Improper Integrals With Finite-Interval Singularities

When a Finite-Interval Integral Is Improper

A finite interval does not guarantee an ordinary definite integral. The interval may have finite endpoints while the integrand becomes unbounded at an endpoint or at a point inside the interval. In that situation, the integral is defined by cutting off the singular point and taking a limit.

For an ordinary definite integral, the integrand must meet the usual conditions for that integral to exist. If it is unbounded where the interval includes a point, an ordinary definite integral across that point is not the appropriate definition. Instead, we integrate over intervals that avoid the singularity and ask whether the resulting values approach a finite limit.

Definition: A finite-interval improper integral is an integral defined as a limit of ordinary definite integrals, where the intervals of integration approach a point at which the integrand is unbounded or not defined. It converges if the required limit or limits exist as finite real numbers; otherwise, it diverges.

This tutorial focuses on unbounded behavior at finite endpoints and at interior points. The key task is to identify the singular point and set up the correct limit or limits. The cutoffs must approach the singularity from the relevant side or sides.

A Singular Endpoint

Suppose \(f\) is continuous on \((a,b]\), but is unbounded or not defined at the left endpoint \(a\). For each \(\varepsilon>0\), the truncated integral \(\int_{a+\varepsilon}^{b} f(x)\,dx\) avoids that endpoint. The improper integral is defined by the limit as \(\varepsilon\) decreases to zero.

\int_a^b f(x)\,dx := \lim_{\varepsilon\to 0^+} \int_{a+\varepsilon}^{b} f(x)\,dx.

The definition requires the displayed limit to exist as a finite real number. If the limit is infinite or does not exist, the improper integral diverges. The integral sign with endpoint \(a\) is shorthand for this limiting process; it does not mean that \(f(a)\) has been assigned a finite value.

For an integrand singular at the right endpoint \(b\), the cutoff approaches from the left:

\int_a^b f(x)\,dx := \lim_{\varepsilon\to 0^+} \int_a^{b-\varepsilon} f(x)\,dx.

When both endpoints are singular, the integral must be split at a point \(c\) strictly between them. Each endpoint then has its own one-sided limit. Neither endpoint cutoff is allowed to stand in for the other.

Worked Example: An Unbounded Integrand at the Left Endpoint

Set up and evaluate \(\int_0^1 \frac{1}{\sqrt{x}}\,dx\) as an improper integral.

The integrand is unbounded as \(x\to0^+\), so introduce a cutoff \(\varepsilon>0\):

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

For \(x>0\), an antiderivative is \(2\sqrt{x}\), since \(\frac{d}{dx}(2\sqrt{x})=x^{-1/2}\). Thus

\int_{\varepsilon}^1 x^{-1/2}\,dx = \left[2\sqrt{x}\right]_{\varepsilon}^1 = 2-2\sqrt{\varepsilon} = 2(1-\sqrt{\varepsilon}).

As \(\varepsilon\to0^+\), \(\sqrt{\varepsilon}\to0\), so the limit is \(2\). The improper integral converges and equals \(2\). The check is that the derivative of the antiderivative used on every truncated interval is \(x^{-1/2}\), and the truncated values approach \(2\).

An Interior Singularity Requires Two Limits

Now suppose \(f\) is unbounded at an interior point \(c\), where \(a<c<b\), and is otherwise continuous on each side of \(c\). There are two approaches to the singularity: from the left and from the right. The integral is defined by considering these sides separately:

\int_a^b f(x)\,dx := \int_a^c f(x)\,dx+\int_c^b f(x)\,dx, $$ $$ \int_a^c f(x)\,dx := \lim_{\delta\to0^+}\int_a^{c-\delta}f(x)\,dx, \qquad \int_c^b f(x)\,dx := \lim_{\eta\to0^+}\int_{c+\eta}^{b}f(x)\,dx.

The symbols \(\delta\) and \(\eta\) represent independent positive cutoffs. The integral across \(c\) converges only if both one-sided integrals converge to finite real numbers. When they do, the improper integral is their sum. If either one diverges, the integral across the singularity diverges.

This separate-limit definition reflects the two distinct parts of the interval. When both one-sided limits exist, their sum is well-defined: choosing cutoffs independently gives two truncated values, each approaching its own finite limit, so their sum approaches the sum of those limits. A single coupled cutoff that approaches \(c\) from both sides does not replace this requirement. In particular, a symmetric cancellation is not ordinary convergence; see the standard subject of Cauchy principal values.

Definition: At an interior singularity \(c\), the improper integral converges if and only if the left-hand and right-hand improper integrals both converge. Its value is the sum of those two finite values.

Worked Example: A Divergent Interior Singularity

Consider \(\int_0^2 \frac{1}{|x-1|}\,dx\). The integrand is unbounded at \(x=1\), so set up separate cutoffs on the two sides.

