Two Endpoints, Two Benchmarks
A power function can have an improper integral because the interval extends without bound, or because the function becomes unbounded near an endpoint. The same family of powers appears in both situations, but the convergence thresholds go in opposite directions. Keeping those two tests distinct is essential.
We will use the form \(x^{-p}\), where \(p\) is a real number and \(x>0\). For an interval extending to infinity, the benchmark is \(p>1\). For an interval approaching zero, the benchmark is \(p<1\). At \(p=1\), the function is \(1/x\), and the logarithm shows that both types of integral diverge.
What Convergence Means Here
An improper integral is defined using a limit of ordinary definite integrals. For example, if \(a>0\), then
provided this limit exists as a finite real number. If \(b>0\), then an integral approaching zero is defined by
again requiring a finite real limit for convergence. In these examples the integrand is positive, so a limit that grows without bound means the integral diverges to \(+\infty\).
For \(p\ne1\), the power rule gives an antiderivative of \(x^{-p}\):
This formula applies because differentiating \(x^{1-p}/(1-p)\) gives \(x^{-p}\). When \(p=1\), the power rule cannot be used: its denominator would be zero. The corresponding antiderivative is \(\ln x\), since \(\frac{d}{dx}\ln x=1/x\) for \(x>0\).
Deriving the Threshold at Infinity
Take \(a>0\) and first suppose \(p\ne1\). Evaluate the truncated integral using the power antiderivative:
If \(p>1\), then \(1-p<0\), so \(R^{1-p}=1/R^{p-1}\) tends to zero as \(R\to\infty\). The limit is finite:
If \(p<1\), then \(1-p>0\), so \(R^{1-p}\) grows without bound. The denominator \(1-p\) is positive, and the truncated integral tends to \(+\infty\). At the remaining value \(p=1\), the integral is \(\int_a^R 1/x\,dx=\ln R-\ln a\), which also tends to \(+\infty\). This proves the benchmark for every real \(p\).
Worked Example: A Convergent Power Integral at Infinity
Evaluate \(\int_2^\infty \frac{1}{x^2}\,dx\). Write the integrand as \(x^{-2}\), so \(p=2\). Since \(2>1\), the benchmark predicts convergence.
First evaluate with a finite upper limit \(R\):
Now take the limit:
Check the antiderivative: \(\frac{d}{dx}(-x^{-1})=x^{-2}\). The finite-limit evaluation and the convergence test agree.
Worked Example: The Logarithmic Boundary at Infinity
Determine whether \(\int_1^\infty \frac{1}{x}\,dx\) converges. Here \(p=1\), exactly the boundary value, so use the logarithmic antiderivative rather than the power rule.
For a finite upper limit \(R\),
Since \(\ln R\to+\infty\) as \(R\to\infty\), the improper integral diverges. The check is \(\frac{d}{dx}\ln x=1/x\); the antiderivative is correct, and its limit is not finite.
Deriving the Threshold Near Zero
Now let \(b>0\), and truncate the interval at \(\varepsilon>0\). For \(p\ne1\),
If \(p<1\), then \(1-p>0\), and \(\varepsilon^{1-p}\to0\) as \(\varepsilon\to0^+\). The limit is finite:
If \(p>1\), then \(1-p<0\), and \(\varepsilon^{1-p}=1/\varepsilon^{p-1}\) grows without bound as \(\varepsilon\to0^+\). The truncated integral tends to \(+\infty\). For \(p=1\), the logarithmic calculation gives \(\int_\varepsilon^b 1/x\,dx=\ln b-\ln\varepsilon\), which also tends to \(+\infty\). Thus convergence at zero requires \(p<1\).
Worked Example: A Convergent Power Integral Near Zero
Evaluate \(\int_0^4 x^{-1/2}\,dx\). Here \(p=\frac12\), and \(\frac12<1\), so the benchmark predicts convergence.
Use a positive lower cutoff \(\varepsilon\). An antiderivative is \(2x^{1/2}=2\sqrt{x}\), because its derivative is \(x^{-1/2}\). Therefore,
Taking the endpoint limit gives
As a check, \(\frac{d}{dx}(2\sqrt{x})=2\cdot\frac12x^{-1/2}=x^{-1/2}\) for \(x>0\). The limit exists and the antiderivative differentiates to the integrand.
Worked Example: The Logarithmic Boundary Near Zero
Determine whether \(\int_0^1 \frac{1}{x}\,dx\) converges. The exponent is \(p=1\), the boundary value for the zero-endpoint test.
For \(0<\varepsilon<1\),
As \(\varepsilon\to0^+\), \(-\ln\varepsilon\to+\infty\). Therefore the improper integral diverges. The derivative check is \(\frac{d}{dx}\ln x=1/x\) on \(x>0\); it is the limit, not an antiderivative error, that causes divergence.
Using Other Exponent Forms
The same benchmarks can be stated for \(x^r\), where \(r\) is any real exponent. Since \(x^r=x^{-p}\) when \(p=-r\), the test at infinity becomes \(r<-1\), while the test at zero becomes \(r>-1\).
| Improper endpoint | Power written as \(x^{-p}\) | Power written as \(x^r\) |
|---|---|---|
| Infinity, from a positive lower bound | Converges when \(p>1\) | Converges when \(r<-1\) |
| Zero, up to a positive upper bound | Converges when \(p<1\) | Converges when \(r>-1\) |
| Either endpoint, boundary power | \(p=1\) | \(r=-1\), giving \(1/x\) |
The interval's fixed positive endpoint affects the numerical value of a convergent integral, but not the threshold. For example, the infinity formula includes \(a\), and the zero formula includes \(b\); the conditions \(p>1\) and \(p<1\) remain unchanged.
A Common Mistake: Reversing the Tests
A frequent error is to remember only that “the exponent must be greater than one” and apply that rule at both endpoints. For \(x^{-p}\), that is incorrect near zero. For example, taking \(p=2\) and claiming that \(\int_0^1 x^{-2}\,dx\) converges because \(2>1\) applies the infinity test to the wrong endpoint.
The cutoff calculation exposes the error:
So this integral diverges. The correction is to match the threshold to the endpoint: for \(x^{-p}\), require \(p>1\) at infinity, but \(p<1\) at zero. A quick check is to test the borderline function \(1/x\): it diverges at both endpoints, separating the two strict inequalities.
Why These Benchmarks Matter
Power integrals provide exact reference cases for deciding whether an improper integral has a finite value. The essential work is not merely finding an antiderivative. One must also use the correct cutoff, evaluate the limit, and verify that the limit is finite. The exponent threshold summarizes that limit analysis without removing the need to identify which endpoint is improper.
For a reliable decision, first rewrite the power in the form \(x^{-p}\), then identify whether the interval extends to infinity or approaches zero. Compare \(p\) with one using the corresponding strict inequality. If \(p=1\), use the logarithm and conclude divergence. When the integral converges, evaluate its value by taking the endpoint limit of the antiderivative.
Check Your Understanding
Use endpoint limits, not just an antiderivative, to justify each conclusion.
- Does \(\int_3^\infty x^{-3/2}\,dx\) converge? If it does, find its value.
- Does \(\int_0^2 x^{-3/2}\,dx\) converge? State the relevant threshold.
- For what values of \(r\) does \(\int_1^\infty x^r\,dx\) converge?
- For what values of \(r\) does \(\int_0^1 x^r\,dx\) converge?
- Why must \(p=1\) be checked with a logarithmic antiderivative rather than the power rule?
