Why the Exponent Matters
A function can become unbounded at a finite endpoint and still have a finite improper integral. Whether that happens depends on how rapidly the function grows as the endpoint is approached. For a power singularity, the exponent alone determines the answer.
The key comparison is between powers with exponent less than, equal to, or greater than \(1\). The threshold case \(p=1\) produces a logarithm, so it must be considered separately from the other powers.
Improper Integrals at a Finite Endpoint
An integral is improper at a finite endpoint if its integrand is unbounded there, or is not defined there, and the integral is defined by a one-sided limit. For example, if \(f\) is continuous on \((a,b]\) but may be unbounded as \(x\) approaches \(a\), then
provided this limit exists as a finite number. If the limit is infinite or does not exist, the improper integral diverges. The value assigned to the integrand at the single point \(a\) does not affect this limit.
For a singularity at a right endpoint \(b\), the corresponding definition is
In both cases, convergence is determined by approaching the endpoint from inside the interval. Directly substituting the singular endpoint into the integrand is not a test for convergence.
The Finite-Endpoint Power Test
Consider the model integral on the right of \(a\), over a segment of length \(\delta>0\):
Set \(t=x-a\). The integral becomes \(\int_0^\delta t^{-p}\,dt\), interpreted as a limit at \(t=0\). The same model appears on the left of \(a\), because the positive distance there is \(t=a-x\). Thus it is enough to evaluate this one limit.
Why the Threshold Is \(p=1\)
For \(p\ne 1\), an antiderivative of \(t^{-p}\) on \(t>0\) is \(\frac{t^{1-p}}{1-p}\). For \(p=1\), an antiderivative is \(\ln t\). Evaluate the integral first with the singular endpoint cut off at \(\varepsilon>0\):
If \(p<1\), then \(1-p>0\), so \(\varepsilon^{1-p}\) tends to \(0\) as \(\varepsilon\to0^+\). The limit is the finite value \(\frac{\delta^{1-p}}{1-p}\). This reasoning includes negative values of \(p\): in that case the integrand tends to zero at the endpoint, and the formula still gives a finite integral.
If \(p>1\), then \(1-p<0\), and \(\varepsilon^{1-p}=1/\varepsilon^{p-1}\) grows without bound as \(\varepsilon\to0^+\). The displayed quotient therefore tends to \(+\infty\). If \(p=1\), then \(\ln\delta-\ln\varepsilon\) also tends to \(+\infty\), since \(\ln\varepsilon\to-\infty\).
Replace the distance \(0\) by a positive \(\varepsilon\), then integrate over \([\varepsilon,\delta]\).
Use a power antiderivative when \(p\ne1\), and the logarithm when \(p=1\).
The limit is finite exactly when \(p<1\); otherwise it is infinite.
Applying the Test on Either Side
For a right-hand endpoint singularity at \(a\), the relevant expression is \((x-a)^{-p}\) for \(x>a\). For a left-hand endpoint singularity, use \((a-x)^{-p}\) for \(x<a\). On each side, the variable \(t\) can be chosen as the positive distance to \(a\). The resulting integral is \(\int_0^\delta t^{-p}\,dt\) in either case.
Consequently, the threshold does not change with the direction of approach. An exponent \(p<1\) gives convergence on either side; an exponent \(p\ge1\) gives divergence on either side. The length \(\delta\) affects the finite value when the integral converges, but it does not affect the threshold.
Worked Example: A Right-Hand Singularity with \(p<1\)
Evaluate \(\int_0^4 x^{-1/2}\,dx\). The endpoint \(0\) is singular, so begin with a cutoff:
Evaluating at both bounds gives \(2\sqrt{4}-2\sqrt{\varepsilon}=4-2\sqrt{\varepsilon}\). Since \(\sqrt{\varepsilon}\to0\), the limit is \(4\). The exponent is \(p=\frac12<1\), as the test predicts.
Check: the antiderivative differentiates to \(2\cdot\frac12 x^{-1/2}=x^{-1/2}\). Also, the cutoff evaluation tends to \(4\), confirming the stated value.
Worked Example: A Left-Hand Singularity with \(p<1\)
Evaluate \(\int_{-3}^0 ( -x)^{-2/3}\,dx\). The singular endpoint is \(0\), approached from the left. Introduce a cutoff at \(-\varepsilon\), then set \(u=-x\), so \(dx=-du\):
Here \(p=\frac23<1\), so convergence is expected. The value is \(3\sqrt[3]{3}\). To check the change of variable, the reversed bounds \(\varepsilon\) to \(3\) account for the negative sign in \(dx=-du\). Differentiating \(3u^{1/3}\) gives \(u^{-2/3}\), as required.
Worked Example: The Critical Exponent \(p=1\)
Determine whether \(\int_1^5 \frac{1}{x-1}\,dx\) converges. The endpoint \(x=1\) is singular, so use a cutoff \(1+\varepsilon\):
The exponent is \(p=1\), the critical case, so the integral diverges. Check: \(\frac{d}{dx}\ln(x-1)=\frac{1}{x-1}\) for \(x>1\), and the lower-bound term \(-\ln\varepsilon\) increases without bound as \(\varepsilon\to0^+\).
Worked Example: A Power Stronger Than the Threshold
Determine whether \(\int_0^2 x^{-3/2}\,dx\) converges. Cut off the endpoint and evaluate:
Since \(p=\frac32>1\), the integral diverges. Check: differentiating \(-2x^{-1/2}\) gives \((-2)(-\frac12)x^{-3/2}=x^{-3/2}\). The cutoff value diverges because \(2/\sqrt{\varepsilon}\to+\infty\).
A Common Mistake: Ignoring the Limit
A frequent error is to use an antiderivative and substitute the singular endpoint as though it were an ordinary bound. For example, one might write
This is incorrect: the expression \(-1/x\) is not defined at \(x=0\). The integral must be evaluated by a limit:
The correction is not merely a different way to write the same calculation. The limit is part of the definition of the improper integral, and it determines whether the integral has a finite value at all.
How to Use the Threshold
When you encounter an exact power singularity at a finite endpoint, first identify the positive distance to the endpoint and read off its exponent \(p\). Then compare that exponent with \(1\). You do not need to know the length of the interval to decide convergence. If you also need the value, evaluate the cutoff integral and take its limit.
| Exponent | Behavior of \(t^{-p}\) near \(0\) | Improper integral |
|---|---|---|
| \(p<1\) | Integrable power, including powers that remain bounded | Converges |
| \(p=1\) | Reciprocal power; logarithmic antiderivative | Diverges |
| \(p>1\) | Stronger power singularity | Diverges |
This test is a useful first step whenever a finite endpoint has a power singularity. It also clarifies why the reciprocal power is a genuine boundary case: its integral grows logarithmically, while every power with exponent above \(1\) has an antiderivative that grows without bound even faster as the cutoff approaches zero.
Check Your Understanding
Use a one-sided limit in each case. Decide whether the integral converges, and evaluate it when it does.
- Does \(\int_0^9 x^{-2/3}\,dx\) converge? What is its value?
- Does \(\int_0^1 x^{-5/4}\,dx\) converge or diverge?
- What exponent separates convergence from divergence for \(\int_a^{a+\delta}(x-a)^{-p}\,dx\), where \(\delta>0\)?
- Does approaching \(a\) from the left change the threshold for \((a-x)^{-p}\)? Explain using the positive distance to \(a\).
- Why is it invalid to substitute \(x=0\) directly into an antiderivative when evaluating \(\int_0^1 x^{-1}\,dx\)?
