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Piecewise Integrands With Singular Points

Learn to organize piecewise improper integrals by their breakpoints and singular points, then test each one-sided limit separately.

Intermediate 11 min read

What You'll Learn

  • Identify which branch applies on each side of a singular point.
  • Set up endpoint and interior improper integrals using one-sided limits.
  • Distinguish a formula change from an unbounded singularity.
  • Decide convergence by checking every required one-sided limit separately.
  • Recognize why opposite divergences cannot be combined to establish convergence.

Why Piecewise Formulas Need Careful Splitting

A finite-interval integrand may be described by different formulas on different parts of the interval. A formula can become unbounded at an endpoint or at an interior point, and a formula change can occur at the same point. To define the integral correctly, it is not enough to find an antiderivative on each branch and combine the resulting expressions: every one-sided limit required by the singular points must be checked.

The key rule is to divide the interval wherever the formula changes or the integrand is singular, then treat each side of each singular point separately. The integral converges only when every required improper limit exists as a finite real number. A finite value on one side cannot compensate for a divergent limit on the other.

Definition: An improper integral at an endpoint or interior point is defined by replacing that endpoint or point with a one-sided limit. For an interior singular point \(c\), the ordinary improper integral across \(c\) converges only if the integrals from the left and from the right each have finite limits. If they do, the integral is their sum.

Set Up the One-Sided Limits

Suppose \(f\) is defined on \([a,b]\), except possibly at a finite number of points. An endpoint singularity at \(a\) requires a right-hand limit; one at \(b\) requires a left-hand limit. An interior singularity \(c\) requires both sides to be examined:

$$ \int_a^c f(x)\,dx = \lim_{u\to c^-}\int_a^u f(x)\,dx, \qquad \int_c^b f(x)\,dx = \lim_{v\to c^+}\int_v^b f(x)\,dx. $$

When \(f\) has several formula changes or singular points, list them in order and split the interval at each one. Use the branch that applies on each open subinterval. At a singular point, the branch on the left determines the left-hand integral, and the branch on the right determines the right-hand integral. A value assigned to the function at the single point does not determine either one-sided limit.

A change of formula does not, by itself, mean that the integral is improper. If the relevant branch stays bounded near a change point, that side does not acquire a singularity there. Still, splitting at the change point helps ensure that each part is evaluated using the correct formula. If a branch becomes unbounded there, the corresponding one-sided integral must be defined by a limit.

Worked Example: An Endpoint Singularity and a One-Sided Interior Singularity

Worked Example: An Endpoint Singularity and a One-Sided Interior Singularity

Consider the piecewise function on \([0,2]\):

$$ f(x)= \begin{cases} x^{-1/2}, & 0<x<1,\\ (x-1)^{-1/2}, & 1<x\leq 2. \end{cases} $$

The formula on the left is unbounded at \(0\), but it is bounded near \(1\): as \(x\to1^-\), \(x^{-1/2}\to1\). The formula on the right is unbounded as \(x\to1^+\). Thus the integral has an endpoint limit at \(0\) and a right-hand improper limit at \(1\). The left-hand contribution up to \(1\) also includes the endpoint singularity at \(0\).

Split at \(1\) and evaluate the first contribution using a lower cutoff \(\varepsilon>0\):

$$ \begin{aligned} \int_0^1 x^{-1/2}\,dx &=\lim_{\varepsilon\to0^+}\int_\varepsilon^1 x^{-1/2}\,dx\\ &=\lim_{\varepsilon\to0^+}\left[2\sqrt{x}\right]_\varepsilon^1\\ &=\lim_{\varepsilon\to0^+}(2-2\sqrt{\varepsilon})\\ &=2. \end{aligned} $$

The left-hand formula does not cause another divergence at \(1\). For example, its contribution on \([1-h,1]\), where \(0<h<1\), is

$$ \int_{1-h}^1 x^{-1/2}\,dx =2\bigl(1-\sqrt{1-h}\bigr), $$

which is finite and tends to \(0\) as \(h\to0^+\).

