One-Sided Convergence at Singular Endpoints

Truncating a Singular Endpoint

Suppose \(f\) is continuous on \((a,b]\), where \(a<b\) are finite real numbers, but is not assumed to have a finite limit or a defined value at \(a\). Every truncated integral \(\int_t^b f(x)\,dx\), for \(a<t<b\), is an ordinary Riemann integral on a compact interval. The one-sided improper integral asks whether these finite numbers approach a finite limit as the cutoff \(t\) decreases to \(a\). Existence of each truncated integral is not enough: it is their behavior as the endpoint is approached that matters.

Definition: For real-valued \(f\) continuous on \((a,b]\), the one-sided improper Riemann integral at \(a\) is said to converge to a finite value \(L\) if \(\lim_{t\downarrow a}\int_t^b f(x)\,dx=L\), with \(L\in\mathbb R\). If the limit is \(+\infty\), \(-\infty\), or does not exist, the improper integral does not converge to a finite value. The function is locally integrable away from \(a\) here if its restriction to every compact interval \([c,b]\), \(a<c<b\), is integrable there. Continuity on \((a,b]\) guarantees this local property, but says nothing by itself about convergence at \(a\).

The endpoint value, if one assigns any value at \(a\), has no effect on the Lebesgue integral with respect to Lebesgue measure \(m\) on \(\mathbb R\), because a singleton has measure zero. It also does not determine whether the improper Riemann limit exists. Both questions concern the behavior on intervals approaching the endpoint, not the value at one point.

The Endpoint Cauchy Criterion

Theorem: Let \(f:(a,b]\to\mathbb R\) be continuous, with \(a<b\) finite. The limit \(\lim_{t\downarrow a}\int_t^b f(x)\,dx\) exists as a finite real number if and only if for every \(\varepsilon>0\) there is a \(\delta>0\) such that \(\left|\int_u^v f(x)\,dx\right|<\varepsilon\) whenever \(a<u<v<a+\delta\) and \(v\le b\).

Proof. Put \(F(t)=\int_t^b f(x)\,dx\) for \(a<t\le b\). If \(u<v\), additivity of the ordinary integral gives \(F(u)-F(v)=\int_u^v f(x)\,dx\). Suppose first that \(F(t)\to L\in\mathbb R\) as \(t\downarrow a\). Given \(\varepsilon>0\), choose \(\delta>0\) so that \(a<t<a+\delta\) implies \(|F(t)-L|<\varepsilon/2\). For \(a<u<v<a+\delta\) with \(v\le b\), the triangle inequality yields \(\left|\int_u^v f(x)\,dx\right|=|F(u)-F(v)|\le |F(u)-L|+|F(v)-L|<\varepsilon\).

Conversely, suppose the stated Cauchy condition holds. For any two cutoffs \(u,v\) sufficiently close to \(a\), if \(u<v\), then \(|F(u)-F(v)|=|\int_u^v f(x)\,dx|<\varepsilon\); if \(v<u\), reversing their roles gives the same bound. Thus \(F(t)\) satisfies the Cauchy criterion as \(t\downarrow a\). More explicitly, take any sequence \(t_n\downarrow a\). The condition implies that \(F(t_n)\) is a Cauchy sequence in \(\mathbb R\), so it has a finite limit \(L\). To see that the full cutoff limit is \(L\), fix \(\varepsilon>0\) and use the criterion with \(\varepsilon/2\). For any \(t\) close enough to \(a\), compare it with \(t_n\) close enough to \(a\), obtaining \(|F(t)-F(t_n)|<\varepsilon/2\). Taking \(n\to\infty\) gives \(|F(t)-L|\le\varepsilon/2<\varepsilon\). Hence \(F(t)\to L\) as \(t\downarrow a\). \(\square\)

This criterion tests cancellation across every sufficiently short interval near the endpoint. It is stronger than checking that a chosen sequence of truncated integrals has a limit: the cutoff is a real variable, so all cutoffs near \(a\) must be controlled.

