Why the Logarithmic Factor Matters
A power singularity near zero has a clear threshold: changing the exponent can switch an improper integral from convergent to divergent. A logarithmic factor can affect the outcome too, but its influence depends on the power exponent. In particular, when the power is exactly at its threshold, the logarithmic exponent determines convergence.
We will classify the integral of \(x^{-p}(\ln(1/x))^q\) near zero for real parameters \(p\) and \(q\). Taking the upper endpoint below \(1\) ensures that \(\ln(1/x)\) is positive throughout the interval, so its real power is well-defined.
The integrand is positive for \(0<x<a\). Thus, convergence means the truncated integrals approach a finite value; divergence means they grow without bound.
Transforming the Endpoint
Use the substitution \(t=\ln(1/x)\), or equivalently \(x=e^{-t}\). Write \(T=\ln(1/a)>0\). For a positive cutoff \(\varepsilon<a\), the substitution takes \(x=a\) to \(t=T\) and \(x=\varepsilon\) to \(t=R=\ln(1/\varepsilon)\). Also, \(dx=-e^{-t}\,dt\) and \(x^{-p}=e^{pt}\). Therefore the truncated integral becomes
As \(\varepsilon\to0^+\), \(R\to\infty\). Consequently, the original endpoint question is equivalent to asking whether \(\int_T^\infty e^{(p-1)t}t^q\,dt\) converges. This transformed integral makes the distinct roles of \(p\) and \(q\) visible: \(p\) controls an exponential factor, while \(q\) controls a power of \(t\).
The Convergence Classification
The cases with \(p\ne1\) follow from the fact that exponential growth or decay eventually dominates every fixed real power. Here is a direct justification in the transformed variable.
Set \(c=1-p>0\). The transformed integrand is \(e^{-ct}t^q\). For \(t\ge1\), if \(q\le0\), then \(t^q\le1\), so the integrand is at most \(e^{-ct}\). If \(q>0\), then \(q\ln t\le ct/2\) for all sufficiently large \(t\), since \(\ln t/t\to0\). Hence \(t^q\le e^{ct/2}\), and \(e^{-ct}t^q\le e^{-ct/2}\) there. Both exponential upper bounds have finite integrals on a tail. The remaining finite interval contributes a finite amount because the integrand is continuous there. Thus the integral converges for every real \(q\).
Set \(c=p-1>0\). The transformed integrand is \(e^{ct}t^q\). For \(q\ge0\) and \(t\ge1\), \(t^q\ge1\). For \(q<0\), the limit \(\ln t/t\to0\) implies \(q\ln t\ge-ct/2\) for all sufficiently large \(t\), so \(t^q\ge e^{-ct/2}\). In either case, eventually \(e^{ct}t^q\ge e^{ct/2}\) (in the first case, \(e^{ct}\ge e^{ct/2}\)). Since the integral of \(e^{ct/2}\) over an infinite tail diverges, so does the transformed integral.
The exponential factor is \(1\), leaving \(\int_T^\infty t^q\,dt\). For \(q\ne-1\), an antiderivative is \(t^{q+1}/(q+1)\); for \(q=-1\), it is \(\ln t\). Taking the upper limit to infinity gives convergence precisely when \(q+1<0\), that is, \(q<-1\).
The power threshold for \(p\) is consistent with the earlier material on the power singularity threshold. The additional feature here is the borderline \(p=1\): at that value, there is no exponential decay or growth after substitution, so the logarithmic factor decides the result.
Worked Examples at the Borderline
Worked Example: A Convergent Logarithmic Borderline
Evaluate \(\int_0^{e^{-2}}\frac{1}{x(\ln(1/x))^2}\,dx\). This has \(p=1\), \(q=-2\), and \(T=\ln(1/e^{-2})=2\). After the substitution, the truncated integral is
As \(\varepsilon\to0^+\), \(R\to\infty\), so \(1/R\to0\) and the integral equals \(1/2\). To check this limit in the original variable, an antiderivative is \(F(x)=1/\ln(1/x)\), because
Also, \(\lim_{x\to0^+}F(x)=0\) and \(F(e^{-2})=1/2\), confirming the value.
