Why Compare Improper Integrals?
An improper integral can be difficult to evaluate directly, even when its convergence is easier to determine. A useful strategy is to compare its integrand with another nonnegative function whose integral is already understood. The comparison must use the right inequality: an upper bound by a convergent integral controls the integral from above, while a lower bound by a divergent integral forces the integral to grow without bound.
The requirement that the integrands be nonnegative is central. It ensures that inequalities between functions remain inequalities after integration, and that truncated integrals increase as the interval approaches the improper endpoint.
Improper Integrals as Limits of Proper Integrals
Consider an integral with a possible singularity at the left endpoint \(a\), where \(a<b\). Suppose \(f\) is nonnegative and continuous on every closed interval \([t,b]\) with \(a<t<b\). The improper integral is defined by the limit of proper integrals
If this limit is finite, the improper integral converges. For nonnegative \(f\), if the truncated integrals tend to \(+\infty\), the improper integral diverges to \(+\infty\). The same comparison argument applies to other improper intervals by taking proper integrals over intervals that approach the problematic endpoint.
For each \(t>a\), \(\int_t^b f(x)\,dx\) is an ordinary proper integral. The comparison theorem starts with inequalities between these proper integrals and then takes limits.
The Direct Comparison Theorem
- If \(0\le f(x)\le g(x)\) and \(\int g\) converges, then \(\int f\) converges.
- If \(0\le g(x)\le f(x)\) and \(\int g\) diverges to \(+\infty\), then \(\int f\) diverges to \(+\infty\).
For a left-endpoint singularity, the first criterion says that if \(f(x)\le g(x)\) and \(\int_a^b g(x)\,dx\) is finite, then \(\int_a^b f(x)\,dx\) is finite. The second says that if \(f(x)\ge g(x)\) and \(\int_a^b g(x)\,dx=+\infty\), then \(\int_a^b f(x)\,dx=+\infty\). The inequalities must hold on the interval being compared, not merely at a few selected points.
Why the Theorem Works
The proof is an application of the order property of integration on proper intervals, followed by the definition of the improper integral. Here is the complete argument for a left-endpoint singularity.
For every \(t>a\), the functions are integrable on \([t,b]\). If \(0\le f(x)\le g(x)\), monotonicity of the proper integral gives \(0\le\int_t^b f(x)\,dx\le\int_t^b g(x)\,dx\).
If \(\int_a^b g(x)\,dx=L<\infty\), then \(\int_t^b g(x)\,dx\to L\) as \(t\to a^+\). The truncated integrals of \(f\) are nondecreasing as \(t\) decreases, and they are bounded above by \(L\). Therefore they have a finite limit, so \(\int_a^b f(x)\,dx\) converges.
If \(0\le g(x)\le f(x)\), then \(\int_t^b f(x)\,dx\ge\int_t^b g(x)\,dx\) for every \(t>a\). If the latter tends to \(+\infty\) as \(t\to a^+\), the former is at least as large and also tends to \(+\infty\). Thus \(\int_a^b f(x)\,dx\) diverges to \(+\infty\).
In the convergence argument, nonnegativity ensures the truncated integrals of \(f\) have a limit as the truncation approaches the endpoint; the finite upper bound prevents that limit from being infinite. In the divergence argument, the lower bound forces the truncated integrals of \(f\) beyond every finite threshold. These are different uses of the same order property.
Worked Examples
Worked Example: An Upper Bound at a Finite Endpoint
Determine whether \(\int_0^1 \frac{1}{\sqrt{x}(1+x)}\,dx\) converges. For \(0<x\le1\), \(1+x\ge1\), so
The comparison integral is finite. Its truncated value is
By the upper-bound criterion, the given improper integral converges. As a direct check, substitute \(u=\sqrt{x}\), so \(x=u^2\), \(dx=2u\,du\), and \(\sqrt{x}=u\). Then
The finite value confirms the convergence conclusion.
Worked Example: A Lower Bound Forces Divergence
Determine whether \(\int_0^1 \frac{1+x}{x}\,dx\) converges. For \(0<x\le1\),
The lower comparison integral diverges, since
Therefore the direct comparison theorem implies that the given integral diverges to \(+\infty\). Checking the truncated integral directly gives the same result:
Both the inequality and the limiting calculation confirm divergence.
Worked Example: Bounding a Sum of Singular Terms
Determine whether \(\int_0^1 \frac{1+\sqrt{x}}{x^{3/4}}\,dx\) converges. Since \(\sqrt{x}\le1\) for \(0<x\le1\),
The upper comparison has a finite improper integral. For \(t>0\),
The direct comparison theorem proves convergence. We can also check the given integral by splitting its two terms and evaluating:
As \(t\to0^+\), this tends to \(4+\frac{4}{3}=\frac{16}{3}\), a finite value. This check agrees with the comparison conclusion.
A Common Comparison Mistake
A frequent error is to use a divergent upper bound to claim that a smaller integral diverges. For instance, on \(0<x\le1\), \(1\le1/x\), and \(\int_0^1 1/x\,dx\) diverges. The incorrect conclusion would be: “Since \(1\le1/x\) and the larger integral diverges, \(\int_0^1 1\,dx\) diverges.” But the calculation is
So the smaller integrand has a convergent integral. The divergence criterion requires the integrand under investigation to be greater than or equal to a divergent comparison function, not less than or equal to one. Similarly, being greater than a function with a convergent integral does not prove convergence: an upper bound, not a lower bound, is needed for that conclusion.
What the Criteria Do—and Do Not—Say
The direct comparison theorem gives sufficient conditions, not a complete test for every pair of functions. If an upper bound has a divergent integral, that fact alone gives no conclusion about the smaller integrand. If a lower bound has a convergent integral, that fact alone gives no conclusion about the larger integrand. In either case, the comparison may simply be too weak to decide the question.
For example, \(1\le1/x\) on \((0,1]\), but the integral of \(1\) converges even though the integral of \(1/x\) diverges. The inequality is valid; it just points in the direction that cannot prove convergence of the smaller function. A useful comparison must satisfy both requirements: the pointwise inequality must have the appropriate direction, and the comparison integral must have the appropriate convergence behavior.
Check Your Understanding
For each question, identify the inequality direction and the behavior of the comparison integral that would justify the stated conclusion.
- If \(0\le f(x)\le g(x)\) near an improper endpoint and \(\int g\) converges, what can you conclude about \(\int f\)?
- If \(0\le g(x)\le f(x)\) and \(\int g\) diverges to \(+\infty\), what can you conclude about \(\int f\)?
- Why are inequalities between truncated proper integrals available when the integrands are nonnegative?
- On \((0,1]\), \(1\le1/x\), and \(\int_0^1 1/x\,dx\) diverges. Does this prove that \(\int_0^1 1\,dx\) diverges? Explain.
- What is missing if an integrand is known to be smaller than a function whose improper integral also diverges?