On the right of \(1\), use a cutoff \(\delta>0\):

$$ \begin{aligned} \int_1^2 (x-1)^{-1/2}\,dx &=\lim_{\delta\to0^+}\int_{1+\delta}^2 (x-1)^{-1/2}\,dx\\ &=\lim_{\delta\to0^+}\left[2\sqrt{x-1}\right]_{1+\delta}^2\\ &=\lim_{\delta\to0^+}(2-2\sqrt{\delta})\\ &=2. \end{aligned} $$

Both contributions are finite, so the improper integral converges and equals \(2+2=4\). Differentiating \(2\sqrt{x}\) gives \(x^{-1/2}\) for \(x>0\), and differentiating \(2\sqrt{x-1}\) gives \((x-1)^{-1/2}\) for \(x>1\), confirming the antiderivatives used.

Why Each Side Must Be Tested Separately

At an interior singular point, the left and right parts are separate limits in the definition. Their cutoffs can approach the point independently. This is why a cancellation produced by choosing matching cutoffs is not enough to establish convergence.

Theorem: If \(c\) is an interior singular point, \(\int_a^b f(x)\,dx\) converges across \(c\) only if both \(\lim_{u\to c^-}\int_a^u f(x)\,dx\) and \(\lim_{v\to c^+}\int_v^b f(x)\,dx\) exist as finite real numbers. When both limits are finite, the integral is their sum.
1
Locate the points that require a split.
Record every formula change and every endpoint or interior point where a branch is unbounded.
2
Choose the branch on each side.
Use the left-hand formula to evaluate a left contribution and the right-hand formula to evaluate a right contribution.
3
Introduce separate cutoffs.
At a singular point, use independent one-sided limits rather than a single cutoff that approaches from both sides.
4
Check every limit.
The full improper integral converges only if each required limit is finite. Add the resulting values only after that has been established.

To justify the rule, let \(c\) be an interior singularity. For cutoffs \(u<c<v\), split the truncated integral at \(c\) using the two sides:

$$ \int_a^u f(x)\,dx+\int_v^b f(x)\,dx. $$

If the left and right limits both exist finitely, then the first term approaches the left limit and the second term approaches the right limit. The limit of their sum is therefore the sum of those two finite limits. Conversely, the definition of an ordinary improper integral across \(c\) requires each one-sided integral to converge separately. A limit obtained only by forcing \(u\) and \(v\) to approach \(c\) in a particular relationship does not meet that requirement.

Worked Example: One Side Diverges

Worked Example: One Side Diverges

On \([0,2]\), let

$$ f(x)= \begin{cases} \dfrac{1}{1-x}, & 0\leq x<1,\\[4pt] \dfrac{1}{\sqrt{x-1}}, & 1<x\leq2. \end{cases} $$

The interior point \(1\) requires separate left- and right-hand tests. On the left, for \(\varepsilon>0\),

$$ \begin{aligned} \int_0^{1-\varepsilon}\frac{1}{1-x}\,dx &=\left[-\ln(1-x)\right]_0^{1-\varepsilon}\\ &=-\ln(\varepsilon)-0\\ &=-\ln(\varepsilon). \end{aligned} $$

As \(\varepsilon\to0^+\), \(-\ln(\varepsilon)\to+\infty\). Thus the left-hand integral does not have a finite limit.

The right-hand contribution does converge:

$$ \begin{aligned} \int_1^2\frac{1}{\sqrt{x-1}}\,dx &=\lim_{\delta\to0^+}\int_{1+\delta}^2 (x-1)^{-1/2}\,dx\\ &=\lim_{\delta\to0^+}\left[2\sqrt{x-1}\right]_{1+\delta}^2\\ &=\lim_{\delta\to0^+}(2-2\sqrt{\delta})\\ &=2. \end{aligned} $$

Since the left-hand limit diverges, the full integral does not converge, even though its right-hand contribution is finite. Differentiation confirms the antiderivatives: \(\frac{d}{dx}[-\ln(1-x)]=1/(1-x)\) for \(x<1\), and \(\frac{d}{dx}[2\sqrt{x-1}]=(x-1)^{-1/2}\) for \(x>1\).