For \(0\le x<1\), \(|x-1|=1-x\). For \(1<x\le2\), \(|x-1|=x-1\). Therefore the two truncated integrals are

\int_0^{1-\delta}\frac{1}{1-x}\,dx = \left[-\ln(1-x)\right]_0^{1-\delta} = -\ln\delta, $$ $$ \int_{1+\eta}^{2}\frac{1}{x-1}\,dx = \left[\ln(x-1)\right]_{1+\eta}^{2} = -\ln\eta.

The antiderivatives check directly: \(\frac{d}{dx}[-\ln(1-x)]=\frac{1}{1-x}\) on the left, and \(\frac{d}{dx}\ln(x-1)=\frac{1}{x-1}\) on the right. As \(\delta\to0^+\), \(-\ln\delta\to+\infty\); as \(\eta\to0^+\), \(-\ln\eta\to+\infty\). Both one-sided integrals diverge, so the integral across \(x=1\) diverges.

A Complete Interior Setup and Evaluation

An interior singularity does not automatically make an integral divergent. What matters for the definition is whether each one-sided truncated integral approaches a finite value. The next example illustrates a singular integrand for which both limits exist.

Worked Example: A Convergent Interior Singularity

Evaluate \(\int_0^2 \frac{1}{\sqrt{|x-1|}}\,dx\) using the definition at the interior point \(x=1\).

Split at \(1\). On the left, \(|x-1|=1-x\), and on the right, \(|x-1|=x-1\). The two limits are

\int_0^1\frac{1}{\sqrt{|x-1|}}\,dx = \lim_{\delta\to0^+} \int_0^{1-\delta}\frac{1}{\sqrt{1-x}}\,dx, $$ $$ \int_1^2\frac{1}{\sqrt{|x-1|}}\,dx = \lim_{\eta\to0^+} \int_{1+\eta}^{2}\frac{1}{\sqrt{x-1}}\,dx.

On the left, \(-2\sqrt{1-x}\) is an antiderivative because its derivative is \((1-x)^{-1/2}\). On the right, \(2\sqrt{x-1}\) is an antiderivative because its derivative is \((x-1)^{-1/2}\). Evaluating gives

\int_0^{1-\delta}\frac{1}{\sqrt{1-x}}\,dx = \left[-2\sqrt{1-x}\right]_0^{1-\delta} = 2(1-\sqrt{\delta}), $$ $$ \int_{1+\eta}^{2}\frac{1}{\sqrt{x-1}}\,dx = \left[2\sqrt{x-1}\right]_{1+\eta}^{2} = 2(1-\sqrt{\eta}).

Both limits are finite: the left value tends to \(2\), and the right value also tends to \(2\). By the definition for an interior singularity, the integral converges and equals \(2+2=4\). The derivative checks above verify the antiderivatives used, and the two finite limits verify convergence.

Why the Cutoffs Must Be Chosen Carefully

The ordinary definite integrals inside the definitions are always taken over intervals that avoid the singular point. For an endpoint singularity, the interval approaches the endpoint from within the original interval. For an interior singularity, the left truncated interval ends before \(c\), while the right truncated interval starts after \(c\).

A reliable setup follows these steps:

Locate each singular point.
Identify every endpoint or interior point where the integrand is unbounded.

Choose a cutoff on each affected side.
Move the cutoff toward the singular point while keeping the truncated interval away from it.

Write the required limit or limits.
Use one limit for a single singular endpoint and separate one-sided limits for an interior singularity.

Check finiteness before combining values.
An interior integral converges only when both one-sided limits are finite; then add them.

Common Pitfall: Combining the Two Sides Too Soon

A common error is to use a single symmetric cutoff around an interior singularity and conclude that the integral converges if the two sides cancel. For the integrand \(1/(x-1)\) on \([0,2]\), the symmetric truncation gives

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

That calculation does not establish convergence of the improper integral. It ties the two cutoffs together and cancels the values before checking either side. The actual one-sided integrals are \(\ln\varepsilon\to-\infty\) on the left and \(-\ln\varepsilon\to+\infty\) on the right. Each side fails to have a finite limit, so the improper integral diverges. The correction is to set up and test the one-sided limits independently.

What the Definition Does—and Does Not—Say

The definition supplies a precise test for convergence; it does not say that every unbounded integrand has a divergent integral. The convergent interior example shows otherwise. Nor does the definition permit one divergent side to be offset by the other. Each required limit must exist on its own before the resulting values can be added.

Keep the distinction clear: the interval can be finite while the integral is still improper, because the difficulty lies in the integrand's behavior at a point of the interval. The finite cutoff integrals are ordinary definite integrals; the improper integral is the limit of those values.

Check Your Understanding

For each question, identify the cutoff or limits required before deciding whether the improper integral converges.

How is an integral defined when the integrand is unbounded at the left endpoint?
What limit is used for an integrand unbounded at the right endpoint?
At an interior singularity \(c\), how many one-sided limits must be checked?
Why does a finite value from a symmetric cutoff not by itself establish ordinary convergence?
If one of the two one-sided integrals at an interior singularity diverges, what can be concluded about the integral across the singularity?