Power and Logarithmic Factors at an Endpoint
For a power \(x^{-p}\), changing the exponent across a single threshold changes whether the integral is finite. A logarithmic factor can affect the answer precisely at that threshold: it can turn a divergent power integral into a convergent one, or make a convergent power integral diverge if it grows sufficiently quickly. This tutorial gives the classifications near zero and infinity, including the borderline cases.
All integrands below are real-valued, positive and continuous on the open intervals where they are considered. Near zero we use \(\log(1/x)\), which is positive for \(0<x<1\); at infinity we use \(\log x\), positive for \(x>1\). The exponents \(p,q\) may be any real numbers. For \(0<a<1\), the underlying measure space \(((0,a),\mathcal L|_{(0,a)},m)\) has finite measure. For \(b>1\), \(((b,\infty),\mathcal L|_{(b,\infty)},m)\) is sigma-finite and has infinite measure. In each case, the nonnegative integral is finite exactly when the corresponding improper integral converges; this is the standard equivalence between nonnegative improper integrability and Lebesgue integrability.
The theorem concerns the behavior at the indicated endpoint. The particular cutoffs \(a\) and \(b\) do not affect the classification: on any closed interval bounded away from zero and infinity, respectively, the integrand is continuous and bounded, so its integral there is finite.
Why the Logarithmic Substitution Gives the Thresholds
We first establish the comparison that handles powers of the new variable against exponentials.
Proof. If \(c\ne0\), then \(\log t/t\to0\) as \(t\to\infty\). Therefore there is \(T_1\ge T\) such that \(|q|\log t\le |c|t/2\) for every \(t\ge T_1\). If \(c<0\), then \[ e^{ct}t^{-q}=\exp(ct-q\log t) \le \exp(-|c|t+|q|\log t) \le e^{-|c|t/2} \] for \(t\ge T_1\). The last function has finite integral on \([T_1,\infty)\), so comparison proves convergence. If \(c>0\), the same bound on \(|q|\log t\) gives \[ e^{ct}t^{-q}=\exp(ct-q\log t) \ge \exp(ct-|q|\log t) \ge e^{ct/2}\ge1 \] for \(t\ge T_1\). Its integral is therefore infinite. If \(c=0\), the integrand is \(t^{-q}\). When \(q\ne1\), an antiderivative is \(t^{1-q}/(1-q)\), which has a finite limit at infinity exactly when \(q>1\). When \(q=1\), the antiderivative is \(\log t\), which tends to infinity. These cases prove the lemma. \(\square\)
Proof of the theorem. For the integral near zero, first truncate at \(\varepsilon\in(0,a)\) and substitute \(t=\log(1/x)\). Since \(x=e^{-t}\) and \(dx=-e^{-t}\,dt\), the limits \(x=\varepsilon\) and \(x=a\) become \(t=\log(1/\varepsilon)\) and \(t=\log(1/a)\), respectively. Thus \[ \int_\varepsilon^a x^{-p}(\log(1/x))^{-q}\,dx =\int_{\log(1/a)}^{\log(1/\varepsilon)}e^{(p-1)t}t^{-q}\,dt. \] As \(\varepsilon\downarrow0\), the upper limit tends to infinity. The lemma applies with \(c=p-1\): the integral converges for \(p<1\), diverges for \(p>1\), and when \(p=1\) converges exactly for \(q>1\).
For the integral at infinity, truncate at \(R>b\) and substitute \(t=\log x\). Now \(x=e^t\) and \(dx=e^t\,dt\), so \[ \int_b^R x^{-p}(\log x)^{-q}\,dx =\int_{\log b}^{\log R}e^{(1-p)t}t^{-q}\,dt. \] As \(R\to\infty\), the upper limit tends to infinity. The lemma applies with \(c=1-p\): the integral converges for \(p>1\), diverges for \(p<1\), and when \(p=1\) converges exactly for \(q>1\). Since the integrands are nonnegative, taking these increasing truncation limits gives their improper integrals, with \(+\infty\) allowed. This proves both classifications. \(\square\)
Worked Example: A Convergent Borderline Integral Near Zero
Evaluate \(\displaystyle\int_0^{1/2}\frac{dx}{x(\log(1/x))^2}\), as a nonnegative Lebesgue integral on \(((0,1/2),\mathcal L|_{(0,1/2)},m)\).
Here \(p=1\) and \(q=2\). The threshold theorem says the integral is finite because \(q>1\). To compute it, truncate at \(\varepsilon\in(0,1/2)\) and set \(t=\log(1/x)\). The lower endpoint \(x=\varepsilon\) gives \(t=\log(1/\varepsilon)\), and the upper endpoint \(x=1/2\) gives \(t=\log 2\). Hence \[ \int_\varepsilon^{1/2}\frac{dx}{x(\log(1/x))^2} =\int_{\log 2}^{\log(1/\varepsilon)}t^{-2}\,dt =\left[-\frac1t\right]_{\log 2}^{\log(1/\varepsilon)} =\frac1{\log 2}-\frac1{\log(1/\varepsilon)}. \] As \(\varepsilon\downarrow0\), \(\log(1/\varepsilon)\to\infty\), so the second term tends to zero. The value is \(\boxed{1/\log 2}\). Check: differentiating the antiderivative \(-1/t\) gives \(t^{-2}\), and the endpoint contribution at infinity is zero.
Worked Example: A Divergent Borderline Integral Near Zero
Determine whether \(\displaystyle\int_0^{1/e}\frac{dx}{x\log(1/x)}\) is finite on \(((0,1/e),\mathcal L|_{(0,1/e)},m)\). The integrand is positive on this interval. Its exponents are \(p=1\) and \(q=1\), so the theorem predicts divergence because the borderline condition \(q>1\) fails.
