Defining Integrals Across An Interior Singularity

Why an Interior Singularity Requires Two Limits

An integrand may be defined and continuous on either side of a point \(c\), yet become unbounded or fail to be defined at \(c\). In that situation, the ordinary definite integral across the entire interval is not defined by integrating straight through the singular point. Instead, we examine the behavior as the variable approaches \(c\) from the left and from the right.

The key is to keep those approaches separate. The left-hand integral and the right-hand integral each have their own limit. The improper integral across the whole interval is defined as the sum of those two limits, provided both are finite.

Definition: Let \(a<c<b\), and suppose \(f\) is continuous on \([a,c)\) and on \((c,b]\). The improper integral of \(f\) across the interior singularity \(c\) is defined if both limits below exist as finite real numbers. Its value is their sum:

\int_a^b f(x)\,dx = \lim_{t\to c^-}\int_a^t f(x)\,dx + \lim_{s\to c^+}\int_s^b f(x)\,dx.

Here \(t\) approaches \(c\) from below and \(s\) approaches \(c\) from above. The two limits are taken separately.

The letters \(t\) and \(s\) are deliberately different: the two sides need not be truncated by the same distance from \(c\). A limit on just one side does not, by itself, define the integral across the full interval.

How the Separate Limits Determine the Sum

Write \(L\) for the left-hand limit and \(R\) for the right-hand limit. If both are finite, then the value assigned by the definition is \(L+R\). The sum is well-defined even when the left and right truncations approach \(c\) at different rates.

To see why, set \(A(t)=\int_a^t f(x)\,dx\) and \(B(s)=\int_s^b f(x)\,dx\). If \(A(t)\to L\) as \(t\to c^-\) and \(B(s)\to R\) as \(s\to c^+\), then for any \(\varepsilon>0\), we can choose \(t\) sufficiently close to \(c\) that \(|A(t)-L|<\varepsilon/2\), and independently choose \(s\) sufficiently close to \(c\) that \(|B(s)-R|<\varepsilon/2\). Consequently,

|A(t)+B(s)-(L+R)| \leq |A(t)-L|+|B(s)-R| <\varepsilon.

This verifies that the sum of the two one-sided limits gives a single value, without requiring a coordinated approach from both sides. It does not mean that we can omit the region between arbitrary truncation points and claim to have split one ordinary integral there.

Worked Example: A Rational Singularity with Divergent Side Limits

Worked Example: \(1/(x-1)\) on \([0,2]\)

Consider \(f(x)=1/(x-1)\), which has a singularity at \(c=1\). The definition requires us to compute the two limits separately. An antiderivative on either side of \(1\) is \(\ln|x-1|\), since

\frac{d}{dx}\ln|x-1|=\frac{1}{x-1}.

For the left-hand limit, let \(t\to1^-\). Then

\int_0^t \frac{1}{x-1}\,dx = \ln|t-1|-\ln|-1| = \ln(1-t).

As \(t\to1^-\), \(1-t\to0^+\), so \(\ln(1-t)\to-\infty\). The left-hand limit is not finite.

For the right-hand limit, let \(s\to1^+\). Then

\int_s^2 \frac{1}{x-1}\,dx = \ln|2-1|-\ln|s-1| = -\ln(s-1).

As \(s\to1^+\), \(s-1\to0^+\), so \(-\ln(s-1)\to+\infty\). This limit is also not finite. Therefore the improper integral across \([0,2]\) is not defined as a finite real number. The antiderivative check above confirms that the evaluations used on both sides have the correct derivative.

This example shows why the definition asks for limits, rather than simply substituting the singular point into an antiderivative. The endpoint expressions have to be examined as their truncation points approach \(c\).

Worked Example: A Logarithmic Interior Singularity

Worked Example: \(-\ln|x-1|\) on \([0,2]\)

Now consider \(f(x)=-\ln|x-1|\), which is unbounded as \(x\to1\). On each side of \(1\), an antiderivative is

F(x)=-(x-1)\ln|x-1|+(x-1).

Indeed, differentiating gives

F'(x) = -\ln|x-1|-(x-1)\frac{1}{x-1}+1 = -\ln|x-1|.

For the left side, evaluate from \(0\) to \(t<1\). Since \(F(0)=-1\),

\int_0^t -\ln|x-1|\,dx = F(t)-F(0) = -(t-1)\ln|t-1|+(t-1)+1.

As \(t\to1^-\), \(t-1\to0\). The term \((t-1)\ln|t-1|\) tends to \(0\), and \(t-1\) tends to \(0\), so the left-hand limit is \(1\).

