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Convergence Comparisons · Tutorial 345 of 998

Local Comparisons at Multiple Singularities

Identify every problematic location, compare the integrand locally on each side, and determine whether all pieces converge.

Intermediate 9 min read

What You'll Learn

  • Split an improper integral into local pieces near each singularity and endpoint.
  • Distinguish the two one-sided tests required at an interior singularity.
  • Apply power-integral and limit-comparison tests separately on each piece.
  • Combine local convergence results to decide whether the full integral converges.
  • Recognize why cancellation across a singularity does not establish ordinary convergence.

Why Multiple Singularities Require Separate Tests

An improper integral can have more than one source of difficulty. Its integrand may be unbounded at an endpoint, at one or more interior points, or along an infinite tail. A comparison that describes the integrand near one location says nothing by itself about its behavior at another.

The reliable approach is local: identify every problematic location, examine the integrand on each side where necessary, and use an appropriate comparison for each piece. The full integral converges only when every required improper piece converges.

Key principle: Test each singular endpoint and each one-sided approach to an interior singularity separately. Then combine the results; a convergent piece cannot compensate for a divergent one.

What Counts as a Separate Piece?

Suppose an integrand \(f\) is considered on a finite interval \([a,b]\), and it has interior singularities at \(c_1,\ldots,c_k\), where \(a<c_1<\cdots<c_k<b\). The endpoints \(a\) and \(b\) may also be singular. At an interior point \(c_j\), the integral must be examined from the left and from the right as separate one-sided limits.

One way to organize the work is to choose small, disjoint neighborhoods around all the singular points, and then cut the remaining regular portions at ordinary points. Each resulting piece has at most one problematic endpoint. Another equivalent approach is to keep the regular portions together and explicitly test both one-sided contributions at every interior singularity.

Definition: An improper integral with finitely many singular locations is considered convergent when each required one-sided improper integral has a finite limit. At an interior singularity, both one-sided limits are required. For a nonnegative integrand, this is equivalent to requiring each local contribution to have finite integral.

This definition rules out cancellation between opposite infinite contributions. A symmetric cancellation can produce a Cauchy principal value, but that is a different notion from convergence of an ordinary improper integral.

The Local Comparison Tools

Near a finite endpoint, the power model \(t^{-p}\), for \(t>0\), is integrable at \(0\) exactly when \(p<1\). At an interior singularity \(c\), the corresponding model is \(|x-c|^{-p}\); the same condition \(p<1\) applies separately on each side. For an infinite tail, the power model \(x^{-p}\) is integrable at infinity exactly when \(p>1\).

These model tests can be used through direct comparison or limit comparison. For nonnegative functions \(f\) and \(g\), if \(f(x)/g(x)\) tends to a finite positive number at the location under consideration, then their integrals have the same convergence behavior there. The condition is local: the ratio limit needs to hold only near the endpoint, singularity, or tail being tested.

A bounded positive factor does not change the local power threshold. More precisely, if \(h\) is bounded above and below by positive constants near a location, then \(h(x)g(x)\) and \(g(x)\) have the same local convergence behavior. This lets us isolate the factor that actually causes the singularity.

Worked Example: Three Integrable Singular Locations

Worked Example: Three Integrable Singular Locations

Determine whether the following integral converges:

$$ \int_0^3 \frac{dx}{x^{2/3}|x-1|^{1/4}(3-x)^{3/4}}. $$

The integrand is nonnegative. Its possible singular locations are \(x=0\), \(x=1\), and \(x=3\). We test them separately.

At \(x=0\): The factors \(|x-1|^{-1/4}\) and \((3-x)^{-3/4}\) approach \(1\) and \(3^{-3/4}\), respectively. They are bounded above and below by positive constants in a sufficiently small neighborhood of \(0\). Thus the local model is \(x^{-2/3}\), and \(2/3<1\), so the endpoint contribution converges.

At \(x=1\): The factors \(x^{-2/3}\) and \((3-x)^{-3/4}\) are bounded above and below by positive constants near \(1\). On either side, the local model is \(|x-1|^{-1/4}\). Since \(1/4<1\), both one-sided contributions converge.

At \(x=3\): The factors \(x^{-2/3}\) and \(|x-1|^{-1/4}\) remain bounded above and below by positive constants near \(3\). The local model is \((3-x)^{-3/4}\), and \(3/4<1\), so this endpoint contribution converges.

Every required local piece converges. The portion away from the three singular locations is continuous on a closed, bounded interval and therefore has a finite integral. Hence the full improper integral converges. The check is that the local power exponents are \(2/3\), \(1/4\), and \(3/4\), each below the finite-endpoint threshold \(1\).

Why Local Results Combine

For a nonnegative integrand, split the interval into finitely many local pieces around its singularities and a regular remainder. The integral over the regular remainder is finite when the integrand is continuous there. The total integral is then the sum of the nonnegative local contributions and the finite remainder. A finite sum of finite contributions is finite; if even one local contribution is infinite, the total is infinite.

1
Locate every difficulty.
List all singular endpoints, interior singularities, and infinite tails.
2
Separate the approaches.
At an interior singularity, treat the left and right sides as distinct local pieces.
3
Choose a model for each piece.
Use a power model or another comparison whose behavior is known at that location.
4
Combine the conclusions.
The full integral converges only if every required local piece converges.

The argument depends on checking all the pieces, not on how small a neighborhood is chosen. Changing a neighborhood boundary at a regular point only moves a finite, ordinary integral from one piece to another; it cannot change convergence at a singular location.

