1}{2}

Converges (p = 3).

Answer: converges

1 + e^{-x} \gt 1, \text{ so } \dfrac{1 + e^{-x}}{x} \gt \dfrac{1}{x}

e⁻ˣ is positive.

\int_1^{\infty} \dfrac{1}{x}\,dx = \infty

Diverges (p = 1).

Answer: diverges

0 \le \dfrac{\cos^2 x}{x^2 + 1} \le \dfrac{1}{x^2 + 1} \lt \dfrac{1}{x^2}

cos²x is at most 1, and a bigger bottom gives a smaller fraction.

\int_1^{\infty} \dfrac{1}{x^2}\,dx = 1

Converges.

Answer: converges

0 \lt \ln x \lt x \text{ for } x \ge 2

ln x grows more slowly than x.

\dfrac{1}{\ln x} \gt \dfrac{1}{x}

Flip both sides of a positive inequality.

\int_2^{\infty} \dfrac{1}{x}\,dx = \infty

Diverges.

Answer: diverges

\lim_{x \to \infty} \dfrac{x/(x^3 + 2)}{1/x^2} = \lim_{x \to \infty} \dfrac{x^3}{x^3 + 2} = 1

Compare with 1/x²; divide by x³ to find the limit.

\int_1^{\infty} \dfrac{1}{x^2}\,dx \text{ converges}

L = 1 is finite and positive.

Answer: converges

\lim_{x \to \infty} \dfrac{\sqrt{x}/(x + 1)}{1/\sqrt{x}} = \lim_{x \to \infty} \dfrac{x}{x + 1} = 1

Compare with 1/√x.

\int_1^{\infty} \dfrac{1}{\sqrt{x}}\,dx \text{ diverges}

p = 1/2.

Answer: diverges

0 \lt \dfrac{e^{-x}}{x} \le e^{-x} \text{ for } x \ge 1

Dividing by x ≥ 1 makes it no bigger.

\int_1^{\infty} e^{-x}\,dx = e^{-1}

Converges.

Answer: converges

0 \lt \dfrac{e^{x}}{\sqrt{x}} \le \dfrac{e}{\sqrt{x}} \text{ on } (0, 1]

eˣ ≤ e when x ≤ 1.

\int_0^{1} \dfrac{e}{\sqrt{x}}\,dx = 2e

Converges (p = 1/2 near 0).

Answer: converges

x^2 + x \le 2x \text{ on } (0, 1]

x² ≤ x when x ≤ 1.

\dfrac{1}{x^2 + x} \ge \dfrac{1}{2x}

A smaller positive bottom gives a bigger fraction.

\int_0^{1} \dfrac{1}{2x}\,dx = \infty

Diverges (p = 1 near 0).

Answer: diverges

10.

\dfrac{1}{x^4 + x} \lt \dfrac{1}{x^4}

A bigger bottom gives a smaller fraction.

\int_1^{\infty} \dfrac{1}{x^4}\,dx = \dfrac{1}{4 - 1} = \dfrac{1}{3}

p-rule with p = 4.

Answer: converges, to a number less than 1/3

Comparison test FAQ

What is the comparison test for improper integrals?

It compares a positive function with one whose integral you know. Below a convergent integral means convergent; above a divergent integral means divergent.

What functions should I compare with?

Use 1/xᵖ or e⁻ᵃˣ. Look at what the function acts like for large x, or near the bad point, and keep only the strongest terms.

What is the limit comparison test for integrals?

If f/g approaches a finite, positive number, the integrals of f and g both converge or both diverge. It saves you from proving an exact inequality.

What if my inequality goes the wrong way?

Then the test tells you nothing. Pick a different comparison function, or try limit comparison.

Summary

The comparison test needs 0 ≤ g(x) ≤ f(x).
If the bigger integral converges, so does the smaller one.
If the smaller integral diverges, so does the bigger one.
Compare with 1/xᵖ or e⁻ᵃˣ.
Limit comparison: a finite, positive limit of f/g means both act the same.

Previous lesson: Improper Integrals with Infinite Discontinuities

Next lesson: Shell Method

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