Why Compare Improper Integrals?
Some improper integrals are difficult to evaluate directly, but their convergence can still be determined. The key is to compare the integrand with a positive function whose integral is already known. The P-integral convergence benchmarks provide especially useful comparisons: for example, \(\int_1^\infty \frac{1}{x^p}\,dx\) converges when \(p>1\) and diverges when \(p\leq 1\). Comparison tests let us use such facts without finding an antiderivative for the integrand under study.
The inequalities and positivity conditions matter. A comparison test does not say that any two functions with roughly similar-looking formulas have the same behavior. It gives a conclusion only when a particular inequality, or a suitable positive ratio limit, has been established near the improper endpoint.
Set Up the Comparison Carefully
Suppose an integral is improper at an endpoint. First identify which endpoint causes the issue, then compare the integrands near that endpoint. The functions should be continuous on every compact part of the interval away from the improper endpoint, and the comparison functions should be positive there. If an integral has more than one improper endpoint, or an interior singularity, treat the relevant one-sided integrals separately.
For an infinite interval, “near the improper endpoint” means for all sufficiently large \(x\). For an endpoint at \(a\), it means for all \(x\) sufficiently close to \(a\) from within the interval. Behavior on an initial compact segment does not affect convergence, provided the functions are integrable there.
The Direct Comparison Test
- If \(\int g(x)\,dx\) converges at that endpoint, then \(\int f(x)\,dx\) also converges there.
- If \(\int f(x)\,dx\) diverges at that endpoint, then \(\int g(x)\,dx\) also diverges there.
The first conclusion follows from the fact that a nonnegative integrand bounded above by one with finite accumulated area cannot accumulate infinite area. The second follows by contraposition: if \(\int g\) converged, the first conclusion would force \(\int f\) to converge too. Thus divergence passes upward from a smaller nonnegative function to a larger one, not downward.
Worked Example: Convergence at a Finite Singular Endpoint
Determine whether \(\int_0^1 \frac{1}{\sqrt{x(1+x)}}\,dx\) converges.
For \(0<x\leq 1\), both \(x\) and \(1+x\) are positive. Also, \(1+x\geq 1\), so \(\sqrt{x(1+x)}\geq\sqrt{x}\). Taking reciprocals of positive quantities reverses the inequality:
The comparison integral is a P-integral at \(0\). Its truncated value is
It converges, so the direct comparison test proves that the original integral converges. The positivity and inequality were checked throughout \(0<x\leq1\), including arbitrarily close to the singular endpoint.
Worked Example: Divergence by a Lower Bound
Determine whether \(\int_1^\infty \frac{2x+1}{x^2+x}\,dx\) converges.
For \(x\geq1\), the numerator and denominator are positive. Compare the integrand with \(1/x\). Since \(x>0\), cross-multiplication preserves the inequality:
The benchmark diverges, since for \(R>1\),
The original integrand is at least this divergent positive benchmark, so its improper integral diverges. The direction of the comparison is essential: a larger nonnegative function cannot have finite total area if a smaller one already has infinite total area.
Why the Direct Test Works
The comparison theorem follows from the ordinary order property of definite integrals. On a truncated interval, if \(0\leq f\leq g\), then integrating preserves the inequality. For an infinite endpoint, take the truncation at \(R\) and then let \(R\) increase. For a finite singular endpoint \(a\), truncate at \(a+\varepsilon\) or \(a-\varepsilon\), as appropriate, and take the corresponding one-sided limit.
For example, for an endpoint at infinity, choose \(R\) large enough that \(0\leq f(x)\leq g(x)\) on \([A,R]\).
For every such \(R\), \(\int_A^R f(x)\,dx\leq\int_A^R g(x)\,dx\).
If the right-hand integrals approach a finite value, the nondecreasing left-hand integrals are bounded above and therefore approach a finite limit too.
This argument also explains why nonnegativity is important: the truncated integrals of a nonnegative function increase as the interval expands toward the improper endpoint. For finite endpoints the same argument uses truncated intervals approaching the endpoint. The comparison is applied only where the inequality holds; any remaining compact segment contributes a finite amount under the stated regularity conditions.
Limit Comparison
Sometimes neither integrand is obviously bounded by the other with a convenient constant. If their ratio approaches a finite positive number, then near the endpoint they are bounded above and below by constant multiples of one another.
Here the limit is taken toward the improper endpoint: \(x\to\infty\), \(x\to a^+\), or \(x\to a^-\), as applicable. Positivity ensures the ratio is defined near the endpoint and allows multiplication of the resulting inequalities without changing their directions.
Worked Example: Limit Comparison for Convergence
Determine whether \(\int_1^\infty \frac{2x+7}{x^3+4}\,dx\) converges.
Both the integrand and \(g(x)=1/x^2\) are positive for \(x\geq1\). Compute the ratio and divide its numerator and denominator by \(x^3\):
The limit is finite and positive, so limit comparison applies. The benchmark converges because
Therefore the original integral converges. As a check on the hypotheses, for \(x\geq1\), its numerator and denominator are positive, and the ratio calculation has a nonzero finite limit, exactly the condition required by this test.
Justification of Limit Comparison
Let \(L=\lim f(x)/g(x)\), with \(0<L<\infty\). By the definition of a limit, choose a positive tolerance smaller than \(L\), such as \(L/2\). Sufficiently near the improper endpoint, the ratio then lies between \(L/2\) and \(3L/2\). Since \(g(x)>0\), multiplying through by \(g(x)\) gives
The constants \(L/2\) and \(3L/2\) are positive and finite. Multiplying a nonnegative function by such a constant does not change whether its improper integral is finite or infinite. The direct comparison test applied to both inequalities therefore gives the same convergence behavior for \(f\) and \(g\). This proves the limit comparison result.
Choosing a Benchmark and Avoiding Pitfalls
Use a benchmark whose behavior is known and whose formula matches the dominant behavior of the integrand near the endpoint. Near infinity, powers of \(x\) often make useful comparisons. Near a finite singularity, powers of the distance to the endpoint are common benchmarks. The comparison need only be valid sufficiently near the endpoint, but it must be proved there.
For a concrete counterexample to the incorrect inference, on \(x\geq1\) we have \(0<1/x^2\leq1/x\), while
The larger function has a divergent integral, but the smaller one converges. The correction is to use the test in its stated direction: an integrable upper bound proves convergence; a divergent lower bound proves divergence.
Limit comparison also has a specific condition. A ratio limit of zero or infinity does not meet the theorem’s requirement \(0<L<\infty\); it cannot, by itself, establish that both integrals have the same behavior. In such cases, use a direct inequality if one can be proved, or find another valid benchmark.
Check Your Understanding
For each question, identify the endpoint and verify that the comparison conditions hold before stating a conclusion.
- If \(0\leq f(x)\leq g(x)\) near infinity and \(\int g\) converges, what can you conclude about \(\int f\)?
- If \(0\leq g(x)\leq f(x)\) near a finite singular endpoint and \(\int g\) diverges, what can you conclude about \(\int f\)?
- For which values of \(p\) does \(\int_1^\infty x^{-p}\,dx\) converge, and how could this benchmark help classify another positive integrand?
- Why does a finite, positive limit of \(f(x)/g(x)\) imply that \(f\) and \(g\) have the same convergence behavior?
- Give a reason that \(f\leq g\) and divergence of \(\int g\) are not enough to prove divergence of \(\int f\).