Worked Example: Local Integrability but Divergence

On \((0,1]\), let \(f(x)=1/x\). The function is continuous and bounded on every \([c,1]\) with \(c>0\), so it is integrable on each such interval. For \(0<t<1\), compute the truncation:

\int_t^1\frac{1}{x}\,dx=[\ln x]_t^1=\ln 1-\ln t=-\ln t.

As \(t\downarrow0\), \(-\ln t\to+\infty\), so the one-sided improper integral does not converge to a finite value. In particular, choosing \(t=e^{-N}\), where \(N>0\), gives \(\int_{e^{-N}}^1 dx/x=N\), which is unbounded as \(N\to\infty\). Check: differentiating \(\ln x\) gives \(1/x\), and substituting \(x=1\) and \(x=t\) into that antiderivative reproduces \(-\ln t\). Thus integrability on every compact interval away from the endpoint is not endpoint convergence.

Nonnegative Truncations and the Lebesgue Integral

For a nonnegative function the truncated integrals have additional structure: making the cutoff smaller can only increase the integral. This rules out cancellation and gives a direct criterion for a finite limit. The Lebesgue formulation makes this precise. In the statement below, \(\mathcal L\) is the Lebesgue sigma-algebra on \(\mathbb R\), and \(m\) is Lebesgue measure.

Proposition: Let \(f:(a,b]\to[0,\infty]\) be \(\mathcal L\)-measurable, and suppose its restriction to every \([c,b]\), \(a<c<b\), is Riemann integrable. For \(t\in(a,b)\), set \(I(t)=\int_t^b f(x)\,dx\). Then \(I(t)\) has a limit in \([0,\infty]\) as \(t\downarrow a\), and \(\int_{(a,b]} f\,dm=\lim_{t\downarrow a} I(t)\). The limit is finite if and only if the Lebesgue integral on \((a,b]\) is finite.

Proof. If \(a<u<v<b\), nonnegativity and additivity give \(I(u)=\int_u^v f(x)\,dx+I(v)\ge I(v)\). Thus \(I(t)\) increases as \(t\) decreases, and it has a limit in \([0,\infty]\), equal to the supremum of its values. Choose \(t_n=a+(b-a)/n\) for integers \(n\ge2\); then \(t_n\downarrow a\). On \((a,b]\), define \(f_n=f\chi_{[t_n,b]}\). These measurable functions are nonnegative, satisfy \(f_n\le f_{n+1}\), and converge pointwise to \(f\): for every \(x>a\), eventually \(t_n\le x\), so \(f_n(x)=f(x)\). The Monotone Convergence Theorem therefore gives \(\int_{(a,b]}f\,dm=\lim_n\int_{(a,b]}f_n\,dm\). On each compact interval \([t_n,b]\), equality of the Riemann and Lebesgue integrals for an integrable function gives \(\int f_n\,dm=I(t_n)\). Hence \(\int_{(a,b]}f\,dm=\lim_n I(t_n)\). Since \(I(t)\) is monotone as the cutoff decreases, its full limit equals its limit along \(t_n\). Finally, a value in \([0,\infty]\) is finite exactly when it is less than \(+\infty\), proving the last assertion. \(\square\)

The proposition uses nonnegativity essentially: it ensures monotonicity of the truncated integrals and the pointwise increase of the truncations. It does not provide a corresponding monotonicity argument for signed functions, where positive and negative contributions can cancel.

Worked Example: A Finite Nonnegative Endpoint Integral

On \((0,1]\), let \(f(x)=x^{-1/2}\). This is nonnegative and continuous away from \(0\). For each \(t>0\), direct integration gives

\int_t^1 x^{-1/2}\,dx=[2\sqrt{x}]_t^1=2-2\sqrt{t}.

As \(t\downarrow0\), \(\sqrt t\downarrow0\), so the truncated integrals tend to \(2\). The proposition applies: \(f\) is measurable and nonnegative, and it is Riemann integrable on every \([c,1]\) with \(c>0\). It follows that \(\int_{(0,1]}x^{-1/2}\,dm=2\). Check: at \(t=1/4\), the formula gives \(2-2(1/2)=1\); integrating the derivative of \(2\sqrt{x}\), which is \(x^{-1/2}\) for \(x>0\), gives the same truncated value. The limit \(2\) is consistent with the explicit truncation formula.

