A Classification Starts with Truncations
An improper integral can fail to be defined as an ordinary proper integral because its integrand is unbounded at an endpoint, its domain is unbounded, or both. Oscillation adds another possibility: positive and negative contributions may cancel enough for the improper integral to converge even when the integral of the absolute value diverges. Classification therefore begins by identifying each problematic endpoint and testing its contribution separately.
In what follows, functions are real-valued and locally Riemann integrable on the open intervals under discussion. The associated Lebesgue measure on \(\mathbb{R}\) is \(m\). A finite improper integral is called absolutely convergent when the corresponding improper integral of \(|f|\) is finite. For measurable \(f\), this is equivalent to Lebesgue integrability, \(\int |f|\,dm<\infty\); this connection between absolute improper convergence and Lebesgue integrability was established earlier.
A practical classification records three things for each piece: the truncation, a bound or convergence test, and whether the absolute-value integral converges. A finite split point is useful because a continuous or bounded piece over a compact interval cannot cause improper divergence.
Two Tools for Checking the Classification
Proof. Set \(F(T)=\int_a^T f(x)\,dx\) for \(T\ge a\). If \(F(T)\) has a finite limit as \(T\to\infty\), it is a Cauchy family: given \(\varepsilon>0\), choose \(A\) so that \(|F(T)-L|<\varepsilon/2\) for all \(T\ge A\), where \(L=\lim_{T\to\infty}F(T)\). Then for \(S>R\ge A\), additivity of the proper integral gives \(\left|\int_R^S f\right|=|F(S)-F(R)|\le |F(S)-L|+|F(R)-L|<\varepsilon\). Conversely, the stated condition says exactly that \(F(T)\) is Cauchy as \(T\to\infty\): for \(S,R\ge A\), order the two endpoints and use the absolute value. Completeness of \(\mathbb{R}\) gives a finite limit of \(F(T)\), which is the improper integral by definition. \(\square\)
For oscillatory tails, a useful result established earlier is Dirichlet’s test for improper integrals: if \(u\) is nonnegative, nonincreasing, and tends to zero at infinity, and the partial integrals \(B(T)=\int_a^T g(x)\,dx\) are bounded uniformly in \(T\ge a\), then \(\int_a^\infty u(x)g(x)\,dx\) converges. It need not converge absolutely. We will apply the test only after checking boundedness of the partial integrals and the conditions on \(u\).
Worked Example: A Logarithmic Singularity at Zero
Classify \(\int_0^1 \frac{\ln x}{\sqrt{x}}\,dx\) on \((0,1)\), with Lebesgue measure \(m\). The function is real-valued and continuous on every closed subinterval of \((0,1]\), but it is unbounded near zero. The truncation is \(\int_\varepsilon^1 \frac{\ln x}{\sqrt{x}}\,dx\), with \(\varepsilon\downarrow0\).
To test absolute convergence, the integrand has negative sign for \(0<x<1\), so \(\left|\frac{\ln x}{\sqrt{x}}\right|=\frac{-\ln x}{\sqrt{x}}\). Substitute \(x=u^2\), so \(dx=2u\,du\), \(\sqrt{x}=u\), and \(-\ln x=-2\ln u\). For \(0<\varepsilon<1\), this gives
At the upper endpoint, \(1\ln1-1=-1\). At the lower endpoint, \(\sqrt{\varepsilon}\ln\sqrt{\varepsilon}-\sqrt{\varepsilon}\to0\), since \(u\ln u\to0\) as \(u\downarrow0\). Thus the absolute integral is \(4\). The signed integral is \(-4\). Check: an antiderivative of \(\ln x/\sqrt{x}\) is \(2\sqrt{x}\ln x-4\sqrt{x}\); its value tends to \(0\) at zero and equals \(-4\) at one, again giving \(-4\). Hence this improper integral converges absolutely and its integrand is Lebesgue integrable.
Worked Example: An Oscillatory Infinite Tail
Classify \(\int_1^\infty \frac{\sin x}{x}\,dx\) on \([1,\infty)\) with Lebesgue measure \(m\). Its truncations are \(\int_1^R \sin x/x\,dx\). For \(1\le A<B\), integration by parts with \(u=1/x\) and \(dv=\sin x\,dx\) yields
Since \(|\cos x|\le1\), the absolute value of the right side is at most \(1/A+1/B+\int_A^B x^{-2}\,dx\). The last integral is \(1/A-1/B\), so the bound is \(2/A\). In particular, for any \(\varepsilon>0\), taking \(A>2/\varepsilon\) makes every tail integral from \(A\) to \(B>A\) smaller than \(\varepsilon\). The Cauchy criterion proves convergence.
