Direct Comparisons Near Singular Endpoints

Comparing an Integrand Near a Finite Endpoint

A finite endpoint singularity is controlled by what the integrand does arbitrarily close to the endpoint. The value of the integrand at the endpoint itself is not the issue: an improper integral is defined by taking a limit of proper integrals that stop short of the singular point.

A useful way to classify many such integrals is to compare the integrand with a power of the distance to the endpoint. At a left endpoint \(a\), that distance is \(x-a\); at a right endpoint \(b\), it is \(b-x\). The key is to choose the comparison direction correctly: an upper bound by an integrable function proves convergence, while a lower bound by a nonintegrable function proves divergence.

Definition: Suppose \(f\) is nonnegative and continuous on \((a,a+\delta]\), where \(\delta>0\). The improper integral \(\int_a^{a+\delta}f(x)\,dx\) is the limit \(\lim_{\varepsilon\to0^+}\int_{a+\varepsilon}^{a+\delta}f(x)\,dx\), if that limit is finite. At a right endpoint \(b\), define \(\int_{b-\delta}^b f(x)\,dx\) using \(\lim_{\varepsilon\to0^+}\int_{b-\delta}^{b-\varepsilon}f(x)\,dx\).

In this tutorial, comparisons are made for nonnegative integrands. If an integrand changes sign, the corresponding question of absolute convergence concerns the nonnegative function \(|f|\).

The Power Threshold at a Finite Endpoint

The basic reference integral at a left endpoint is \(\int_0^\delta t^{-p}\,dt\), where \(t\) is the distance to the endpoint and \(\delta>0\). Truncating at \(\varepsilon>0\), for \(p\ne1\) we have

\int_\varepsilon^\delta t^{-p}\,dt = \frac{\delta^{1-p}-\varepsilon^{1-p}}{1-p}.

When \(p<1\), the term \(\varepsilon^{1-p}\) tends to zero, so the limit is finite. When \(p>1\), \(\varepsilon^{1-p}\) grows without bound as \(\varepsilon\to0^+\), and the integral diverges. At \(p=1\), direct integration gives a logarithm that also diverges.

\int_\varepsilon^\delta \frac{1}{t}\,dt = \ln\delta-\ln\varepsilon \longrightarrow+\infty \quad\text{as }\varepsilon\to0^+.

Thus the threshold is \(p=1\). A power with exponent below \(1\) is integrable near the endpoint; a power with exponent at least \(1\) is not. For a right endpoint \(b\), the substitution \(t=b-x\) gives the same threshold for powers of \(b-x\).

Theorem: Let \(f\) be nonnegative and continuous on \((a,a+\delta]\). If \(f(x)\le C(x-a)^{-p}\) there, for a constant \(C>0\) and an exponent \(p<1\), then \(\int_a^{a+\delta}f(x)\,dx\) converges. If \(f(x)\ge c(x-a)^{-p}\) there, for \(c>0\) and \(p\ge1\), then \(\int_a^{a+\delta}f(x)\,dx\) diverges. The corresponding statements at a right endpoint use \(b-x\) in place of \(x-a\).

Why the Comparison Rules Work

The argument comes from comparing truncated integrals, where all the integrals are proper. For convergence, suppose \(0\le f(x)\le C(x-a)^{-p}\) and \(p<1\). For every \(\varepsilon>0\),

0\le \int_{a+\varepsilon}^{a+\delta}f(x)\,dx \le C\int_{a+\varepsilon}^{a+\delta}(x-a)^{-p}\,dx.

The comparison integral has a finite limit as \(\varepsilon\to0^+\), by the power calculation above. The truncated integrals of \(f\) increase as \(\varepsilon\) decreases, because \(f\ge0\), and they are bounded above by a quantity with a finite limit. Therefore their limit is finite.

For divergence, suppose instead that \(f(x)\ge c(x-a)^{-p}\) with \(p\ge1\). Then

\int_{a+\varepsilon}^{a+\delta}f(x)\,dx \ge c\int_{a+\varepsilon}^{a+\delta}(x-a)^{-p}\,dx.