Worked Example: An Oscillating Truncation with No Limit

Let \(f(x)=\cos(\ln x)/x\) on \((0,1]\). This real-valued function is continuous on every \([c,1]\), so every truncated integral exists. Since \(\frac{d}{dx}\sin(\ln x)=\cos(\ln x)/x\),

\int_t^1\frac{\cos(\ln x)}{x}\,dx =\sin(\ln 1)-\sin(\ln t) =-\sin(\ln t).

Consider the two cutoff sequences \(t_n=e^{-2\pi n}\) and \(s_n=e^{-(\pi/2+2\pi n)}\), for integers \(n\ge1\). Both tend to \(0\). For the first, \(\ln t_n=-2\pi n\), so the truncated integral is \(-\sin(-2\pi n)=0\). For the second, \(\ln s_n=-\pi/2-2\pi n\), so the truncated integral is \(-\sin(-\pi/2-2\pi n)=1\). The two sequences of truncated values have different limits, so the improper integral has no finite limit. Check: the antiderivative calculation differentiates back to \(f\), and the two cutoff sequences produce exactly \(0\) and \(1\), respectively; therefore no single cutoff limit is possible.

Worked Example: Convergence Through an Explicit Primitive

On \((0,1]\), define the signed continuous function \(g(x)=\sin(1/x)-\cos(1/x)/x\). Direct differentiation gives \(\frac{d}{dx}\bigl(x\sin(1/x)\bigr)=\sin(1/x)-\cos(1/x)/x=g(x)\). Therefore, for \(0<t<1\),

\int_t^1 g(x)\,dx =\sin(1)-t\sin(1/t).

Since \(|\sin(1/t)|\le1\), the magnitude of \(t\sin(1/t)\) is at most \(t\), which tends to \(0\). Thus the truncated integral tends to \(\sin(1)\), a finite value. Check: substituting \(t=1/2\) gives \(\sin(1)-\frac12\sin(2)\), exactly the difference \(1\cdot\sin(1)-\frac12\sin(2)\) of the primitive at the two endpoints. The bound by \(t\) verifies the endpoint limit without interchanging a limit and an integral.

A Common Error and What Truncation Really Shows

A common mistake is to infer convergence from the fact that every truncated integral is finite. For \(f(x)=1/x\), each integral \(\int_t^1 f(x)\,dx=-\ln t\) is finite for \(t>0\), but these values grow without bound as \(t\downarrow0\). The correction is to keep the cutoff variable in view and test its limit—or, equivalently, use the endpoint Cauchy criterion. Individual truncated integrals only establish that the problem is well-defined away from the endpoint.

There are two distinct questions. First, is the function integrable on each compact interval that stays away from the endpoint? Second, do the truncated integrals converge as the endpoint is approached? Continuity on \((a,b]\) answers the first question, but Examples 1 and 3 show that it does not settle the second. For nonnegative functions, monotone truncation gives an extended limit and identifies it with the Lebesgue integral. For signed functions, truncations must be analyzed as signed quantities; the existence of a finite limit depends on their net behavior near the endpoint.

Check Your Understanding

For each question, distinguish the behavior on compact intervals away from the endpoint from the limit of the truncated integrals.

State the endpoint Cauchy criterion using \(\int_u^v f(x)\,dx\), including the order of \(u\) and \(v\).
For \(f(x)=x^{-1/2}\) on \((0,1]\), compute \(\int_t^1 f(x)\,dx\) and identify its limit as \(t\downarrow0\).
Why does the finiteness of \(\int_t^1 dx/x\) for each fixed \(t>0\) not imply convergence of the improper integral?
For a nonnegative measurable \(f\), explain why the truncations \(f\chi_{[t_n,b]}\), with \(t_n\downarrow a\), increase pointwise to \(f\) on \((a,b]\).
For \(f(x)=\cos(\ln x)/x\), give two cutoff sequences tending to \(0\) along which the truncated integrals have different limits.