To test absolute convergence, for each integer \(k\ge1\) use the disjoint intervals \(J_k=[k\pi+\pi/6,k\pi+5\pi/6]\). On \(J_k\), \(|\sin x|\ge1/2\), the interval length is \(2\pi/3\), and \(x\le k\pi+5\pi/6\). Therefore
The last inequality follows from \(k\pi+5\pi/6\le11k\pi/6\) for \(k\ge1\). The sum of these lower bounds diverges, so \(\int_1^\infty|\sin x|/x\,dx=\infty\). Thus the original integral converges conditionally, not absolutely, and its integrand is not Lebesgue integrable on \([1,\infty)\). Check: the tail estimate tends to zero as \(A\to\infty\), while the disjoint-interval lower bounds have a divergent harmonic sum; the two checks establish convergence and failure of absolute convergence separately.
Putting Endpoint and Tail Tests Together
Worked Example: One Singular Endpoint and One Oscillatory Tail
Define the real-valued function \(f\) on \((0,\infty)\) by \(f(x)=x^{-1/2}\) for \(0<x<1\), and \(f(x)=\sin x/x\) for \(x\ge1\). This function is locally Riemann integrable away from zero. Classify its improper integral on \((0,\infty)\) with Lebesgue measure \(m\), taking the endpoint truncation \(\varepsilon\downarrow0\) and the tail truncation \(R\to\infty\) independently.
For \(0<\varepsilon<1\), the contribution from the singular endpoint is computed directly:
The tail contribution \(\int_1^R\sin x/x\,dx\) converges as \(R\to\infty\) by the preceding example. Consequently the separately truncated integral is \(2-2\sqrt{\varepsilon}+\int_1^R\sin x/x\,dx\), and it has a finite limit as \(\varepsilon\downarrow0\) and \(R\to\infty\). Its value is \(2+\int_1^\infty\sin x/x\,dx\); no evaluation of the oscillatory constant is needed to classify the integral.
For absolute convergence, the endpoint piece contributes \(\int_0^1x^{-1/2}\,dx=2\), but the tail has \(\int_1^\infty|\sin x|/x\,dx=\infty\) by the interval estimate above. Thus the full improper integral is conditionally convergent, and \(f\) is not Lebesgue integrable on \((0,\infty)\). Check: the endpoint limit is \(2\), whereas the absolute tail is infinite; splitting the domain at \(1\) prevents the finite endpoint contribution from concealing the tail’s failure of absolute integrability.
A Common Classification Error
A tempting but incorrect calculation is: “Since the positive and negative lobes of \(\sin x\) cancel, \(\int_1^\infty|\sin x|/x\,dx\) converges.” The absolute value removes the signs, so those lobes do not cancel. The concrete intervals \(J_k\) above give contributions at least \(2/(11k)\), whose sum diverges. The correction is to test the signed integral and the absolute-value integral separately: Dirichlet’s test establishes signed convergence here, while the interval lower bounds rule out absolute convergence.
Another error is to merge endpoint limits before testing them. For an integral with two singular endpoints, define the improper integral as the sum of the two one-sided limits, each required to exist finitely. A cancellation between two divergent quantities is not a valid endpoint classification. Likewise, for an infinite interval, split at a finite point if needed and test the tail through its own truncations.
Mark every singular endpoint and every infinite tail; choose a finite split point between separate difficulties.
Use a small positive cutoff at a singular endpoint and a large finite cutoff for an infinite tail.
Use comparison for nonnegative bounds, and an oscillatory test such as Dirichlet’s test only after checking its hypotheses.
The full improper integral converges only when every independently defined piece has a finite limit; it is absolutely convergent only when the absolute-value integral is finite on every piece.
Check Your Understanding
For each question, specify the truncation and distinguish convergence of the signed integral from absolute convergence.
- For \(\int_0^1 x^{-1/3}\,dx\), compute the truncated integral and decide whether the convergence is absolute.
- In the oscillatory-tail example, why does the bound \(2/A\) establish convergence of \(\int_1^\infty\sin x/x\,dx\), while it does not establish absolute convergence?
- For the piecewise function in the final worked example, write its integral over \((\varepsilon,R)\) when \(0<\varepsilon<1<R\), and take the two limits independently.
- What fails if two singular endpoint contributions are allowed to cancel before their separate improper limits are checked?