Comparison of Improper Integrals

Why Compare Improper Integrals?

In the previous tutorial, improper integrals were defined through limits of proper integrals, and the Cauchy criterion characterized convergence by the smallness of distant or short truncated pieces. Often, however, the limit of a truncated integral is difficult to compute. Comparison offers another route: if an integrand is bounded above or below by one whose behavior is known, the bound can transfer convergence or divergence.

The direction of the inequality matters. A nonnegative function smaller than a convergent comparison function must have a convergent integral. A nonnegative function larger than a divergent comparison function must have a divergent integral. The reverse implications do not generally hold. We will also use ratios to compare functions that are not ordered by a simple inequality everywhere but have the same size near the problematic endpoint.

The Direct Comparison Test

Begin with an infinite interval. The functions in the comparison must be nonnegative; otherwise, large positive and negative contributions can cancel, and pointwise order alone does not control the size of the integral. Local Riemann integrability is also needed so that every truncated integral is proper.

Theorem (Direct Comparison Test on an Infinite Interval): Suppose \(f\) and \(g\) are nonnegative and Riemann integrable on every closed, bounded subinterval of \([a,\infty)\). If \(0\leq f(x)\leq g(x)\) for all sufficiently large \(x\), then:

If \(\int_a^\infty g(x)\,dx\) converges, then \(\int_a^\infty f(x)\,dx\) converges.
If \(\int_a^\infty f(x)\,dx\) diverges, then \(\int_a^\infty g(x)\,dx\) diverges.

The same conclusions hold if the inequalities are assumed for every \(x\geq a\). If \(f(x)\geq g(x)\geq0\) for all sufficiently large \(x\), then divergence of \(\int_a^\infty g(x)\,dx\) implies divergence of \(\int_a^\infty f(x)\,dx\).

Proof. Choose \(A>a\) so that \(0\leq f(x)\leq g(x)\) for every \(x\geq A\). The integrals over the finite interval \([a,A]\) are proper, so adding or removing those initial pieces does not affect convergence at infinity. It is enough to consider the integrals from \(A\) onward.

For \(R\geq A\), set \(F(R)=\int_A^R f(x)\,dx\) and \(G(R)=\int_A^R g(x)\,dx\). Order preservation for proper Riemann integrals gives

0\leq F(R)\leq G(R).

If \(\int_A^\infty g(x)\,dx\) converges, then \(G(R)\) has a finite limit and is therefore bounded above. Thus \(F(R)\) is nondecreasing and bounded above. By completeness of the real numbers, \(F(R)\) has a finite limit as \(R\to\infty\): specifically, its limit is the supremum of its values. This proves convergence for \(f\).

For the divergence statements, use the same order inequality on each truncated interval. If \(f\geq g\geq0\) on \([A,\infty)\) and the truncated integrals of \(g\) grow without bound, then

\int_A^R f(x)\,dx\geq\int_A^R g(x)\,dx

for every \(R\geq A\), so the truncated integrals of \(f\) grow without bound as well. This proves that divergence of the smaller function's integral forces divergence of the larger function's integral. Equivalently, when \(f\leq g\), convergence of \(f\) does not force convergence of \(g\), but divergence of \(f\) does force divergence of \(g\). \(\square\)

At a finite singular endpoint, the same order argument applies to truncated integrals. For example, suppose \(f\) and \(g\) are nonnegative and Riemann integrable on every \([a+\varepsilon,b]\), and \(f(x)\leq g(x)\) near \(a\). If \(\int_a^b g(x)\,dx\) converges, then \(\int_a^b f(x)\,dx\) converges. Indeed, the truncated integrals satisfy the corresponding order inequality, and as \(\varepsilon\downarrow0\), their nonnegative values increase while remaining bounded by the convergent comparison integrals together with any proper portion away from \(a\). The analogous divergence conclusion holds when a function is bounded below by a nonnegative function with divergent integral.

Worked Example: A Rational Function Compared with a Power

Consider \(\int_1^\infty \frac{2x+3}{x^3+1}\,dx\). For \(x\geq1\), the numerator satisfies \(2x+3\leq5x\), and the denominator satisfies \(x^3+1\geq x^3\). Both are positive, so

0\leq\frac{2x+3}{x^3+1} \leq\frac{5x}{x^3} =\frac{5}{x^2}.

