Why the Endpoints Must Be Tested Separately
When an integrand is unbounded at both ends of a finite interval, the integral has two potential sources of divergence. It is not enough to check that a calculation with cutoffs gives a finite-looking answer. The behavior near the left endpoint and the behavior near the right endpoint must each be examined.
This separation is essential to the definition of an ordinary improper integral. A cancellation between the two ends, even if it occurs when cutoffs are tied together, cannot establish convergence at either endpoint by itself.
Definition Using a Split Point
Suppose \(f\) is continuous on the open interval \((a,b)\), where \(a<b\), but may be unbounded as \(x\) approaches \(a\) or \(b\). Choose any fixed point \(c\) with \(a<c<b\). The interval is divided at \(c\), away from both singular endpoints.
The two limits are independent. In the left-hand integral, only \(u\) approaches \(a\); the right endpoint of that piece remains fixed at \(c\). In the right-hand integral, only \(v\) approaches \(b\); its left endpoint remains fixed at \(c\). There is no requirement to make the two cutoffs approach their endpoints at the same rate.
Why This Definition Does Not Depend on the Split Point
The point \(c\) is a convenient way to separate the endpoint questions, not an additional feature of the integral's value. Here is why a different split point gives the same result whenever both endpoint limits converge.
Let \(d\) be another point with \(a<d<b\), and first suppose \(c<d\). For cutoffs \(u<c\) and \(v>d\), additivity of ordinary definite integrals gives
and
The finite middle integral \(\int_c^d f(x)\,dx\) exists because \(f\) is continuous on \([c,d]\). Taking the left-endpoint limit in the first equality shows that changing the left piece from ending at \(c\) to ending at \(d\) adds exactly \(\int_c^d f(x)\,dx\). Taking the right-endpoint limit in the second equality shows that changing the right piece from starting at \(c\) to starting at \(d\) removes exactly that same finite quantity. Thus the sum is unchanged. If \(d<c\), the same argument applies with the roles of \(c\) and \(d\) reversed.
This also explains the convergence test: changing the split point only changes a piece by a finite integral over an interval on which the integrand is continuous. It cannot repair a divergent endpoint limit.
Using an Antiderivative
If \(F'(x)=f(x)\) on \((a,b)\), the antiderivative rule for ordinary definite integrals applies on every closed subinterval \([u,c]\) and \([c,v]\) contained in \((a,b)\). Consequently, the endpoint pieces can be evaluated by taking their limits separately:
provided both limits are finite. The \(F(c)\) terms then cancel, leaving the difference between the two endpoint limits of \(F\). This cancellation is justified only after each one-sided limit has been checked.
Worked Example: A Square-Root Singularity at Each Endpoint
Evaluate \(\int_0^1 \frac{1}{\sqrt{x(1-x)}}\,dx\) as an ordinary improper integral.
Choose \(c=\frac12\). On \(0<x<1\), define \(F(x)=\arcsin(2x-1)\). By the chain rule,
For \(0<u<\frac12\), the left piece is
As \(u\to0^+\), \(2u-1\to-1\), so the left piece tends to \(0-(-\frac{\pi}{2})=\frac{\pi}{2}\). For \(\frac12<v<1\), the right piece is
As \(v\to1^-\), \(2v-1\to1\), so the right piece tends to \(\frac{\pi}{2}-0=\frac{\pi}{2}\). Each endpoint limit is finite, so the integral converges, and their sum is
As a check, the endpoint limits of the antiderivative are \(F(0^+)=-\frac{\pi}{2}\) and \(F(1^-)=\frac{\pi}{2}\). Their difference is \(\pi\), matching the separately computed pieces.
Power Behavior at Both Ends
The power singularity threshold gives a useful test when an integrand behaves like a power near each endpoint. For \(p>0\), the integral \(\int_0^r x^{-p}\,dx\), where \(r>0\), converges exactly when \(p<1\). The same threshold applies to \((b-x)^{-q}\) near a right endpoint \(b\): its integral converges there exactly when \(q<1\).
For instance, suppose an integrand has the form \(x^{-p}(1-x)^{-q}\) on \((0,1)\), with \(p,q>0\). Near \(x=0\), the factor \((1-x)^{-q}\) stays bounded and approaches \(1\), so the left-endpoint behavior is determined by \(x^{-p}\). Near \(x=1\), the factor \(x^{-p}\) stays bounded and approaches \(1\), so the right-endpoint behavior is determined by \((1-x)^{-q}\). Thus the integral converges exactly when \(p<1\) and \(q<1\). Both conditions matter: passing one endpoint test does not imply passing the other.
