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One-Sided Limits At A Singular Endpoint

Use one-sided limits of truncated integrals to classify endpoint integrals as finite, divergent to positive or negative infinity, or without a limit.

Intermediate 9 min read

What You'll Learn

  • Define an improper integral at a singular lower or upper endpoint using the appropriate one-sided limit.
  • Distinguish a finite integral value from divergence to positive or negative infinity.
  • Recognize when oscillation prevents the truncated integrals from having a limit.
  • Use truncation and sequences to verify whether a limit exists.
  • Avoid confusing bounded oscillation or cancellation with convergence.

Why a Singular Endpoint Requires a Limit

A definite integral is ordinarily defined when its integrand is well behaved throughout the interval, including at its endpoints. If the integrand is undefined or unbounded at an endpoint, the usual definite integral cannot be applied directly across the entire interval. Instead, we integrate over intervals that stay away from the singular endpoint and examine a one-sided limit.

The endpoint may be singular even when the integral has a finite value. Conversely, the fact that an integrand is defined at every interior point, or that its oscillations appear to balance, does not guarantee that the integral exists. The limit of the truncated integrals decides.

Definition: Suppose \(f\) is integrable on every closed interval \([t,b]\) with \(a<t<b\), but may be singular at \(a\). The improper integral at the lower endpoint is defined by
$$ \int_a^b f(x)\,dx = \lim_{t\to a^+}\int_t^b f(x)\,dx, $$
provided this limit exists. If it exists as a finite real number, the integral converges and its value is that number. If the limit is \(+\infty\) or \(-\infty\), the integral diverges to the corresponding infinity. If the limit does not exist, the integral diverges without having either of those infinite limits.

The inner integral is an ordinary integral because its interval stays away from \(a\). The endpoint integral is not defined by substituting \(a\) into the integrand or by simply ignoring the endpoint. It is defined by the behavior of the entire truncated integral as \(t\) approaches \(a\) from within the interval.

For a singular upper endpoint, the direction of approach is reversed:

$$ \int_a^b f(x)\,dx = \lim_{t\to b^-}\int_a^t f(x)\,dx. $$

Here \(t\to b^-\) means that \(t\) approaches \(b\) from values less than \(b\). In either case, the limit is one-sided because the integration takes place on only one side of the endpoint.

What the Limit Can Do

There are three distinct outcomes to check. A finite limit gives a finite integral value. A limit of \(+\infty\) or \(-\infty\) records unbounded growth in that direction. If the truncated integrals do not approach any value, including either infinity, the integral has no limit. It is important not to merge the last two outcomes: an oscillating quantity can fail to approach a limit without tending to either infinity.

Behavior of the truncated integral Conclusion
Approaches a finite real number \(L\) Converges; the integral equals \(L\)
Grows without bound in the positive direction Diverges to \(+\infty\)
Decreases without bound in the negative direction Diverges to \(-\infty\)
Does not approach a finite value or either infinity Diverges without a limit

Why Nonnegative Integrands Have a Simple Test

For a nonnegative integrand, shrinking the omitted endpoint interval cannot make the truncated integral smaller. This gives a useful result that also explains why positive endpoint singularities cannot cancel themselves.

Theorem: Suppose \(f(x)\geq 0\) on \((a,b]\), and each truncated integral \(\int_t^b f(x)\,dx\) is defined for \(a<t<b\). As \(t\to a^+\), these integrals increase or stay the same. Consequently, their limit is finite if they remain bounded above, and is \(+\infty\) if they are unbounded above.

To justify the result, take \(a<t_1<t_2<b\). Splitting the ordinary integral at \(t_2\) gives

$$ \int_{t_1}^b f(x)\,dx = \int_{t_1}^{t_2} f(x)\,dx+\int_{t_2}^b f(x)\,dx \geq \int_{t_2}^b f(x)\,dx, $$

because \(f\geq 0\) implies \(\int_{t_1}^{t_2} f(x)\,dx\geq 0\). Thus moving the truncation point closer to \(a\) makes the accumulated integral no smaller. If all these integrals are bounded above, let \(S\) be their supremum. For every \(\delta>0\), the definition of supremum gives a truncation point \(t_0\) whose integral exceeds \(S-\delta\). Every truncation point closer to \(a\) has an integral between that value and \(S\), so its distance from \(S\) is less than \(\delta\). The limit is therefore \(S\), a finite number. If the integrals are not bounded above, monotonicity means that, for every real \(M\), they eventually exceed \(M\); this is exactly the statement that their limit is \(+\infty\).

This result applies to nonnegative integrands, not to arbitrary signed ones. A signed integrand can have truncated integrals that move up and down, so its limit must be checked directly. In particular, boundedness of those integrals alone does not establish convergence.

