Removable Discontinuities Versus Improper Integrals

Why a Hole Is Not Necessarily an Obstruction

A rational expression can be undefined at a point even when its values nearby approach a finite number. This happens when the numerator and denominator share a factor that vanishes there. After canceling that factor, the simplified expression may have a finite value at the point, although the original formula still has a hole.

For integration, the important question is not simply whether the original formula is defined at every point. We must determine what happens as the variable approaches the missing point. A finite limiting value leads to a removable discontinuity; an unbounded limit signals a different kind of singularity and may make the improper integral diverge.

Key distinction: A factor that cancels completely can leave a finite hole. A denominator zero that remains after cancellation may produce an unbounded singularity. Simplify the expression, but also keep track of the original domain and check the limit.

Removable Discontinuities and Improper Integrals

Suppose \(f\) is defined near \(c\), except possibly at \(c\), and the limit \(\lim_{x\to c} f(x)\) exists as a finite real number \(L\). The discontinuity at \(c\) is removable: defining \(f(c)=L\) makes the function continuous at \(c\). If \(f(c)\) was already assigned a different value, changing it to \(L\) also removes the discontinuity.

Definition: An isolated missing point \(c\) is a removable discontinuity of \(f\) if \(\lim_{x\to c} f(x)=L\) exists and is finite. The continuous extension is the function that agrees with \(f\) away from \(c\) and has value \(L\) at \(c\).

To define an integral across an interior point \(c\) where the integrand is not defined, split the interval there and take the one-sided limits separately. For \(a<c<b\), the improper integral converges when both limits exist as finite numbers, and its value is their sum:

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

At an endpoint, only the corresponding one-sided limit is needed. This definition makes the distinction important: a finite limit at a hole can give a convergent integral, while an unbounded singularity can prevent convergence.

Why Filling a Removable Hole Does Not Change the Integral

Suppose \(f\) has a removable hole at \(c\), and let \(g\) be its continuous extension, so \(g(x)=f(x)\) for \(x\ne c\) and \(g(c)=L\). The integral across the hole, defined by one-sided limits, equals the ordinary integral of \(g\). Here is a justification.

Use the extension.
Since \(g\) is continuous near \(c\), it is bounded on a sufficiently small closed interval around \(c\). Choose \(M\) such that \(|g(x)|\le M\) there.

Bound the omitted piece.
For a small \(\delta>0\), the absolute value of the integral of \(g\) from \(c-\delta\) to \(c+\delta\) is at most \(2M\delta\).

Let the gap shrink.
As \(\delta\to0^+\), the bound \(2M\delta\) tends to zero. Thus the omitted interval contributes no residual amount, and the two one-sided limits sum to the integral of the continuous extension over the whole interval.

This reasoning explains why the missing value itself does not change the integral. It does not say that every singularity can be ignored: the boundedness of the extension is essential.

Worked Example: A Simple Removable Hole

Worked Example: Integrate After Canceling a Common Factor

Evaluate \(\int_0^2 \frac{x^2-1}{x-1}\,dx\), interpreting the integrand at its missing point \(x=1\) by one-sided limits.

At \(x=1\), both numerator and denominator vanish. Factor the numerator:

$$ x^2-1=(x-1)(x+1). $$

For \(x\ne1\), cancel the common factor. The original expression is undefined at \(x=1\), but away from that point it equals \(x+1\), whose limit at \(1\) is \(2\). Therefore the hole is removable, and the continuous extension has value \(2\).

Compute the two one-sided integral limits using the simplified expression:

\begin{aligned} \lim_{u\to1^-}\int_0^u (x+1)\,dx &=\left[\frac{x^2}{2}+x\right]_0^1 =\frac12+1 =\frac32,\\ \lim_{v\to1^+}\int_v^2 (x+1)\,dx &=\left[\frac{x^2}{2}+x\right]_1^2 =(2+2)-\left(\frac12+1\right) =\frac52. \end{aligned}

Add the two finite contributions:

\frac32+\frac52=4.

Check: An antiderivative of the extension \(x+1\) is \(\frac{x^2}{2}+x\), since its derivative is \(x+1\). Evaluating over the whole interval gives \((2+2)-(0+0)=4\), agreeing with the sum of the one-sided limits.

Reading a Rational Expression After Cancellation

For a rational function, factor the numerator and denominator near the point in question, then cancel all common factors. Cancellation is valid only where the canceled factor is nonzero, so the simplified formula describes the original function away from the missing point; it does not retroactively put that point in the original domain.

If cancellation leaves a denominator that is nonzero at \(c\), the simplified expression has a finite value there, and the original hole is removable. If a denominator factor still vanishes at \(c\), the limit must be checked; cancellation alone does not establish that the singularity is removable. A remaining pole can make one or both one-sided integrals diverge.

Worked Example: A Removable Hole with a Remaining Denominator

Evaluate \(\int_0^2 \frac{x^2-1}{(x-1)(x+2)}\,dx\), treating the missing point \(x=1\) by limits.

