Why Mixed Improper Integrals Need a Local Analysis
An improper integral may be difficult to classify because its integrand behaves differently in different parts of the domain. It might have an endpoint singularity, an infinite tail, or several singular points inside a finite interval. A comparison that works in one region may say nothing about another.
The reliable approach is to identify each problematic region, analyze the integrand there, and combine the local conclusions. The integral converges only if every improper piece converges. A single divergent piece makes the whole integral divergent.
For a nonnegative integrand, the main comparison rules are these. If \(0\le f(x)\le g(x)\) near a problematic point and the corresponding integral of \(g\) converges, then the integral of \(f\) converges there. If \(0\le g(x)\le f(x)\) and the integral of \(g\) diverges, then the integral of \(f\) diverges there. In a limit comparison, if \(f\) and \(g\) are positive near the point and \(\lim f(x)/g(x)=L\) for some \(0<L<\infty\), their integrals have the same convergence behavior locally.
A Workshop Procedure
Before choosing a comparison function, mark every place where the integral is improper. Then analyze each marked region on its own. The useful reference cases are power integrals, logarithmic integrals, and functions with a clear dominant power. The comparison must be valid near the problematic point; behavior elsewhere does not establish local convergence.
Include infinite endpoints and every point where the integrand is unbounded or undefined.
Use separate one-sided integrals at each interior singularity, and isolate finite pieces on which the integrand is continuous.
Identify the dominant power, logarithmic factor, or radical behavior at each problematic point.
Check the inequality direction or the limit-comparison ratio, classify each piece, and then combine the results.
Worked Examples
Worked Example: An Endpoint Singularity and an Infinite Tail
Classify \[ \int_0^\infty \frac{\ln(1+x)}{\sqrt{x}(1+x)^2}\,dx. \]
The integrand is nonnegative. The two problematic regions are \(x=0\) and \(x\to\infty\). On every closed interval bounded away from zero and infinity, the integrand is continuous, so it remains to check those two regions.
Near zero, compare with \(g(x)=x^{1/2}\). The ratio is \[ \frac{\ln(1+x)}{\sqrt{x}(1+x)^2}\cdot\frac{1}{x^{1/2}} = \frac{\ln(1+x)}{x(1+x)^2} \longrightarrow 1 \quad (x\to0^+). \] The limit follows from \(\ln(1+x)/x\to1\) and \((1+x)^2\to1\). Since \(0<1<\infty\), limit comparison applies. The reference integral is \[ \int_0^1 x^{1/2}\,dx = \left[\frac{2}{3}x^{3/2}\right]_0^1 = \frac{2}{3}, \] which is finite. Thus the original integral converges near zero.
At infinity, compare with \(h(x)=\ln(x)/x^{5/2}\), taking \(x\ge2\). The ratio is \[ \frac{\ln(1+x)}{\sqrt{x}(1+x)^2}\cdot\frac{x^{5/2}}{\ln x} = \frac{\ln(1+x)}{\ln x}\cdot\frac{x^2}{(1+x)^2} \longrightarrow 1. \] Here \(\ln(1+x)/\ln x\to1\), and \(x^2/(1+x)^2\to1\). The reference tail converges. In fact, integration by parts gives \[ \int_2^\infty \frac{\ln x}{x^{5/2}}\,dx = \left[ -\frac{2}{3}x^{-3/2}\ln x-\frac{4}{9}x^{-3/2} \right]_2^\infty = 2^{-3/2}\left(\frac{2}{3}\ln2+\frac{4}{9}\right), \] a finite value. Therefore the original tail converges by limit comparison.
Check: The endpoint ratio and tail ratio both tend to \(1\), so each local comparison has a finite, positive limit. Both problematic pieces converge, and the integral over the intervening compact interval is finite. Hence the full integral converges.
Worked Example: A Convergent Logarithmic Endpoint
Classify \[ \int_0^{e^{-1}}\frac{dx}{x\left(1+\ln(1/x)\right)^{3/2}}. \]
The integrand is unbounded as \(x\to0^+\). Substitute \(t=1+\ln(1/x)\). Then \(dt=-dx/x\). At the upper endpoint \(x=e^{-1}\), \(t=2\); as \(x\to0^+\), \(t\to\infty\). Reversing the limits to account for the negative differential gives \[ \int_0^{e^{-1}}\frac{dx}{x\left(1+\ln(1/x)\right)^{3/2}} = \int_2^\infty \frac{dt}{t^{3/2}}. \] Evaluate the tail: \[ \int_2^\infty t^{-3/2}\,dt = \left[-2t^{-1/2}\right]_2^\infty = 0-\left(-\frac{2}{\sqrt{2}}\right) = \sqrt{2}. \] This value is finite, so the improper integral converges.
Check: Differentiating the antiderivative \(-2t^{-1/2}\) gives \(t^{-3/2}\), and the transformed limits are \(2\) and \(\infty\). The evaluation therefore checks the convergence at the original endpoint \(x=0\).