Worked Example: Multiple Breakpoints and Endpoint Singularities

Worked Example: Multiple Breakpoints and Endpoint Singularities

Define \(f\) on \([0,3]\), except at the endpoints where the formulas are interpreted by limits, by

$$ f(x)= \begin{cases} x^{-1/2}, & 0<x<1,\\ 2, & 1\leq x<2,\\ (3-x)^{-1/2}, & 2\leq x<3. \end{cases} $$

The branch \(x^{-1/2}\) is unbounded at the left endpoint \(0\); the branch \((3-x)^{-1/2}\) is unbounded at the right endpoint \(3\). At \(1\) and \(2\), the formula changes, but the adjacent branch values stay bounded near those points. Split at both breakpoints so each formula is integrated on its own interval:

$$ \int_0^3 f(x)\,dx = \int_0^1 x^{-1/2}\,dx +\int_1^2 2\,dx +\int_2^3 (3-x)^{-1/2}\,dx, $$

where the first and third integrals are interpreted as endpoint limits. The first is

$$ \begin{aligned} \int_0^1 x^{-1/2}\,dx &=\lim_{\varepsilon\to0^+}\left[2\sqrt{x}\right]_\varepsilon^1\\ &=\lim_{\varepsilon\to0^+}(2-2\sqrt{\varepsilon})\\ &=2. \end{aligned} $$

The middle contribution is \( \int_1^2 2\,dx=2(2-1)=2\). For the right endpoint, substitute the antiderivative \(-2\sqrt{3-x}\), whose derivative is \((3-x)^{-1/2}\):

$$ \begin{aligned} \int_2^3 (3-x)^{-1/2}\,dx &=\lim_{\delta\to0^+}\int_2^{3-\delta}(3-x)^{-1/2}\,dx\\ &=\lim_{\delta\to0^+}\left[-2\sqrt{3-x}\right]_2^{3-\delta}\\ &=\lim_{\delta\to0^+}(2-2\sqrt{\delta})\\ &=2. \end{aligned} $$

All required endpoint limits are finite, so the integral converges and equals \(2+2+2=6\). This example illustrates why a list of breakpoints is useful: it keeps the endpoint singularities and the ordinary formula changes distinct while ensuring every branch is integrated on the correct interval.

A Common Mistake: Combining Opposite Divergences

Consider the piecewise integrand on \([0,2]\) given by \(1/(1-x)\) for \(x<1\) and \(-1/(x-1)\) for \(x>1\). A tempting but incorrect calculation uses the same cutoff \(\varepsilon\) on each side and adds the truncated integrals:

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

This does not show convergence. The left-hand integral tends to \(+\infty\), while the right-hand integral tends to \(-\infty\). Neither has a finite limit on its own, so the ordinary improper integral diverges. The correction is to test the two one-sided limits independently and to add their values only when both are finite.

Reliable Setup for Piecewise Singular Integrals

For a finite interval, the practical task is not to find one formula that covers every part of the domain. It is to make the domain structure explicit. Mark the endpoints, all formula changes, and all singular points. Then write one integral for each interval on which a single branch applies. Put a one-sided limit exactly where that branch approaches a singular endpoint, and verify its value before assembling the total.

This organization also prevents a subtler error: mistaking a branch’s behavior at one point for its behavior on the other side. A branch may be bounded as it approaches an interior point from the left, while the branch on the right is unbounded; the reverse can also happen. Determine each behavior from the formula valid on that side. A formula written for \(x<c\) gives no information about the right-hand limit at \(c\).

Check Your Understanding

For each question, identify the relevant branches and state which limits must be checked.

  1. For a piecewise integrand on \([a,b]\) with an interior singularity at \(c\), why must the left and right contributions be tested separately?
  2. In Worked Example 1, which formula is unbounded as \(x\to1^-\), and which is unbounded as \(x\to1^+\)?
  3. If one one-sided integral is finite and the other diverges, does the ordinary improper integral converge? Explain.
  4. What is the purpose of splitting at a formula change where both branches remain bounded?
  5. Why does a cancellation obtained by using matching cutoffs on opposite sides of a singularity fail to establish ordinary convergence?

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