Factor and cancel the common factor \(x-1\):

\frac{x^2-1}{(x-1)(x+2)} = \frac{(x-1)(x+1)}{(x-1)(x+2)} = \frac{x+1}{x+2}, \qquad x\ne1.

At \(x=1\), the reduced expression has the finite value \(\frac{2}{3}\). The original expression therefore has a removable hole there. Rewrite the reduced expression to integrate it:

\frac{x+1}{x+2} = 1-\frac{1}{x+2}.

An antiderivative on \([0,2]\) is \(F(x)=x-\ln(x+2)\), because \(F'(x)=1-\frac{1}{x+2}\). The integral is therefore

\begin{aligned} \int_0^2 \frac{x+1}{x+2}\,dx &=F(2)-F(0)\\ &=(2-\ln4)-(0-\ln2)\\ &=2-\ln2. \end{aligned}

To check using the improper definition, evaluate the contributions on either side of \(1\):

\begin{aligned} \int_0^1 \frac{x+1}{x+2}\,dx &=F(1)-F(0) =1+\ln2-\ln3,\\ \int_1^2 \frac{x+1}{x+2}\,dx &=F(2)-F(1) =1+\ln3-\ln4. \end{aligned}

Both are finite, and their sum is \(2+\ln2-\ln4=2-\ln2\). This agrees with the whole-interval evaluation. The derivative check above confirms the antiderivative used.

When a Denominator Zero Remains

A finite limit cannot be assumed just because some cancellation is possible. For instance, in \(\frac{x+1}{x-1}\), the denominator still vanishes at \(x=1\), and there is no common factor to cancel. In fact, the expression can be written \(1+\frac{2}{x-1}\), which becomes unbounded near \(1\).

Worked Example: A Pole That Makes the Integral Diverge

Determine whether \(\int_0^2 \frac{x+1}{x-1}\,dx\) converges. The point \(x=1\) is an interior singularity, so evaluate the two sides separately.

Rewrite the integrand and find an antiderivative away from \(1\):

\frac{x+1}{x-1}=1+\frac{2}{x-1}, \qquad F(x)=x+2\ln|x-1|.

For \(0<u<1\), the left-hand truncated integral is

\int_0^u \frac{x+1}{x-1}\,dx = u+2\ln(1-u).

As \(u\to1^-\), \(u\to1\) and \(\ln(1-u)\to-\infty\), so this contribution tends to \(-\infty\).

For \(1<v<2\), the right-hand truncated integral is

\int_v^2 \frac{x+1}{x-1}\,dx = 2-v-2\ln(v-1).

As \(v\to1^+\), \(2-v\to1\) and \(-2\ln(v-1)\to+\infty\), so this contribution tends to \(+\infty\). Neither side has a finite limit. Consequently the improper integral does not converge.

Check: Differentiating \(F(x)=x+2\ln|x-1|\) gives \(1+\frac{2}{x-1}=\frac{x+1}{x-1}\) for \(x\ne1\), as required. The divergent one-sided limits, not the antiderivative formula alone, decide convergence.

Common Mistake: Canceling More Than the Common Factor

Consider \(\frac{x^2-1}{(x-1)^2}\). A common error is to cancel both copies of \(x-1\) from the denominator against the numerator, then claim that the result is \(x+1\). But the numerator contains only one such factor:

\frac{x^2-1}{(x-1)^2} = \frac{(x-1)(x+1)}{(x-1)(x-1)} = \frac{x+1}{x-1}, \qquad x\ne1.

One denominator factor remains. Near \(x=1\), the reduced expression equals \(1+\frac{2}{x-1}\), which is unbounded. Thus the hole is not removable, and it is not valid to evaluate the integral as though the integrand were \(x+1\). Cancel common factors only in matching numerator and denominator pairs, then inspect what remains.

Why the Distinction Matters

When you meet a rational integrand with a missing point, classify it before integrating:

After simplifying near \(c\)	Limit at \(c\)	Integral consequence
The denominator is nonzero at \(c\)	Finite	The hole is removable; evaluate the one-sided integral limits.
A denominator factor still vanishes at \(c\)	Must be checked	If a one-sided integral limit is not finite, the improper integral diverges.
The original expression is defined and continuous at \(c\)	Its value	There is no hole at \(c\) requiring an improper limit.

The practical sequence is straightforward: factor, cancel only common factors, retain the original missing point, compute the limit of the reduced expression, and then use one-sided integral limits if the point lies in the interval. This prevents a removable hole from being mistaken for a pole—and prevents an uncanceled pole from being treated as harmless.

Check Your Understanding

For each question, identify the limiting behavior before deciding how the integral should be evaluated.

For \(\frac{x^2-9}{x-3}\), what is the limit as \(x\to3\), and what value fills the removable hole?
Why must the original missing point still be treated as a limit after canceling a factor?
For \(\frac{x^2-4}{(x-2)(x+1)}\), simplify for \(x\ne2\) and find the limit as \(x\to2\).
For \(\frac{x^2-1}{(x-1)^2}\), how many factors of \(x-1\) remain in the denominator after cancellation?
What separate limits must be finite for an improper integral with an interior singularity to converge?