Comparison Test for Integrals

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Textbook Reference

Opening Scenario

Some improper integrals, like $\displaystyle\int_1^{\infty} e^{-x^2}\,dx$, have no elementary antiderivative. We cannot evaluate them directly using FTC. But we can still determine whether they converge by comparing them to simpler integrals whose convergence is known.

The Comparison Test for integrals works on the same geometric principle as squeezing: if one function is larger than another and the larger one has a finite integral, then the smaller one certainly does too.

Quick Reference

Direct Comparison Test. Suppose $0 \leq f(x) \leq g(x)$ for all $x \geq a$.

• If $\displaystyle\int_a^{\infty} g(x)\,dx$ converges, then $\displaystyle\int_a^{\infty} f(x)\,dx$ converges.
• If $\displaystyle\int_a^{\infty} f(x)\,dx$ diverges, then $\displaystyle\int_a^{\infty} g(x)\,dx$ diverges.

Limit Comparison Test. If $f(x), g(x) > 0$ and $\displaystyle\lim_{x\to\infty}\frac{f(x)}{g(x)} = L$ where $0 < L < \infty$, then $\int_a^{\infty} f$ and $\int_a^{\infty} g$ either both converge or both diverge.

Most useful comparison functions: $\displaystyle\frac{1}{x^p}$ (use the $p$-test for reference), $e^{-x}$, and $\dfrac{1}{\sqrt{x}}$.

Key Concepts

1. The Geometric Idea

If $f(x) \leq g(x) \geq 0$ on $[a, \infty)$, then the area under $f$ is at most the area under $g$. If the larger area is finite, the smaller area is certainly finite. If the smaller area is infinite, the larger area is certainly infinite.

The comparison test turns a question about an unfamiliar function into a question about a familiar one. The skill is choosing the right familiar function.

2. Choosing the Comparison Function

For functions that behave like $1/x^p$ for large $x$:

• Drop lower-order terms: $\dfrac{1}{x^2+1} \approx \dfrac{1}{x^2}$ for large $x$, so compare with $\dfrac{1}{x^2}$.
• Keep dominant terms in products: $\dfrac{x+1}{x^3+x^2} \approx \dfrac{x}{x^3} = \dfrac{1}{x^2}$ for large $x$.

For functions involving exponentials: $e^{-x^2} \leq e^{-x}$ for $x \geq 1$ (since $x^2 \geq x$), and $\int_1^{\infty} e^{-x}\,dx$ converges.

3. Limit Comparison

When the direct inequality is hard to establish but the functions behave similarly for large $x$, the Limit Comparison Test is more convenient. If $f(x)/g(x) \to L$ with $0 < L < \infty$, then $f$ and $g$ “look the same at infinity” and share the same convergence/divergence.

Example. Does $\displaystyle\int_1^{\infty}\frac{3x+2}{x^3+1}\,dx$ converge?

Compare with $g(x) = \dfrac{1}{x^2}$. Limit: $\lim_{x\to\infty}\dfrac{(3x+2)/(x^3+1)}{1/x^2} = \lim_{x\to\infty}\dfrac{x^2(3x+2)}{x^3+1} = \lim_{x\to\infty}\dfrac{3x^3}{x^3} = 3$. Finite, nonzero. Since $\int_1^{\infty}\frac{1}{x^2}\,dx$ converges ($p=2>1$), the original integral also converges.

Worked Example

Determine whether $\displaystyle\int_1^{\infty} e^{-x^2}\,dx$ converges.

For $x \geq 1$: $x^2 \geq x$, so $e^{-x^2} \leq e^{-x}$.

Both functions are positive, and $\displaystyle\int_1^{\infty} e^{-x}\,dx = \left[-e^{-x}\right]_1^{\infty} = 0 + e^{-1} = \frac{1}{e}$ converges.

By the Direct Comparison Test, $\displaystyle\int_1^{\infty} e^{-x^2}\,dx$ converges.

Determine whether $\displaystyle\int_2^{\infty}\frac{1}{\sqrt{x}-1}\,dx$ converges.

For $x \geq 4$: $\sqrt{x} - 1 \geq \frac{\sqrt{x}}{2}$ (since $\sqrt{x} \geq 2$). So $\dfrac{1}{\sqrt{x}-1} \leq \dfrac{2}{\sqrt{x}} = 2x^{-1/2}$.

$\int_4^{\infty} 2x^{-1/2}\,dx$ diverges ($p = 1/2 \leq 1$). However, $f \leq 2x^{-1/2}$ and the larger function diverges, so direct comparison gives no conclusion.

Try from the other direction: $\sqrt{x} - 1 \leq \sqrt{x}$, so $\dfrac{1}{\sqrt{x}-1} \geq \dfrac{1}{\sqrt{x}} = x^{-1/2}$.

Now $f(x) \geq x^{-1/2}$ and $\int_2^{\infty} x^{-1/2}\,dx$ diverges. The smaller function $x^{-1/2}$ diverges, so the larger function $1/(\sqrt{x}-1)$ also diverges.

Common Errors Summary

Leveled Practice

Level 1 -- Direct Comparison

Problem 1. Use direct comparison to determine whether $\displaystyle\int_1^{\infty}\frac{\cos^2 x}{x^2}\,dx$ converges.

Level 2 -- Limit Comparison

Problem 2. Determine whether $\displaystyle\int_1^{\infty}\frac{2+\sin x}{x}\,dx$ converges.

Mastery Checklist

• State the Direct Comparison Test and identify which direction of inequality gives which conclusion
• Apply the Direct Comparison Test using $1/x^p$ as the comparison function
• State the Limit Comparison Test and apply it when direct inequality is difficult
• Choose an appropriate comparison function by identifying the dominant behavior of $f(x)$ for large $x$
• Avoid the common error of applying divergence information in the wrong direction

Mental Model

Think of the comparison test like a budget analogy. If your expenses (the smaller function) are always less than your income (the larger function), and your income is finite, then your expenses are certainly finite. Conversely, if your expenses are always more than some known minimum that already exceeds your income, you are overspending.

The comparison function is your “reference budget” -- something you know everything about (a power function or exponential) that can be compared to the unfamiliar function.

Connections

Looking back

• $p$-test (Section 7.8): The reference result for comparison; most comparisons reduce to a $p$-test.
• Squeeze Theorem (Section 2.3): The same trapping principle applied to sequences/functions.

Looking ahead

• Comparison test for series (Chapter 11): The exact same principle applied to sums instead of integrals; the series versions are direct translations.

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