The right-hand side tends to \(+\infty\), so the left-hand side cannot have a finite limit. This proves divergence. The proof at a right endpoint is identical after replacing \(x-a\) by \(b-x\).

A Practical Comparison Process

First isolate the endpoint that may cause trouble. If the integrand is continuous on every closed subinterval away from that endpoint, the part of the integral away from it is finite and does not change the convergence classification. Then express the distance to the endpoint as a variable or factor, and bound the remaining expression using inequalities valid throughout a sufficiently small interval.

Locate the endpoint and its distance.
For a left endpoint \(a\), use \(x-a\); for a right endpoint \(b\), use \(b-x\).

Find a one-sided bound.
For convergence, seek an upper bound by a power with exponent below \(1\). For divergence, seek a lower bound by a power with exponent at least \(1\).

Check the inequality and apply the threshold.
Verify the bound on the whole interval being compared, then use the power-integral result.

Worked Examples

Worked Example: A Shifted Numerator at the Left Endpoint

Classify \(\int_0^1 \frac{2+x}{\sqrt{x}(1+x)}\,dx\).

The possible singularity is at \(x=0\), and the distance to this endpoint is \(x\). For \(0<x\le1\), \(2+x\le3\) and \(1+x\ge1\). All quantities are positive, so

0\le \frac{2+x}{\sqrt{x}(1+x)} \le \frac{3}{\sqrt{x}} =3x^{-1/2}.

The comparison exponent is \(p=\frac12<1\). In fact,

\int_0^1 3x^{-1/2}\,dx = \lim_{\varepsilon\to0^+}3\left[2\sqrt{x}\right]_{\varepsilon}^{1} = \lim_{\varepsilon\to0^+}6(1-\sqrt{\varepsilon}) =6.

The comparison theorem therefore proves convergence. As a direct check, evaluate the original integral using \(x=u^2\), so \(dx=2u\,du\) and \(\sqrt{x}=u\). Its truncated endpoint can be handled by taking the corresponding limit \(u\to0^+\):

\int_0^1 \frac{2+x}{\sqrt{x}(1+x)}\,dx = \int_0^1 2\frac{2+u^2}{1+u^2}\,du = \int_0^1\left(2+\frac{2}{1+u^2}\right)\,du = 2+\frac{\pi}{2}.

Differentiating \(2u+2\arctan u\) gives \(2+2/(1+u^2)\), confirming the antiderivative used in the evaluation. The exact value is finite, as the comparison predicted.

Worked Example: A Shifted Denominator at the Left Endpoint

Classify \(\int_0^1 \frac{1+x}{x(2+x)}\,dx\).

For \(0<x\le1\), \(1+x\ge1\) and \(2+x\le3\). Since \(x>0\), these inequalities give a lower bound:

\frac{1+x}{x(2+x)} \ge \frac{1}{3x}.

The comparison power is \(x^{-1}\), whose exponent \(p=1\) is at the divergent threshold. Indeed,

\int_\varepsilon^1 \frac{1}{3x}\,dx = \frac13\bigl(\ln 1-\ln\varepsilon\bigr) = -\frac13\ln\varepsilon \longrightarrow+\infty.

Therefore the original integral diverges. We can verify this conclusion directly. Partial fractions give

\frac{1+x}{x(x+2)} = \frac{1}{2x}+\frac{1}{2(x+2)},

because the numerator on the right is \(\frac12(x+2)+\frac12x=x+1\). Thus

\int_\varepsilon^1\frac{1+x}{x(2+x)}\,dx = \frac12\left[-\ln\varepsilon+\ln3-\ln(2+\varepsilon)\right].

As \(\varepsilon\to0^+\), the term \(-\frac12\ln\varepsilon\) tends to \(+\infty\), while the other logarithmic terms have finite limits. This confirms divergence.

