Recipe
Prove an improper integral converges or diverges by comparison
Use when: you have a nonnegative integrand with no elementary antiderivative, and only need to know whether the improper integral converges or diverges
Goal: a proof of convergence or divergence, backed by a benchmark integral whose behavior is already known
Taught in Comparison tests for improper integrals
Quick recipe
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1
Check whether the goal is proving convergence or proving divergence; the two goals bound the integrand in opposite directions.
Example
To prove \displaystyle\int_1^\infty \dfrac{1}{x^2+\sqrt{x}}\,dx converges, the goal calls for bounding the integrand above by a convergent benchmark.
- proving convergence — bound the integrand ABOVE by a larger, known-convergent benchmark.
- proving divergence — bound the integrand BELOW by a smaller, known-divergent benchmark.
Why: the comparison only runs in one logical direction per case; a smaller function converging says nothing about a larger one, since the larger one might still diverge.
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2
Pick a benchmark of the form x^{-p} (or another integral whose convergence is already known), then prove the inequality holds for every x past some starting point.
Example
For x\ge1, x^2+\sqrt{x} > x^2, so \dfrac{1}{x^2+\sqrt{x}} < \dfrac{1}{x^2} for every x\ge1.
Why: the comparison must hold for every x from that point on, not just eventually or approximately; one exception anywhere breaks the argument.
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3
Check f,g\ge0 throughout the interval, since the whole argument rests on each integral's running total never decreasing.
Example
Both \dfrac{1}{x^2+\sqrt{x}} and \dfrac{1}{x^2} are positive for every x\ge1.
Why: a negative value could make an integral's running total decrease instead of only grow, which would break the comparison argument entirely.
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4
Cite the known behavior of the benchmark, together with the direction of the inequality from Step 1, to conclude.
Example
\displaystyle\int_1^\infty x^{-2}\,dx has p=2>1, so it converges. Since the original integrand is smaller than this convergent benchmark, the Comparison Test guarantees the original integral converges too.
Why: this is the actual claim being proved; convergence or divergence of the original integral follows from the theorem applied to the inequality, not from the inequality by itself.
Common mistakes
- Bounding the integrand in the wrong direction for the goal, such as bounding above by a convergent benchmark while trying to prove divergence.
- Checking the inequality only near the starting point instead of for every x past it.
- Forgetting to confirm f,g\ge0, especially when the integrand involves a term like \sin x that isn't obviously nonnegative on its own.
- Choosing a benchmark whose own convergence or divergence isn't actually known yet, which proves nothing.