Why an Interior Singularity Changes the Integral
An ordinary definite integral over \([a,b]\) assumes that the integrand is defined and well behaved throughout the interval. If the integrand becomes unbounded or is undefined at an interior point \(c\), where \(a<c<b\), the usual definite-integral notation cannot be interpreted as one uninterrupted accumulation across \(c\).
The interval must be divided at \(c\). The portion to the left approaches the singularity from below, while the portion to the right approaches it from above. These are separate limiting processes, and each one must produce a finite value for the improper integral to converge.
Definition: Split at the Singular Point
Suppose \(f\) is defined and integrable on intervals that stay away from \(c\), but has a singularity at \(c\in(a,b)\). The improper integral is defined by taking one limit from the left and another from the right:
provided that both limits exist as finite real numbers. The notation \(t\to c^-\) means that \(t\) approaches \(c\) from values less than \(c\); \(s\to c^+\) means that \(s\) approaches from values greater than \(c\).
If either limit fails to exist as a finite number, the improper integral diverges. A one-sided limit may diverge to \(+\infty\) or \(-\infty\), or may fail to settle to any limit. In each case, the ordinary improper integral does not converge.
When both limits are finite, their values are added. This is consistent with ordinary additivity of definite integrals: away from the singularity, an interval can be divided into left and right portions. The limiting definition preserves that addition while preventing the calculation from crossing the point where \(f\) is singular.
A Reliable Procedure
Identify the interior point \(c\) where the integrand is undefined or unbounded, and confirm that \(a<c<b\).
Separate the interval into the part ending at \(c\) and the part beginning at \(c\).
Use a left-hand limit for the first integral and a right-hand limit for the second.
Compute each limit independently. Add the results only if both are finite.
A useful safeguard is to keep two distinct cutoff variables, such as \(t\) and \(s\), until both one-sided limits have been evaluated. This makes it harder to accidentally treat the two sides as a single limit that allows their values to cancel.
Worked Examples
Worked Example: A Convergent Singularity at the Center
Determine the value and convergence of \(\displaystyle\int_{-1}^{1}\frac{1}{\sqrt{|x|}}\,dx\). The integrand is unbounded at \(x=0\), so split the integral there.
On the left, \(x<0\), so \(|x|=-x\). In the integral ending at \(t<0\), substitute \(u=-x\), giving \(dx=-du\). When \(x=-1\), \(u=1\); when \(x=t\), \(u=-t\). Thus
As \(t\to0^-\), \(-t\to0^+\), so the left-hand limit is \(2\). On the right, \(|x|=x\), and
Both one-sided integrals converge, so their values may be added. The integral equals \(2+2=4\). As a check on the antiderivative used on the right, \(\frac{d}{dx}(2\sqrt{x})=x^{-1/2}\) for \(x>0\); the substitution on the left gives the same power integral in \(u\).
Worked Example: A Divergence on Only One Side
Let \(f\) be defined on \([-1,1]\setminus\{0\}\) by \(f(x)=1/(-x)\) for \(x<0\), and \(f(x)=1\) for \(x>0\). Test \(\displaystyle\int_{-1}^{1}f(x)\,dx\).
The left-hand integral is
Here \(-t\to0^+\), so \(\ln(-t)\to-\infty\). The right-hand integral is finite:
The right side converges, but the left side diverges to \(+\infty\). Therefore the improper integral diverges; there is no finite value to obtain by adding the two sides. This example illustrates why testing only one side is not enough.
For a check of the left calculation, an antiderivative of \(1/(-x)=-1/x\) on \(x<0\) is \(-\ln(-x)\), since its derivative is \(-1/x=1/(-x)\).
Worked Example: Symmetric Cancellation Is Not Convergence
Consider \(\displaystyle\int_{-1}^{1}\frac{1}{x}\,dx\), which has an interior singularity at \(x=0\). The separate one-sided integrals are
and
Each side diverges, so the improper integral does not converge. Yet if the same cutoff \(\varepsilon>0\) is used on both sides, the paired calculation gives
This zero comes from pairing two divergent contributions with the same cutoff. It is not the value of the improper integral defined by separate one-sided limits. As a derivative check, \(\frac{d}{dx}\ln|x|=1/x\) on both intervals that avoid zero; the failure occurs in the limits, not in the antiderivative.
Why the Limits Must Be Separate
The definition asks whether accumulation remains finite while approaching the singularity from each side. On one side, the variable approaches \(c\) through values below it; on the other, it approaches through values above it. There is no reason the two sides must approach at matching rates, and the definition does not require them to do so.
For this reason, a calculation that forces the cutoffs to move together can hide divergence. In the \(1/x\) example, the left contribution becomes arbitrarily negative and the right contribution arbitrarily positive. Equal cutoffs make these large contributions cancel exactly at every stage, but neither one-sided accumulation has a finite limit. Cancellation under a chosen pairing does not change that fact.
Common Mistake: Treating the Singularity as an Ordinary Split
A frequent error is to write \(\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx\) and then evaluate both pieces as though the integrand were defined at \(c\). If \(f\) is singular there, those ordinary integrals are not defined in that way. Each piece must instead be expressed as a limit that approaches \(c\) from its own side.
Another version of the same error is to use one shared cutoff and declare the integral convergent because the two truncated pieces cancel. For \(\int_{-1}^{1}1/x\,dx\), that produces \(0\), but the separate limits are \(-\infty\) and \(+\infty\). The correction is to evaluate those limits independently and require both to be finite before adding them.
Check Your Understanding
For each question, identify the one-sided limits before deciding whether the integral converges.
- At an interior singularity \(c\), what two limits define \(\int_a^b f(x)\,dx\)?
- What must be true of both one-sided limits for the improper integral to converge?
- Why does a symmetric cutoff calculation that gives a finite number not prove that an interior-singularity integral converges?
- For \(\int_{-1}^{1}1/\sqrt{|x|}\,dx\), what are the values of the left and right one-sided integrals?
- If one one-sided integral converges and the other diverges, what is the status of the full improper integral?
