Logarithmic Singularities at Finite Endpoints

Why Logarithms Change the Endpoint Test

A power singularity such as \(x^{-p}\) has a clear convergence threshold at zero: the integral converges when \(p<1\) and diverges when \(p\geq 1\). A logarithmic factor can affect the outcome, but its influence depends on the power.

The integral considered here is \(\int_0^a 1/[x^p(\log(1/x))^q]\,dx\), where \(0<a<1\) and \(p,q\) are real. Restricting \(a\) to be less than 1 ensures that \(\log(1/x)>0\) throughout the interval, so real powers of that logarithm are well-defined.

The key distinction is this: when \(p\) differs from 1, the power determines convergence for every fixed logarithmic exponent \(q\). Only at the critical power \(p=1\) does the logarithmic exponent decide the result.

Transforming the Endpoint

The substitution \(t=\log(1/x)\), equivalently \(x=e^{-t}\), turns the endpoint \(x=0\) into \(t=\infty\). Since \(dx=-e^{-t}\,dt\) and \(x^{-p}=e^{pt}\), the transformed integrand is \(e^{(p-1)t}t^{-q}\).

To make this precise, fix \(0<\varepsilon<a\), set \(T=\log(1/a)>0\), and set \(R=\log(1/\varepsilon)\). Reversing the bounds after the substitution gives

As \(\varepsilon\) decreases to zero, \(R\) increases without bound. Thus the original improper integral converges precisely when the transformed integral has a finite limit as \(R\to\infty\). The part of the original integral away from zero causes no difficulty: the integrand is continuous on every closed interval \([\varepsilon,a]\) with \(\varepsilon>0\).

The Power Threshold and the Critical Case

The transformed expression makes the cases easy to separate. If \(p=1\), the exponential factor is exactly 1, leaving an ordinary power of \(t\). If \(p\neq1\), an exponential factor remains, and it determines the outcome regardless of the fixed real value of \(q\).

Here is a proof that also explains why the logarithmic factor cannot shift the power threshold. Write \(c=1-p\) when \(p<1\), so \(c>0\). The transformed integrand is then \(e^{-ct}t^{-q}\). For \(q\geq0\) and \(t\geq1\), \(t^{-q}\leq1\), so the integrand is at most \(e^{-ct}\), whose integral over a tail is finite.

For \(q<0\), put \(r=-q>0\), making the integrand \(e^{-ct}t^r\). For \(t\geq1\), \(\log t\leq\sqrt{t}\). Indeed, \(\sqrt{t}-\log t\) has its minimum at \(t=4\), where it is positive. If \(t\) is sufficiently large that \(r\sqrt{t}\leq ct/2\), then

The final exponential has a finite integral on a tail. This proves convergence for \(p<1\), including when the logarithmic factor grows rather than shrinks.

Now suppose \(p>1\), and write \(c=p-1>0\). The transformed integrand is \(e^{ct}t^{-q}\). If \(q<0\), then \(t^{-q}\geq1\) for \(t\geq1\), so the integrand is at least \(e^{ct}\) and its integral diverges. If \(q\geq0\), the same estimate \(\log t\leq\sqrt{t}\) shows that \(t^q\leq e^{ct/2}\) for all sufficiently large \(t\). Consequently,

for all sufficiently large \(t\). This lower bound has a divergent integral, proving divergence for \(p>1\).

It remains to consider \(p=1\). The transformed integral is \(\int_T^\infty t^{-q}\,dt\). For \(q\neq1\), its truncated value is

This has a finite limit as \(R\to\infty\) exactly when \(q>1\). At \(q=1\), the truncated value is \(\log R-\log T\), which tends to infinity. Thus the critical logarithmic threshold is \(q=1\), with convergence only for \(q>1\).

Worked Examples

A Common Misreading of the Borderline Case

A frequent mistake is to see that \(1/t\to0\) and conclude that \(\int_1^\infty 1/t\,dt\) must converge. The fact that an integrand tends to zero does not establish convergence of its improper integral.

For the corresponding endpoint integral, the incorrect reasoning might be: “\(1/\log(1/x)\) tends to zero as \(x\to0^+\), so \(\int_0^{e^{-1}}1/[x\log(1/x)]\,dx\) converges.” But the substitution gives the exact calculation

The correction is to test the accumulated integral, not just the limiting value of the integrand. At \(p=1\), the distinction between \(q=1\) and \(q>1\) matters: \(q=1\) diverges, while \(q=2\) converges.

Using the Classification Reliably

Before applying the test, identify the endpoint and put the integrand into the form \(1/[x^p(\log(1/x))^q]\) near zero. A factor \((\log(1/x))^2\) in the numerator corresponds to \(q=-2\), not \(q=2\). Once the exponents are identified, check \(p\) first; examine \(q\) only if \(p=1\).

This distinction is useful whenever a logarithmic multiplier accompanies a power singularity. It prevents two opposite errors: treating every logarithmic factor as decisive, and treating it as irrelevant even at the critical power. Away from \(p=1\), the power decides; at \(p=1\), the logarithm supplies the final threshold.