Worked Example: A Radical in a Logarithmic Tail
Classify \[ \int_e^\infty \frac{\sqrt{x^2+1}}{x^2(\ln x)^2}\,dx. \]
The only problematic region is the infinite tail. Compare the integrand \(f(x)\) with \(g(x)=1/(x(\ln x)^2)\), which is positive for \(x\ge e\). Their ratio is \[ \frac{f(x)}{g(x)} = \frac{\sqrt{x^2+1}}{x} = \sqrt{1+\frac{1}{x^2}} \longrightarrow 1 \quad (x\to\infty). \] Thus limit comparison is applicable. The reference integral is evaluated using \(u=\ln x\), \(du=dx/x\): \[ \int_e^\infty \frac{dx}{x(\ln x)^2} = \int_1^\infty \frac{du}{u^2} = \left[-\frac{1}{u}\right]_1^\infty = 1. \] It converges, so the original integral converges.
Check: The radical factor divided by \(x\) tends to \(1\), not to \(0\) or infinity. The positive finite ratio and the finite reference integral establish convergence of the tail.
Worked Example: Two Interior Singularities
Classify \[ \int_0^3\frac{dx}{\sqrt{|x-1||x-2|}}. \]
The integrand has singularities at \(x=1\) and \(x=2\). Split at both: \[ \int_0^3 f(x)\,dx = \int_0^1 f(x)\,dx+\int_1^2 f(x)\,dx+\int_2^3 f(x)\,dx, \qquad f(x)=\frac{1}{\sqrt{|x-1||x-2|}}, \] with one-sided improper limits at \(1\) and \(2\).
Near \(x=1\), compare with \(g_1(x)=|x-1|^{-1/2}\). On either side of \(1\), \[ \frac{f(x)}{g_1(x)} = \frac{1}{\sqrt{|x-2|}} \longrightarrow 1 \quad (x\to1). \] The reference integral is finite on either side: \[ \int_0^1(1-x)^{-1/2}\,dx = \left[-2\sqrt{1-x}\right]_0^1 = 2, \qquad \int_1^2(x-1)^{-1/2}\,dx = \left[2\sqrt{x-1}\right]_1^2 = 2. \] Thus the pieces adjacent to \(1\) converge locally at \(1\).
Near \(x=2\), compare with \(g_2(x)=|x-2|^{-1/2}\). This time, \[ \frac{f(x)}{g_2(x)} = \frac{1}{\sqrt{|x-1|}} \longrightarrow 1 \quad (x\to2). \] The reference integrals on either side are finite: \[ \int_1^2(2-x)^{-1/2}\,dx=2, \qquad \int_2^3(x-2)^{-1/2}\,dx=2. \] Each limit comparison therefore gives local convergence at \(2\). The endpoints \(0\) and \(3\) are not singular, and \(f\) is continuous there.
Check: The only singularities are \(1\) and \(2\); each has local power \(-1/2\), with a ratio tending to \(1\) to the corresponding power model. All one-sided pieces converge, so the full integral converges.
Worked Example: A Divergent Logarithmic Endpoint
Classify \[ \int_0^{e^{-1}}\frac{dx}{x\left(1+\ln(1/x)\right)}. \]
Use \(t=1+\ln(1/x)\), so \(dt=-dx/x\). The limits are \(t=2\) at \(x=e^{-1}\) and \(t\to\infty\) as \(x\to0^+\). Thus \[ \int_0^{e^{-1}}\frac{dx}{x\left(1+\ln(1/x)\right)} = \int_2^\infty\frac{dt}{t}. \] For a finite upper limit \(T\), \[ \int_2^T\frac{dt}{t} = [\ln t]_2^T = \ln T-\ln2 \longrightarrow\infty \quad (T\to\infty). \] The transformed integral diverges, so the original improper integral diverges.
Check: The truncated integral grows without bound rather than approaching a finite limit. Therefore convergence fails at \(x=0\).
Common Pitfall: Using the Wrong Comparison Direction
A divergent comparison function does not prove that a smaller function has a divergent integral. For example, for \(x\ge1\), \[ 0\le \frac{1}{x^2}\le\frac{1}{x}. \] It would be incorrect to say that \(\int_1^\infty 1/x^2\,dx\) diverges because \(\int_1^\infty 1/x\,dx\) diverges. The inequality has the wrong direction for that conclusion: an upper bound with a divergent integral gives no result. In fact, \[ \int_1^\infty\frac{dx}{x^2} = \left[-\frac{1}{x}\right]_1^\infty = 1, \] so the smaller function is integrable. To prove divergence by direct comparison, a divergent function must be a lower bound for the function being tested.
Check Your Understanding
For each question, identify the problematic region and state the comparison or method that would justify the classification.
- For \(\int_0^1 x^{-2/3}(1+x)^{-1}\,dx\), what power model is useful near zero, and does that local piece converge?
- What substitution turns \(\int_0^{e^{-1}} dx/[x(1+\ln(1/x))^p]\) into a power integral on an infinite interval?
- For \(\int_e^\infty \sqrt{x^2+4}/[x^2(\ln x)^2]\,dx\), what is the limit of the ratio to \(1/[x(\ln x)^2]\)?
- Why must an integral with an interior singularity be split into two one-sided improper integrals at that point?
- Does \(0\le f(x)\le 1/x\) near infinity prove that \(\int f(x)\,dx\) diverges? Explain why or why not.