Why Both Sides Of An Interior Singularity Must Converge

Why an Interior Singularity Requires Two Tests

An integrand can behave well on either side of a point and still fail to have an ordinary improper integral across that point. The reason is that the integral must have a finite limit as the left and right cutoffs approach the singularity independently. A finite limit obtained only by coordinating those cutoffs is not enough.

For an interval \([a,b]\) with \(a<c<b\), suppose \(f\) is integrable on every closed subinterval that stays away from \(c\), but may be unbounded or undefined at \(c\). To define the integral across \(c\), we must examine what accumulates from the left and what accumulates from the right.

Definition: The left-hand improper integral converges if the finite limit \(\lim_{u\to c^-}\int_a^u f(x)\,dx\) exists. The right-hand improper integral converges if the finite limit \(\lim_{v\to c^+}\int_v^b f(x)\,dx\) exists. The ordinary improper integral across \(c\) converges if and only if both one-sided integrals converge. Its value is the sum of their limits.

The cutoffs \(u\) and \(v\) are independent: \(u\) approaches \(c\) from below, while \(v\) approaches \(c\) from above. This independence is essential. One side cannot make up for a failure of the other.

The Criterion and Why It Works

Set \(A(u)=\int_a^u f(x)\,dx\) and \(B(v)=\int_v^b f(x)\,dx\). Cutting out a small interval around \(c\) gives the truncated integral \(A(u)+B(v)\). Ordinary convergence means this sum has a finite limit as both cutoffs approach \(c\), with no prescribed relationship between them.

Theorem: The limit of \(A(u)+B(v)\) as \(u\to c^-\) and \(v\to c^+\), independently, exists as a finite real number if and only if \(A(u)\) and \(B(v)\) each have finite one-sided limits. When these limits are \(L_-\) and \(L_+\), the two-cutoff limit is \(L_-+L_+\).

If the one-sided limits exist, then \(A(u)\to L_-\) and \(B(v)\to L_+\). Given any tolerance, choose \(u\) close enough to \(c\) to make \(A(u)\) close to \(L_-\), and choose \(v\) close enough to make \(B(v)\) close to \(L_+\). The triangle inequality then makes \(A(u)+B(v)\) close to \(L_-+L_+\). Thus the independent-cutoff limit exists and equals that sum.

The converse requires care. It is not valid to hold \(v\) fixed while \(u\to c^-\): that does not test the independent two-cutoff limit, because \(v\) is not also approaching \(c\). Instead, use the Cauchy criterion for finite limits. We give the argument for \(A\); the same reasoning applies to \(B\).

Use the assumed joint limit.
Suppose \(A(u)+B(v)\to L\) as the cutoffs independently approach \(c\). For any \(\varepsilon>0\), sufficiently near cutoffs satisfy \(|A(u)+B(v)-L|<\varepsilon/2\).

Compare two left cutoffs.
Choose any two sufficiently near values \(u_1,u_2\). Choose one sufficiently near \(v\), so the joint-limit bound holds for both pairs \((u_1,v)\) and \((u_2,v)\). Subtracting cancels the same \(B(v)\), and therefore \(|A(u_1)-A(u_2)|<\varepsilon\).

Conclude that the left limit exists.
The Cauchy criterion says that this bound for all sufficiently near \(u_1,u_2\) makes \(A(u)\) Cauchy as \(u\to c^-\). Since the real numbers are complete, \(A(u)\) has a finite limit. Fixing a sufficiently near \(u\) and comparing two near values \(v_1,v_2\) proves in the same way that \(B(v)\) has a finite limit.

Identify the joint limit.
Let the one-sided limits be \(L_-\) and \(L_+\). Taking the independent limits in \(A(u)+B(v)\) gives \(L=L_-+L_+\).

This establishes both directions: the ordinary integral exists exactly when the two one-sided improper integrals exist separately. The Cauchy argument is what rules out the mistaken idea that a finite combined limit can conceal a nonconvergent side.

Worked Examples

Worked Example: Both Sides Converge

Evaluate the improper integral of \(f(x)=1/\sqrt{|x|}\) across \(c=0\) on \([-1,1]\). The integrand is singular at zero, so calculate the two limits separately.

For \(-1\le x<0\), \(|x|=-x\). For \(-1<u<0\), substitute \(t=-x\), or use the antiderivative \(-2\sqrt{-x}\), whose derivative is \(1/\sqrt{-x}\). Thus

\int_{-1}^{u}\frac{dx}{\sqrt{|x|}} =\int_{-1}^{u}\frac{dx}{\sqrt{-x}} =\left[-2\sqrt{-x}\right]_{-1}^{u} =2-2\sqrt{-u}.

As \(u\to0^-\), \(-u\to0^+\), so the left-hand limit is \(2\). On the right, an antiderivative of \(x^{-1/2}\) is \(2\sqrt{x}\), since its derivative is \(1/\sqrt{x}\). Therefore

\int_v^1\frac{dx}{\sqrt{|x|}} =\int_v^1\frac{dx}{\sqrt{x}} =\left[2\sqrt{x}\right]_v^1 =2-2\sqrt{v} \longrightarrow 2 \qquad (v\to0^+).