Worked Example: The Logarithmic Threshold Diverges
Consider \(\int_0^{e^{-1}}\frac{1}{x\ln(1/x)}\,dx\). Here \(p=1\), \(q=-1\), and \(T=1\). For \(0<\varepsilon<e^{-1}\), the substitution gives
As \(\varepsilon\to0^+\), \(R=\ln(1/\varepsilon)\to\infty\), and therefore \(\ln R\to\infty\). The improper integral diverges. The equality \(\int_1^R dt/t=\ln R\) also checks the antiderivative, since the derivative of \(\ln t\) is \(1/t\). This is the exact boundary between the convergent case \(q<-1\) and the divergent case \(q\ge-1\) when \(p=1\).
Logarithms Cannot Rescue a Stronger Power Singularity
For \(p>1\), even a negative logarithmic power—which makes the integrand smaller—cannot counteract the resulting exponential growth in the transformed variable. A concrete example illustrates why it is important to classify the power exponent before drawing conclusions from the logarithm alone.
Worked Example: A Negative Logarithmic Power Still Diverges
Determine whether \(\int_0^{e^{-1}}x^{-3/2}(\ln(1/x))^{-2}\,dx\) converges. Here \(p=3/2>1\), \(q=-2\), and \(T=1\). The truncated integral transforms to
For \(t\ge32\), \(2\ln t\le t/4\). Indeed, this holds at \(t=32\), and the derivative of \(t/4-2\ln t\) is \(1/4-2/t\ge0\) for \(t\ge32\). Exponentiating gives \(t^{-2}\ge e^{-t/4}\). Thus, on this tail, \(e^{t/2}t^{-2}\ge e^{t/4}\), and
Therefore the original improper integral diverges. The comparison is checked by differentiating the displayed antiderivative \(4e^{t/4}\), which gives \(e^{t/4}\). The theorem predicts the same result because \(p>1\), regardless of \(q\).
A Common Mistake: Looking at the Logarithm in Isolation
A tempting but incorrect argument is to see a negative logarithmic exponent and conclude that the integrand must be sufficiently small near zero. For example, one might reason that \((\ln(1/x))^{-2}\to0\), so \(x^{-3/2}(\ln(1/x))^{-2}\) should be integrable. That conclusion does not follow: the factor \(x^{-3/2}\) grows strongly enough to force divergence.
Another common error is to include \(q=-1\) among the convergent cases because it is “just” a reciprocal power. But its transformed integral is \(\int_T^\infty dt/t\), whose truncated value grows like \(\ln R\). The inequality in the borderline condition is strict: convergence requires \(q<-1\), not \(q\le-1\).
Why the Classification Is Useful
This test gives a quick and precise way to judge common endpoint models without confusing two different effects. A power factor controls the main threshold; logarithmic powers refine the answer only when that power is exactly critical. It is also useful to keep the transformed variable in view: a difficult-looking singularity at \(x=0\) becomes an exponential-power tail, and the cases can then be read from its limiting behavior.
| Power exponent | Logarithmic exponent | Result near zero |
|---|---|---|
| \(p<1\) | Any real \(q\) | Converges |
| \(p=1\) | \(q<-1\) | Converges |
| \(p=1\) | \(q\ge-1\) | Diverges |
| \(p>1\) | Any real \(q\) | Diverges |
In every case, the classification concerns behavior near zero. Changing the upper endpoint to another fixed number in \((0,1)\) changes the finite lower limit \(T\) in the transformed integral, not the convergence of its tail.
Check Your Understanding
Use the substitution \(t=\ln(1/x)\) and the endpoint limit to answer each question.
- Does \(\int_0^a x^{-3/4}(\ln(1/x))^5\,dx\) converge for \(0<a<1\)? State which part of the classification applies.
- For which real values of \(q\) does \(\int_0^a x^{-1}(\ln(1/x))^q\,dx\) converge?
- Does \(\int_0^a x^{-2}(\ln(1/x))^{-100}\,dx\) converge? Explain why the logarithmic factor does or does not change the result.
- For \(q=-3\), evaluate \(\int_0^{e^{-1}}x^{-1}(\ln(1/x))^q\,dx\).
- What limit of the transformed upper endpoint corresponds to \(\varepsilon\to0^+\)?
