How does the comparison test decide convergence or divergence for improper integrals with nonnegative functions?

Conditions

• The functions $f(x)$ and $g(x)$ are nonnegative on the tail interval $[a, \infty)$.
• The inequality $0 \le g(x) \le f(x)$ holds for all $x \ge a$.
• Finite-interval integrability is assumed for the functions involved.

Reasoning, step by step

1. Identify a simpler function $f(x)$ that bounds the target integrand $g(x)$ from above on the tail.
2. Verify that both functions are nonnegative on the interval $[a, \infty)$.
3. Determine the convergence or divergence of the bounding integral $\int_a^\infty f(x)\,dx$.
4. If $\int_a^\infty f(x)\,dx$ converges, conclude that $\int_a^\infty g(x)\,dx$ also converges.
5. If $\int_a^\infty g(x)\,dx$ diverges, conclude that $\int_a^\infty f(x)\,dx$ also diverges.

Example

For $0\le g(x)\le f(x)$ on the tail, with finite-interval integrability, convergence of $\int_a^\infty f(x)\,dx$ forces convergence of the smaller integral. Divergence of the smaller integral forces divergence of the larger one.

Common misconceptions

• Assuming the test works if the functions are not nonnegative on the tail.
• Believing that convergence of the smaller integral proves convergence of the larger one.
• Thinking that divergence of the larger integral proves divergence of the smaller one.

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Connected concepts

Explore next

• What are the convergence conditions for a p-integral?
• How do you apply the comparison test to a rational function?
• What does the comparison test not tell us about the integral?

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