For the right side, evaluate from \(s>1\) to \(2\). We have \(F(2)=1\), and therefore

\int_s^2 -\ln|x-1|\,dx = F(2)-F(s) = 1+(s-1)\ln(s-1)-(s-1).

As \(s\to1^+\), both \((s-1)\ln(s-1)\) and \(s-1\) tend to \(0\). The right-hand limit is \(1\). Both separate limits are finite, so the improper integral is their sum:

\int_0^2 -\ln|x-1|\,dx=1+1=2.

As a check, \(F'(x)=-\ln|x-1|\) on both intervals where the integrand is defined, and the limiting endpoint values used on each side give exactly the two contributions \(1\) and \(1\).

Worked Example: Unequal Distances from the Singularity

Worked Example: \(1/\sqrt{|x-1|}\) on \([0,3]\)

This example illustrates that the two sides are evaluated on their own intervals and need not contribute equally. The singular point is \(c=1\). On the left, \(0\leq x<1\), so \(|x-1|=1-x\). For \(t<1\),

\int_0^t \frac{1}{\sqrt{|x-1|}}\,dx = \int_0^t (1-x)^{-1/2}\,dx = 2-2\sqrt{1-t}.

The antiderivative \(2-2\sqrt{1-x}\) has derivative \((1-x)^{-1/2}\). Taking \(t\to1^-\), the left-hand limit is \(2\).

On the right, \(1<x\leq3\), so \(|x-1|=x-1\). For \(s>1\),

\int_s^3 \frac{1}{\sqrt{|x-1|}}\,dx = \int_s^3 (x-1)^{-1/2}\,dx = 2\sqrt{2}-2\sqrt{s-1}.

The antiderivative \(2\sqrt{x-1}\) has derivative \((x-1)^{-1/2}\). Taking \(s\to1^+\), the right-hand limit is \(2\sqrt{2}\). Thus the improper integral exists and equals

\int_0^3 \frac{1}{\sqrt{|x-1|}}\,dx = 2+2\sqrt{2}.

The check on each side is the derivative of its antiderivative, and the limits at \(c\) are finite. The two contributions differ because the interval extends distance \(1\) to the left of \(c\) and distance \(2\) to its right.

A Common Pitfall: Losing the Gap

Suppose \(a<t<c<s<b\). The expression \(\int_a^t f(x)\,dx+\int_s^b f(x)\,dx\) consists of two separate truncated integrals. It is not, in general, equal to \(\int_a^b f(x)\,dx\): the region from \(t\) to \(s\) has been left out.

For a concrete check, take \(f(x)=1\) on \([0,4]\), \(t=1\), and \(s=3\). The two outer integrals add to

\int_0^1 1\,dx+\int_3^4 1\,dx=1+1=2.

But the full integral is \(\int_0^4 1\,dx=4\). The omitted interval contributes \(\int_1^3 1\,dx=2\). When all the integrals in question are ordinary integrals, the valid additivity statement includes that gap:

\int_0^4 1\,dx = \int_0^1 1\,dx + \int_1^3 1\,dx + \int_3^4 1\,dx.

At a singularity, the purpose of the separate limits is not to split an ordinary integral across the singular point. It is to define the two one-sided contributions independently and then add their finite limiting values.

Why the Definition Matters

The definition specifies exactly what it means to integrate across a point where the integrand is singular. It prevents an apparent cancellation between left and right truncations from being mistaken for two finite one-sided limits. For example, in the rational example, using the same small distance from \(1\) on both sides can make the two truncated values add to zero. But the separate values are \(\ln\varepsilon\) and \(-\ln\varepsilon\), which do not have finite limits as \(\varepsilon\to0^+\). Their finite-truncation sum does not change the one-sided limits required by the definition.

When setting up an interior-singularity integral, identify the singular point, write one truncated integral on each side, and evaluate each limit separately. Only after both limits are finite should they be added to obtain the value of the improper integral.

Check Your Understanding

For each question, keep the two sides of the singularity distinct.

State the definition of \(\int_a^b f(x)\,dx\) when \(f\) has an interior singularity at \(c\).
For \(f(x)=1/(x-2)\) on \([1,3]\), write the separate left-hand and right-hand limits. Do not combine them before evaluating.
For \(f(x)=-\ln|x-2|\) on \([1,3]\), what antiderivative could be used separately on each side?
Why does \(\int_a^t f(x)\,dx+\int_s^b f(x)\,dx\) omit part of the interval when \(t<s\)?
In the definition, must the left and right truncation points approach \(c\) at the same rate?