Worked Example: One Failed Local Test Decides the Result

Worked Example: One Failed Local Test Decides the Result

Determine whether this integral converges:

$$ \int_0^2 \frac{dx}{x^{1/2}|x-1|(2-x)^{1/3}}. $$

The singular locations are \(0\), \(1\), and \(2\). Near \(x=0\), the other factors approach \(1\) and \(2^{-1/3}\), both finite and positive. The local model is \(x^{-1/2}\), which is integrable because \(1/2<1\).

Near \(x=2\), the factors \(x^{-1/2}\) and \(|x-1|^{-1}\) remain bounded and positive. The local model is \((2-x)^{-1/3}\), integrable because \(1/3<1\).

At \(x=1\), both sides must be considered. The ratio of the integrand to the model \(|x-1|^{-1}\) is

$$ \frac{1}{x^{1/2}|x-1|(2-x)^{1/3}} \bigg/\frac{1}{|x-1|} = x^{-1/2}(2-x)^{-1/3} \longrightarrow 1 \quad\text{as }x\to 1. $$

Thus limit comparison gives the same behavior as \(|x-1|^{-1}\) on either side. Since the power exponent is \(1\), the local integral diverges: the finite-endpoint criterion requires an exponent strictly less than \(1\). This single failed test is enough to conclude that the full integral diverges, even though the contributions at \(0\) and \(2\) converge.

As a final check, the three local models are \(x^{-1/2}\), \(|x-1|^{-1}\), and \((2-x)^{-1/3}\). The middle one fails the convergence criterion, so the result is divergence.

Including an Infinite Tail

The same bookkeeping applies when an interval is unbounded. A finite singularity and the tail at infinity are different locations and need different comparisons. For example, an integrand may be integrable near every finite singularity and still fail to have a convergent integral over the full infinite interval.

Worked Example: Finite Singularities Converge but the Tail Does Not

Determine whether

$$ \int_0^\infty \frac{dx}{x^{1/2}|x-2|^{1/2}} $$

converges. At \(x=0\), the factor \(|x-2|^{-1/2}\) approaches \(2^{-1/2}\), so the local model is \(x^{-1/2}\). Its exponent is below \(1\), hence this endpoint contribution converges.

At \(x=2\), the factor \(x^{-1/2}\) approaches \(2^{-1/2}\), so the local model is \(|x-2|^{-1/2}\). Its exponent is again below \(1\), and both one-sided contributions converge.

For the infinite tail, compare with \(1/x\):

$$ \frac{\dfrac{1}{x^{1/2}|x-2|^{1/2}}}{1/x} = \left(\frac{x}{|x-2|}\right)^{1/2} \longrightarrow 1 \quad\text{as }x\to\infty. $$

The model integral \(\int_1^\infty dx/x\) diverges, so limit comparison shows that the tail also diverges. The two finite singular locations pass their tests, but the infinite tail fails. Therefore the full integral diverges. This conclusion checks each distinct source of improper behavior rather than assuming that convergence near the finite locations settles the question.

A Common Pitfall: Cancellation Across a Singularity

Consider \(\int_0^2 dx/(x-1)\), which has an interior singularity at \(x=1\). A tempting but invalid calculation is to remove a symmetric interval around \(1\) and claim cancellation:

$$ \int_0^{1-\varepsilon}\frac{dx}{x-1} + \int_{1+\varepsilon}^{2}\frac{dx}{x-1} = \ln(\varepsilon)-\ln(\varepsilon)=0. $$

This calculation gives the Cauchy principal value, not the ordinary improper integral. The ordinary definition requires the one-sided limits separately. Using the antiderivative \(\ln|x-1|\), they are

$$ \int_0^{1-\varepsilon}\frac{dx}{x-1}=\ln(\varepsilon)\longrightarrow-\infty, \qquad \int_{1+\varepsilon}^{2}\frac{dx}{x-1}=-\ln(\varepsilon)\longrightarrow+\infty. $$

Neither one-sided integral has a finite limit, so the ordinary improper integral diverges. The correction is to test each side independently; cancellation between them cannot establish convergence.

Practical Checklist

Before declaring an improper integral convergent, confirm that every source of improper behavior has been examined. For a nonnegative integrand, every local contribution must be finite. For a signed integrand, ordinary convergence still requires the specified one-sided improper limits to exist; absolute-integrability comparisons can provide a sufficient condition for convergence.

  • Have all singular endpoints and interior singularities been listed?
  • At each interior singularity, have the left and right sides been tested separately?
  • If the interval is infinite, has the tail been tested independently?
  • Does each comparison apply near the particular location being studied?
  • Have you avoided relying on cancellation across a singularity?

The central habit is simple: make the local tests complete before combining them. Multiple singularities do not require one complicated global comparison; they require a collection of correct local comparisons and a conclusion that accounts for every piece.

Check Your Understanding

For each question, identify the relevant local pieces before deciding what comparison applies.

  1. Why must an interior singularity be tested from both the left and the right?
  2. For a local model \(|x-c|^{-p}\), what condition on \(p\) gives convergence on each side of \(c\)?
  3. If all finite singularities pass their local tests but the tail is comparable to \(1/x\), what can you conclude about the full integral?
  4. Why does a finite Cauchy principal value not by itself prove ordinary improper convergence?
  5. If a nonnegative integrand has one divergent local contribution and all other pieces converge, does the full integral converge? Explain.

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