The comparison integral converges:

\int_1^\infty \frac{5}{x^2}\,dx = \lim_{R\to\infty}5\left(1-\frac{1}{R}\right) =5.

The Direct Comparison Test therefore proves that \(\int_1^\infty \frac{2x+3}{x^3+1}\,dx\) converges. No antiderivative for the original rational function is needed.

Worked Example: A Lower Bound Proves Divergence

Consider \(\int_1^\infty \frac{x+1}{x^2+4}\,dx\). For \(x\geq2\), we have \(x+1\geq x\) and \(x^2+4\leq2x^2\). Since the denominators are positive, these inequalities give

\frac{x+1}{x^2+4} \geq\frac{x}{2x^2} =\frac{1}{2x}.

For \(R\geq2\),

\int_2^R\frac{x+1}{x^2+4}\,dx \geq \frac12\int_2^R\frac{1}{x}\,dx = \frac12\ln\left(\frac{R}{2}\right).

The right-hand side tends to \(+\infty\). Thus the truncated integrals of the original nonnegative function are unbounded, and the improper integral diverges. The fact that the numerator and denominator both grow is not enough to decide convergence; the lower bound identifies a divergent tail.

Limit Comparison

Sometimes a useful inequality is not apparent at first, but the ratio of two positive functions has a simple limit. A positive finite ratio limit means that, sufficiently close to the problematic endpoint, each function is bounded above and below by constant multiples of the other. This is the basis of limit comparison.

Theorem (Limit Comparison Test): Suppose \(f\) and \(g\) are positive and Riemann integrable on every closed, bounded subinterval of \([a,\infty)\). If

\lim_{x\to\infty}\frac{f(x)}{g(x)}=c \quad\text{for some }0<c<\infty,

then \(\int_a^\infty f(x)\,dx\) converges if and only if \(\int_a^\infty g(x)\,dx\) converges. The same conclusion holds near a finite singular endpoint if the ratio tends to a positive finite limit as the endpoint is approached.

Proof. Since \(c>0\), the definition of the limit gives \(A>a\) such that for every \(x\geq A\),

\left|\frac{f(x)}{g(x)}-c\right|<\frac{c}{2}.

Because \(g(x)>0\), this implies

\frac{c}{2}g(x)<f(x)<\frac{3c}{2}g(x) \quad\text{for every }x\geq A.

If \(\int_a^\infty g(x)\,dx\) converges, the upper bound and the Direct Comparison Test show that \(\int_A^\infty f(x)\,dx\) converges. The portion from \(a\) to \(A\) is proper, so \(\int_a^\infty f(x)\,dx\) converges. Conversely, the lower bound gives \(g(x)<(2/c)f(x)\) for \(x\geq A\). If the integral of \(f\) converges, direct comparison proves convergence of the integral of \(g\). Thus each integral converges exactly when the other does. The argument near a finite singular endpoint is identical, using inequalities on a sufficiently short interval next to that endpoint; the remaining portion is proper. \(\square\)

Worked Example: Limit Comparison with a Power at Infinity

Consider \(\int_1^\infty \frac{3x+1}{x^3+2}\,dx\). Compare its integrand with \(g(x)=1/x^2\). The ratio is

\frac{\frac{3x+1}{x^3+2}}{\frac{1}{x^2}} = \frac{3x^3+x^2}{x^3+2}.

Dividing numerator and denominator by \(x^3\) shows that this ratio tends to \(3\) as \(x\to\infty\). Since \(3\) is positive and finite, the Limit Comparison Test applies. The integral of \(1/x^2\) on \([1,\infty)\) converges, so the original integral converges as well.

Worked Example: Limit Comparison at a Finite Endpoint

Consider \(\int_0^1 \frac{1}{\sqrt{x(1+x)}}\,dx\), which is improper at \(0\). Compare the integrand with \(g(x)=x^{-1/2}\). For \(x>0\), their ratio is

\frac{\frac{1}{\sqrt{x(1+x)}}}{x^{-1/2}} = \frac{1}{\sqrt{1+x}}.

As \(x\downarrow0\), this ratio tends to \(1\). The comparison integral converges because

\int_0^1 x^{-1/2}\,dx = \lim_{\varepsilon\downarrow0}2(1-\sqrt{\varepsilon}) =2.