For \(0<\varepsilon<1/e\), the same substitution gives \[ \int_\varepsilon^{1/e}\frac{dx}{x\log(1/x)} =\int_1^{\log(1/\varepsilon)}\frac{dt}{t} =\left[\log t\right]_1^{\log(1/\varepsilon)} =\log\bigl(\log(1/\varepsilon)\bigr)-\log 1 =\log\bigl(\log(1/\varepsilon)\bigr). \] As \(\varepsilon\downarrow0\), \(\log(1/\varepsilon)\to\infty\), and therefore \(\log(\log(1/\varepsilon))\to\infty\). The integral is \(+\infty\). Check: the antiderivative \(\log t\) differentiates to \(1/t\), and the lower-end contribution is \(\log 1=0\).
Thresholds at Infinity and for Pure Powers
The same logarithmic factor has the same borderline condition at infinity, but the power threshold is reversed: a power \(x^{-p}\) decays fast enough when \(p>1\). The following example shows that even a growing logarithmic factor does not prevent convergence when the power is sufficiently strong.
Worked Example: A Growing Logarithm at Infinity
Evaluate \(\displaystyle\int_2^\infty\frac{\log x}{x^2}\,dx\) on the sigma-finite measure space \(((2,\infty),\mathcal L|_{(2,\infty)},m)\). This is the case \(p=2\), \(q=-1\), since \((\log x)^{-q}=\log x\). The threshold theorem gives convergence because \(p>1\), regardless of \(q\).
For \(R>2\), integration by parts with \(u=\log x\) and \(dv=x^{-2}\,dx\) gives \(du=dx/x\) and \(v=-1/x\). Therefore \[ \int_2^R\frac{\log x}{x^2}\,dx =\left[-\frac{\log x}{x}\right]_2^R+\int_2^R\frac{1}{x^2}\,dx =-\frac{\log R}{R}+\frac{\log 2}{2}+\left[-\frac1x\right]_2^R =\frac{\log 2+1}{2}-\frac{\log R+1}{R}. \] The final term tends to zero as \(R\to\infty\); for example, \(\log R/R\to0\) and \(1/R\to0\). Thus the integral is \(\boxed{(\log 2+1)/2}\). Check: the antiderivative \(-(\log x+1)/x\) differentiates to \((\log x)/x^2\), and its value at infinity is zero.
Worked Example: A Pure Power Near Zero
Evaluate \(\displaystyle\int_0^{1/2}x^{-1/2}\,dx\) on \(((0,1/2),\mathcal L|_{(0,1/2)},m)\). This is the special case \(p=1/2\), \(q=0\). Since \(p<1\), the theorem predicts a finite integral.
For \(0<\varepsilon<1/2\), \[ \int_\varepsilon^{1/2}x^{-1/2}\,dx =\left[2\sqrt{x}\right]_\varepsilon^{1/2} =2\sqrt{\frac12}-2\sqrt{\varepsilon} =\sqrt{2}-2\sqrt{\varepsilon}. \] Letting \(\varepsilon\downarrow0\) gives \(\boxed{\sqrt{2}}\). Check: the derivative of \(2\sqrt{x}\) is \(x^{-1/2}\) for \(x>0\), and the omitted endpoint term \(2\sqrt{\varepsilon}\) tends to zero.
Reading the Classification Correctly
The power exponent decides the answer except at \(p=1\). At that boundary, the substitution turns the problem into the familiar integral of \(t^{-q}\): it converges precisely when \(q>1\). In particular, a logarithmic factor with \(q=1\) is not strong enough to produce convergence, while any \(q>1\) is sufficient.
| Endpoint | Power range | Convergence condition |
|---|---|---|
| Zero | \(p<1\) | Every \(q\in\mathbb R\) |
| Zero | \(p=1\) | \(q>1\) |
| Zero | \(p>1\) | No \(q\) |
| Infinity | \(p>1\) | Every \(q\in\mathbb R\) |
| Infinity | \(p=1\) | \(q>1\) |
| Infinity | \(p<1\) | No \(q\) |
Common mistake. It is incorrect to classify every integral with a factor \(x^{-1}\) as divergent by looking only at the power. For example, the claim that \(\int_0^{1/2}dx/[x(\log(1/x))^2]\) diverges because \(\int_0^{1/2}dx/x\) diverges ignores a factor that changes the borderline behavior. Substituting \(t=\log(1/x)\) gives \(\int_{\log2}^\infty t^{-2}\,dt=1/\log2\), which is finite. The correction is to check the logarithmic exponent whenever \(p=1\): convergence requires \(q>1\), not merely \(q\ge1\).
These thresholds provide quick tests for nonnegative power-logarithmic integrands, while the substitutions show exactly why the boundary cases differ. For signed or oscillatory integrands, cancellation can change convergence, so this classification alone does not decide their improper integrals.
Check Your Understanding
For each integral, identify the endpoint, substitute its \(p\) and \(q\) values into the threshold, and state whether the integral is finite.
- Does \(\displaystyle\int_0^{1/3}x^{-3/4}(\log(1/x))^{-5}\,dx\) converge?
- Classify \(\displaystyle\int_0^{1/2}x^{-1}(\log(1/x))^{-3/2}\,dx\).
- For which real \(q\) does \(\displaystyle\int_3^\infty x^{-1}(\log x)^{-q}\,dx\) converge?
- Does \(\displaystyle\int_2^\infty x^{-5/4}(\log x)^4\,dx\) converge, and which exponent determines the answer?
- Use \(t=\log x\) to express \(\displaystyle\int_1^R dx/[x(\log x)^2]\) with a positive lower limit, and explain why the integral from \(1\) requires separate care.