Worked Example: A Finite Value Despite an Unbounded Integrand

Worked Example: A Finite Value Despite an Unbounded Integrand

Classify \(\int_0^1 \frac{1}{\sqrt{x}}\,dx\). The integrand is unbounded as \(x\to0^+\), so define the integral by truncation:

$$ \int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{t\to0^+}\int_t^1 x^{-1/2}\,dx. $$

For \(t>0\), an antiderivative of \(x^{-1/2}\) is \(2\sqrt{x}\), since

$$ \frac{d}{dx}\bigl(2\sqrt{x}\bigr) = 2\left(\frac{1}{2\sqrt{x}}\right) = \frac{1}{\sqrt{x}}. $$

Therefore the truncated integral is

$$ \int_t^1 x^{-1/2}\,dx = 2\sqrt{1}-2\sqrt{t} = 2-2\sqrt{t}. $$

As \(t\to0^+\), \(\sqrt{t}\to0\), so \(2-2\sqrt{t}\to2\). Hence the improper integral converges and has the finite value \(2\). Checking by differentiation confirms that the antiderivative used gives the integrand on \(x>0\); the limit of the truncated values confirms the endpoint value of the integral.

Worked Example: Divergence to Positive and Negative Infinity

Worked Example: Divergence to Positive and Negative Infinity

First classify \(\int_0^1 \frac{1}{x}\,dx\). Its truncated integral is

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

As \(t\to0^+\), \(\ln(t)\to-\infty\), so \(-\ln(t)\to+\infty\). Thus

$$ \int_0^1 \frac{1}{x}\,dx=+\infty $$

in the extended-limit sense: the integral diverges to positive infinity and does not have a finite value. The calculation is consistent with the theorem for nonnegative integrands: the truncated integrals increase without bound as the omitted interval shrinks.

Now consider the negative integrand:

$$ \int_0^1 -\frac{1}{x}\,dx = \lim_{t\to0^+}\int_t^1 -\frac{1}{x}\,dx = \lim_{t\to0^+}\ln(t) = -\infty. $$

This integral diverges to negative infinity. For a check, the antiderivative \(\ln(x)\) has derivative \(1/x\), so the antiderivative \(-\ln(x)\) has derivative \(-1/x\). Evaluating it from \(t\) to \(1\) gives \(-\ln(1)+\ln(t)=\ln(t)\), as required.

Worked Example: Oscillation Can Prevent Any Limit

Worked Example: Oscillation Can Prevent Any Limit

Consider

$$ \int_0^1 \frac{\sin(1/x)}{x^2}\,dx. $$

The integrand changes sign near the endpoint, so use the definition and examine the truncated integral:

$$ \int_t^1 \frac{\sin(1/x)}{x^2}\,dx = \cos(1)-\cos(1/t). $$

To check this calculation, differentiate \(\cos(1/x)\):

$$ \frac{d}{dx}\cos(1/x) = -\sin(1/x)\left(-\frac{1}{x^2}\right) = \frac{\sin(1/x)}{x^2}. $$

Now choose two sequences of truncation points tending to zero. For positive integers \(n\) sufficiently large, set \(t_n=1/(2\pi n)\). Then \(\cos(1/t_n)=\cos(2\pi n)=1\), so the truncated integral equals \(\cos(1)-1\). Set \(s_n=1/((2n+1)\pi)\). Then \(\cos(1/s_n)=\cos((2n+1)\pi)=-1\), so the truncated integral equals \(\cos(1)+1\). Both sequences approach zero, but the corresponding truncated integrals approach different values. Therefore the one-sided limit does not exist, and the integral diverges without tending to \(+\infty\) or \(-\infty\).

A Common Mistake: Treating Oscillation as Cancellation

A tempting but incorrect argument for the preceding example is: “The cosine oscillates, so its average is zero; therefore the truncated integral approaches \(\cos(1)\).” That reasoning confuses an informal average over many oscillations with a limit as the single variable \(t\) approaches zero. A limit must have the same value along every sequence approaching the endpoint. The two sequences above give the distinct values \(\cos(1)-1\) and \(\cos(1)+1\), so no such limit exists.

There is a related distinction between a finite integral value and bounded truncated integrals. A bounded quantity can continue oscillating and have no limit. Conversely, a signed integral may converge even when its integrand changes sign, but that conclusion must come from the actual one-sided limit, not from an assumed cancellation. A symmetric balancing procedure is not a replacement for the one-sided endpoint definition.

A Reliable Classification Procedure

1
Identify the singular endpoint.
For a lower endpoint \(a\), truncate at \(t>a\); for an upper endpoint \(b\), truncate at \(t<b\).
2
Form the ordinary truncated integral.
Write the integral over an interval that excludes the singular endpoint.
3
Take the correct one-sided limit.
Track the full truncated integral as the truncation point approaches the endpoint from inside the interval.
4
State the outcome precisely.
Report a finite value, divergence to \(+\infty\), divergence to \(-\infty\), or failure of the limit to exist.

The truncation variable is essential: it records what happens to the accumulated area, not merely what happens to the integrand at points near the endpoint. This definition also keeps the question of convergence separate from techniques for computing the truncated integrals. Whatever calculation is used, the final classification must be justified by the one-sided limit.

Check Your Understanding

For each question, focus on the limit of the truncated integral, not just the pointwise behavior of the integrand.

  1. Write the one-sided-limit definition for an integral whose upper endpoint \(b\) is singular.
  2. What does it mean for an endpoint integral to diverge to \(-\infty\)?
  3. Why does a nonnegative integrand give truncated integrals that cannot decrease as the truncation point moves toward the singular lower endpoint?
  4. Can bounded truncated integrals fail to define a convergent improper integral? Explain.
  5. For \(\int_0^1 \frac{\sin(1/x)}{x^2}\,dx\), what two sequences of truncation points demonstrate that the defining limit does not exist?

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