Worked Example: Testing Two Different Power Singularities
Determine whether \(\int_0^1 \left(x^{-1/2}+(1-x)^{-1/3}\right)\,dx\) converges, and evaluate it if it does.
At \(x=0\), the term \(x^{-1/2}\) is unbounded, but \((1-x)^{-1/3}\) approaches \(1\). At \(x=1\), the term \((1-x)^{-1/3}\) is unbounded, but \(x^{-1/2}\) approaches \(1\). The power thresholds are \(1/2<1\) and \(1/3<1\), so the respective endpoint singularities are integrable.
For an explicit evaluation, an antiderivative on \((0,1)\) is
Indeed, differentiating the first term gives \(x^{-1/2}\), and differentiating the second gives \((1-x)^{-1/3}\). With \(c=\frac12\), the left piece has limit \(F(\frac12)-F(0^+)\), and the right piece has limit \(F(1^-)-F(\frac12)\). Since \(F(0^+)=-\frac32\) and \(F(1^-)=2\), both limits are finite. Adding them gives
The split-point check is visible in the sum: \(F(\frac12)\) occurs once with a positive sign and once with a negative sign, so it cancels. Differentiating the displayed \(F\) recovers the original integrand.
One Failed Endpoint Test Is Enough
Because the definition requires two finite limits, divergence at either endpoint makes the ordinary integral diverge. A finite contribution from the other endpoint cannot compensate for it. If both endpoint pieces diverge, they remain two divergent pieces; a coordinated cutoff does not change that definition.
Worked Example: A Positive Integrand That Diverges at Both Ends
Test \(\int_0^1 \frac{1}{x(1-x)}\,dx\). The integrand is positive on \((0,1)\) and becomes unbounded at both endpoints. The identity
shows directly that each endpoint has a nonintegrable power singularity: the left piece includes \(1/x\), and the right piece includes \(1/(1-x)\). More explicitly, splitting at \(c=\frac12\) and using the antiderivative \(F(x)=\ln x-\ln(1-x)\), the left piece is
The right piece is
Both endpoint limits diverge, so the ordinary improper integral diverges. To see why using a shared cutoff does not help, integrate from \(\varepsilon\) to \(1-\varepsilon\), with \(0<\varepsilon<\frac12\):
This check agrees with the separate tests. More generally, even a finite shared-cutoff result would not replace the two limits in the definition.
A Common Mistake: Subtracting Endpoint Infinities
A common error is to find an antiderivative, substitute both singular endpoints, and treat the result as though it were an ordinary endpoint difference. For the previous example, the antiderivative is \(F(x)=\ln x-\ln(1-x)\). As \(x\to0^+\), \(F(x)\to-\infty\); as \(x\to1^-\), \(F(x)\to+\infty\). Writing “\(+\infty-(-\infty)\)” is not a finite integral evaluation. It signals divergence, which the separate limits confirm.
Why Separate Tests Matter
For an integral with singularities at both ends, the split-point definition gives a dependable procedure:
Fix \(c\) strictly between the endpoints so each singularity belongs to its own one-sided integral.
Let only the lower limit approach \(a\), and check whether that limit is finite.
Let only the upper limit approach \(b\), and check whether that limit is finite.
If both limits are finite, add them. If either fails, the ordinary improper integral diverges.
This procedure applies whether the tests use a known antiderivative or a comparison with a simpler power. It prevents a finite-looking combined calculation from obscuring failure at an individual endpoint.
Check Your Understanding
For each question, keep the left and right endpoint limits distinct.
- State the definition of convergence for an integral whose integrand is unbounded at both endpoints.
- For \(\int_0^1 x^{-3/4}(1-x)^{-2/3}\,dx\), what power threshold applies at each endpoint, and does the integral converge?
- For \(\int_0^1 x^{-1}(1-x)^{-1/2}\,dx\), identify which endpoint test fails.
- Why does a calculation that ties the two endpoint cutoffs together not by itself establish ordinary convergence?
- For the antiderivative method, what must be checked before subtracting its endpoint values?