Limit comparison proves that the integral of \(1/\sqrt{x(1+x)}\) also converges. Notice that the ratio test concerns behavior near the singular endpoint; the integrand is continuous and bounded on every interval separated from zero.

Absolute Convergence as a Comparison Principle

Comparison also handles signed integrands. The integral of \(f\) may converge because positive and negative parts cancel, even when the integral of \(|f|\) diverges. If the integral of \(|f|\) does converge, however, then the integral of \(f\) must converge. This is called absolute convergence.

Theorem (Absolute Convergence Implies Convergence): Suppose \(f\) is Riemann integrable on every closed, bounded subinterval of \([a,\infty)\). If \(\int_a^\infty |f(x)|\,dx\) converges, then \(\int_a^\infty f(x)\,dx\) converges. The statement also holds for an improper integral at a finite endpoint.

Proof. For finite \(S\geq R\geq a\), the absolute-value inequality for proper Riemann integrals gives

\left|\int_R^S f(x)\,dx\right| \leq\int_R^S |f(x)|\,dx.

Since the improper integral of \(|f|\) converges, the Cauchy criterion for improper integrals says that, for every \(\varepsilon>0\), there is \(A>a\) such that

\int_R^S |f(x)|\,dx<\varepsilon \quad\text{whenever }S\geq R\geq A.

The preceding inequality then yields

\left|\int_R^S f(x)\,dx\right|<\varepsilon \quad\text{whenever }S\geq R\geq A.

The Cauchy criterion applied to \(f\) proves that \(\int_a^\infty f(x)\,dx\) converges. At a finite singular endpoint, the same proof uses short truncated intervals in the endpoint form of the Cauchy criterion. \(\square\)

Worked Example: Absolute Convergence by Direct Comparison

Consider \(\int_1^\infty \frac{\sin x}{x^2}\,dx\). For \(x\geq1\), \(|\sin x|\leq1\), and therefore

0\leq\frac{|\sin x|}{x^2}\leq\frac{1}{x^2}.

The comparison integral \(\int_1^\infty 1/x^2\,dx\) converges. The Direct Comparison Test shows that \(\int_1^\infty |\sin x|/x^2\,dx\) converges, so the integral of \(\sin x/x^2\) converges absolutely and hence converges. This argument establishes convergence without needing to calculate the signed integral.

How to Choose and Use a Comparison

For powers, the results from the previous tutorial provide particularly useful benchmarks: \(\int_1^\infty x^{-p}\,dx\) converges exactly when \(p>1\), while \(\int_0^1 x^{-p}\,dx\) converges exactly when \(p<1\). A comparison near infinity should reflect the eventual size of the integrand there; a comparison near a finite singularity should reflect its size as the endpoint is approached. Limit comparison is often convenient when the ratio has a nonzero finite limit.

Two cautions prevent common errors. First, an upper bound by a divergent function proves nothing about convergence: a smaller function may still converge. A lower bound by a convergent function proves nothing about divergence for the same reason. Second, the limit comparison conclusion requires a positive finite limit. If the ratio tends to zero or infinity, the test as stated does not give equivalence; a one-sided direct comparison may still be available, but its direction must be checked.

Finally, absolute convergence is a sufficient condition, not a necessary one. The previous tutorial established that \(\int_1^\infty \sin x/x\,dx\) converges while the integral of its absolute value diverges. Thus comparison with \(|f|\) can prove convergence, but failure of absolute convergence does not by itself prove that the signed integral diverges.

Check Your Understanding

Use the comparison principles to decide what can be concluded in each situation.

If \(0\leq f(x)\leq g(x)\) eventually and \(\int_a^\infty g(x)\,dx\) converges, what can you conclude about \(\int_a^\infty f(x)\,dx\)?
If \(f(x)\geq g(x)\geq0\) eventually and \(\int_a^\infty g(x)\,dx\) diverges, why must the integral of \(f\) diverge?
What condition on \(\lim_{x\to\infty}f(x)/g(x)\) makes the Limit Comparison Test apply as stated?
Near zero, how could you use limit comparison to assess an integrand whose ratio to \(x^{-1/3}\) tends to a positive finite number?
What does convergence of \(\int_a^\infty |f(x)|\,dx\) imply about \(\int_a^\infty f(x)\,dx\), and why?