Both one-sided limits are finite, so the ordinary improper integral converges and equals \(2+2=4\). The answer is checked by the two computed one-sided limits: their sum is exactly \(4\).

Worked Example: The Left Side Converges but the Right Side Diverges

On \([-1,1]\), let \(f(x)=1\) for \(x<0\) and \(f(x)=1/x\) for \(x>0\). The value at \(0\) is irrelevant to these improper integrals. Evaluate both sides of the singularity.

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

\int_{-1}^{u}1\,dx=[x]_{-1}^{u}=u+1 \longrightarrow 1 \qquad (u\to0^-).

The antiderivative \(x\) differentiates to \(1\), confirming the calculation. The left side converges. For \(0<v<1\), the right-hand integral is

\int_v^1\frac{1}{x}\,dx=[\ln x]_v^1 =\ln 1-\ln v=-\ln v.

Since \(\ln v\to-\infty\) as \(v\to0^+\), this integral tends to \(+\infty\). Its antiderivative \(\ln x\) differentiates to \(1/x\) for \(x>0\). The right side diverges, so the ordinary improper integral across zero does not converge, despite the finite left-hand limit. There is no finite sum of the two one-sided improper integrals.

Worked Example: Opposite Divergences Do Not Give Ordinary Convergence

Consider \(f(x)=1/x\) on \([-1,1]\), with a singularity at zero. On the left, for \(-1<u<0\), \(\ln|x|\) is an antiderivative of \(1/x\), since its derivative is \(1/x\) for \(x\ne0\). Hence

\int_{-1}^{u}\frac{dx}{x} =[\ln|x|]_{-1}^{u} =\ln|u|-\ln 1 =\ln|u| \longrightarrow -\infty \qquad (u\to0^-).

On the right, for \(0<v<1\),

\int_v^1\frac{dx}{x} =[\ln x]_v^1 =-\ln v \longrightarrow +\infty \qquad (v\to0^+).

Neither one-sided integral has a finite limit, so the ordinary improper integral does not converge. In particular, setting \(u=-\varepsilon\) and \(v=\varepsilon\) makes the truncated sum equal to \(\ln\varepsilon-\ln\varepsilon=0\) for every \(0<\varepsilon<1\), but that calculation only uses matched cutoffs. It does not establish the independent-cutoff limit required for ordinary convergence.

A Common Mistake: Letting Divergences Cancel

A tempting but incorrect argument is to take the last example, use the same distance \(\varepsilon\) on both sides, and conclude that the integral is zero because the two logarithms cancel. The error is that ordinary convergence requires a finite limit for all sufficiently near pairs \((u,v)\), not just for the special pairs \(u=-\varepsilon,\ v=\varepsilon\).

For example, choose \(u=-\varepsilon\) and \(v=2\varepsilon\), where \(0<2\varepsilon<1\). The truncated sum is then

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

That particular choice gives a different value from the matched-cutoff sum. More importantly, the one-sided calculations show that the left integral tends to \(-\infty\) and the right integral to \(+\infty\). Neither has a finite limit. The correct conclusion is therefore divergence of the ordinary improper integral, not convergence by cancellation.

Takeaway: Always test the left-hand and right-hand improper integrals separately. Their finite limits can be added; divergent one-sided limits cannot be combined to define an ordinary improper integral.

A Reliable Evaluation Procedure

When a singularity lies at an interior point \(c\), keep the two sides distinct throughout the calculation. The procedure is short, but each step prevents a different kind of error.

Locate the singularity.
Identify the interior point \(c\) and verify that the integrand can be integrated on intervals staying away from \(c\).

Introduce separate cutoffs.
Use \(u\to c^-\) for the left and \(v\to c^+\) for the right. Do not require the cutoffs to approach at the same rate.

Evaluate the one-sided limits.
Compute \(\lim_{u\to c^-}\int_a^u f(x)\,dx\) and \(\lim_{v\to c^+}\int_v^b f(x)\,dx\) independently, checking whether each is finite.

State the conclusion.
If both limits are finite, add them. If either fails to be finite, the ordinary improper integral across \(c\) diverges.

Why the Separate Test Matters

The separate test makes the meaning of convergence stable under independent choices of cutoffs. A genuine finite limit must not depend on whether one side is cut off slightly closer to the singularity than the other. This is why a finite contribution from one side cannot repair divergence on the other, and why apparent cancellation along a specially chosen path is insufficient.

The same principle is useful whenever a calculation splits an improper integral into pieces. Each piece with its own singular endpoint must be shown to converge before the pieces can be added as ordinary improper integrals. The basic decision is always local: establish the behavior on each side first, then combine only the finite results.

Check Your Understanding

For each question, keep the left and right limits distinct until you have determined whether they are finite.

State the condition that makes an improper integral across an interior singularity converge.
For \(f(x)=1/\sqrt{|x|}\) on \([-1,1]\), what are the two one-sided limits, and what is their sum?
If the left-hand integral has a finite limit but the right-hand integral tends to \(+\infty\), does the ordinary improper integral converge? Explain.
Why does a finite value obtained by choosing \(u=c-\varepsilon\) and \(v=c+\varepsilon\) not by itself prove ordinary convergence?
In the Cauchy-criterion argument, why must the same right cutoff be used when comparing two left cutoffs?