Worked Example: A Factored Singularity at a Shifted Right Endpoint

Classify \(\int_1^4 \frac{x+2}{(x+1)\sqrt{4-x}}\,dx\).

The singular endpoint is \(x=4\), so the distance to it is \(4-x\). For \(1\le x<4\),

\frac{x+2}{x+1} = 1+\frac{1}{x+1}, \qquad 0<\frac{x+2}{x+1}\le\frac32.

Here the upper bound follows from \(x+1\ge2\), so \(1/(x+1)\le1/2\). Consequently,

0\le \frac{x+2}{(x+1)\sqrt{4-x}} \le \frac32(4-x)^{-1/2}.

With \(t=4-x\), the comparison integral is

\int_1^4 \frac32(4-x)^{-1/2}\,dx = \frac32\int_0^3 t^{-1/2}\,dt = \frac32(2\sqrt3) = 3\sqrt3.

Its exponent is \(1/2<1\), so the original integral converges. For an additional check, an antiderivative of the integrand is

F(x) = -2\sqrt{4-x} -\frac{1}{\sqrt5} \ln\left( \frac{\sqrt5+\sqrt{4-x}}{\sqrt5-\sqrt{4-x}} \right).

To verify it, set \(t=\sqrt{4-x}\). Differentiating the first term gives \(1/t\). Differentiating the negative logarithmic term gives \(1/[t(5-t^2)]\). Their sum is \((6-t^2)/[t(5-t^2)]\), which equals the integrand because \(x+2=6-t^2\), \(x+1=5-t^2\), and \(\sqrt{4-x}=t\). Taking endpoint values gives

\int_1^4 \frac{x+2}{(x+1)\sqrt{4-x}}\,dx = 2\sqrt3+ \frac{1}{\sqrt5} \ln\left(\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\right),

a finite value, in agreement with the comparison.

Bounds That Do Not Decide the Question

A comparison only proves what its direction and reference integral allow. An upper bound by a divergent function does not prove divergence, and a lower bound by a convergent function does not prove convergence. Even a valid bound can be inconclusive if it uses the wrong power threshold.

For example, the calculation

0<x\le1 \quad\Longrightarrow\quad x^{-1/2}\le x^{-1}

is true. But using it to claim that \(\int_0^1x^{-1/2}\,dx\) diverges because \(\int_0^1x^{-1}\,dx\) diverges is an error: it is an upper bound by a divergent function, which gives no divergence conclusion. Direct integration shows the correction:

\int_0^1x^{-1/2}\,dx = \lim_{\varepsilon\to0^+}2(1-\sqrt{\varepsilon}) = 2.

To prove convergence by comparison, the bound must place the integrand below a convergent power. To prove divergence, it must place the integrand above a divergent power. If neither useful bound is available, this direct-comparison method has not classified the integral; that does not mean the integral converges or diverges.

Why Local Bounds Are Enough

Suppose an integrand has an endpoint singularity at \(a\), but is continuous on \([a+\delta,b]\) for some \(\delta>0\). Its integral over \([a+\delta,b]\) is finite, so the convergence question for the whole interval reduces to its behavior on \((a,a+\delta]\). This is why a bound need only hold sufficiently near the singular endpoint, rather than across the entire original interval.

Factored expressions are often especially suited to this method. A factor such as \((x-a)^{-p}\) displays the endpoint power directly; the remaining factors can then be bounded above or below by constants on a short interval. When comparing, check that every factor is positive before multiplying inequalities, and ensure that the resulting bound holds throughout the interval used.

Check Your Understanding

For each question, identify the comparison direction and the endpoint power threshold that apply.

What is the convergence threshold for \(\int_0^\delta t^{-p}\,dt\), and what happens at \(p=1\)?
Near a left endpoint \(a\), what kind of power bound proves convergence for a nonnegative integrand?
Near a right endpoint \(b\), what distance-to-endpoint expression should replace \(x-a\)?
Why does a lower bound by a divergent power prove divergence?
Does an upper bound by a divergent power prove that the integrand's improper integral diverges? Explain.