Comparison Tests for Improper Integrals

Comparing Integrals Near an Improper Endpoint

When an integrand is nonnegative, its integral over a growing interval cannot decrease. This makes inequalities between integrands especially useful: a simpler function can bound the truncated integrals and settle whether they remain finite or grow without bound. The comparison must be made near the endpoint in question. It does not, by itself, determine what happens at another improper endpoint.

We work with Lebesgue measure \(m\) on an interval \(I\subseteq\mathbb R\). Its Lebesgue sigma-algebra is denoted by \(\mathcal L\); on a finite interval this measure is finite, and on an unbounded interval it is sigma-finite. The functions in the comparison tests are real-valued, nonnegative, and continuous in the interior of \(I\). Thus they are measurable and integrable on every compact subinterval of \(I\). On those compact subintervals their Lebesgue integrals agree with their Riemann integrals. All improper integrals below are interpreted as limits of these truncated integrals, and may equal \(+\infty\).

Changing the interior point \(c\) does not affect convergence near the endpoint: the difference between two truncation choices is an integral over a compact interval, which is finite by continuity. A full integral with two improper endpoints converges only if the integral near each endpoint is finite. Indeed, splitting at any interior point expresses the nonnegative full integral as the sum of its two endpoint integrals. This local viewpoint is essential in applying comparison tests.

The Direct Comparison Test

Proof. Choose \(d\) close enough to \(b\) that the stated inequality holds on \((d,b)\). In the upper-bound case, for every \(t\in(d,b)\), monotonicity of the integral gives \(\int_d^t f\,dm\le C\int_d^t g\,dm\). If the integral of \(g\) near \(b\) is finite, the right side is bounded above independently of \(t\). Therefore the nondecreasing truncated integrals of \(f\) have a finite limit. In the lower-bound case, \(\int_d^t f\,dm\ge c\int_d^t g\,dm\). If the truncated integrals of \(g\) tend to \(+\infty\), then those of \(f\) do too. The portions between \(c\) and \(d\), when included, are finite by continuity, so they do not change either conclusion. For a left endpoint, the same argument applies to truncations \((t,d)\) as \(t\) decreases to that endpoint. \(\square\)

The inequality direction matters. An upper bound by a function with a finite integral proves finiteness; a lower bound by a function with an infinite integral proves divergence. The other two combinations do not settle the question.

The Limit Comparison Test

Proof. Choose any \(\varepsilon\) with \(0<\varepsilon<L\). By the definition of the limit, there is a neighborhood of \(b\) on which \(L-\varepsilon< f(x)/g(x)<L+\varepsilon\). Since \(g(x)>0\) there, multiplication by \(g(x)\) gives \((L-\varepsilon)g(x)<f(x)<(L+\varepsilon)g(x)\). If the integral of \(g\) converges near \(b\), the upper inequality and the direct comparison test imply that the integral of \(f\) converges there. If the integral of \(g\) diverges near \(b\), the lower inequality and direct comparison imply that the integral of \(f\) diverges there. These two implications establish the claimed equivalence. The argument takes place only in a neighborhood of \(b\); it gives no bound on an integral near a different endpoint. \(\square\)

When the ratio limit is zero or \(+\infty\), the two-sided conclusion is unavailable. The same inequalities give useful one-sided conclusions: if \(f/g\to0\), then \(f\le g\) sufficiently near the endpoint, so convergence of \(g\) implies convergence of \(f\). If \(f/g\to+\infty\), then \(f\ge g\) sufficiently near the endpoint, so divergence of \(g\) implies divergence of \(f\). In the other cases these bounds alone do not decide convergence.

Endpoint Locality and Common Pitfalls

A ratio limit near one endpoint does not automatically compare full improper integrals. For instance, define \(g(x)=1\) on \((0,\infty)\), and define \(f(x)=1\) for \(0<x\le1\) and \(f(x)=e^{-(x-1)}\) for \(x>1\). Both functions are nonnegative and continuous on \((0,\infty)\). As \(x\downarrow0\), \(f(x)/g(x)=1\), so their integrals agree in convergence status near \(0\). But their full integrals differ: \(\int_0^\infty g(x)\,dx=+\infty\), whereas \(\int_0^\infty f(x)\,dx=\int_0^1 1\,dx+\int_1^\infty e^{-(x-1)}\,dx=1+1=2\). The exponential tail equals \(1\), as is seen by setting \(u=x-1\) and evaluating \(\int_0^\infty e^{-u}\,du=1\). The ratio test near zero says nothing about the distinct behavior near infinity.

Common mistake. A mistaken argument might say, “\(f/g\to1\) at zero, so the full integrals of \(f\) and \(g\) both converge or both diverge.” The example above disproves that conclusion: the full integral of \(f\) is \(2\), while that of \(g\) is infinite. The correction is to state the conclusion locally—both converge or both diverge near zero—and then check the other endpoint separately. Similarly, a principal value or cancellation in a signed integral cannot be established by these tests: they rely on nonnegative integrands and ordinary improper limits, not on cancellation between positive and negative parts.

For an integral with several improper endpoints, apply a comparison near each one. On any compact interior portion, continuity ensures a finite integral. The full nonnegative integral is finite exactly when the contributions from all its improper endpoints are finite. This procedure keeps endpoint behavior separate and prevents a valid local comparison from being used beyond its scope.