Logarithmic Factors in Infinite Tails

Why logarithmic factors need their own test

For many positive tails, the power of \(x\) determines whether the integral converges. A logarithmic factor can change the result at the boundary where the power test is critical. To see exactly where that happens, consider

\int_c^\infty \frac{1}{x^p(\log x)^q}\,dx,

where \(p,q\in\mathbb R\) and \(c>1\). The assumption \(c>1\) keeps \(\log x\) positive throughout the interval, so the integrand is defined and positive for every real \(q\). The integral is improper only because its upper endpoint is infinite.

The main distinction is between \(p\ne 1\) and \(p=1\). Away from \(p=1\), the power of \(x\) controls convergence, regardless of the logarithmic power. At \(p=1\), the logarithm controls the answer, and \(q=1\) is the second critical threshold.

Theorem: For \(c>1\) and real \(p,q\), the integral \(\int_c^\infty \frac{1}{x^p(\log x)^q}\,dx\) converges if and only if either \(p>1\), or \(p=1\) and \(q>1\). It diverges when \(p<1\), and also when \(p=1\) and \(q\le 1\).

The substitution that reveals the thresholds

Set \(u=\log x\). Then \(x=e^u\), \(dx=e^u\,du\), and the lower endpoint \(c\) becomes \(\log c>0\). For a finite upper endpoint \(R>c\), this gives

\int_c^R \frac{1}{x^p(\log x)^q}\,dx = \int_{\log c}^{\log R} e^{(1-p)u}u^{-q}\,du.

The formula displays the two cases. If \(p=1\), the exponential factor is exactly \(1\), leaving a power integral in \(u\). If \(p\ne 1\), an exponential factor remains. Exponential growth or decay eventually outweighs any fixed real power of \(u\), so \(q\) cannot change the convergence decision.

Why the noncritical cases are controlled by p

Here is a justification of the exponential-versus-power comparison used in the theorem. The standard limit \(\frac{\log u}{u}\to 0\) as \(u\to\infty\) means that, for any fixed positive constant \(a\) and any fixed real \(q\), there is a number \(U>1\) such that

|q|\log u\le \frac{a}{2}u \qquad\text{for all }u\ge U.

This inequality gives bounds in both directions:

e^{-au}u^{-q}\le e^{-au/2} \qquad\text{and}\qquad e^{au}u^{-q}\ge e^{au/2} \qquad (u\ge U).

Indeed, taking logarithms of the factor \(u^{-q}\) gives \(-q\log u\), which lies between \(-|q|\log u\) and \(|q|\log u\). The displayed bounds follow from the choice of \(U\). The tail \(\int_U^\infty e^{-au/2}\,du\) is finite, while \(\int_U^\infty e^{au/2}\,du\) is infinite. Since the integrands are nonnegative, comparison therefore determines the two cases.

If \(p>1\).
Take \(a=p-1>0\). The transformed integrand is \(e^{-au}u^{-q}\), which is eventually at most \(e^{-au/2}\). Its tail is bounded by a convergent exponential integral.

If \(p<1\).
Take \(a=1-p>0\). The transformed integrand is \(e^{au}u^{-q}\), which is eventually at least \(e^{au/2}\). Its tail is bounded below by a divergent exponential integral.

If \(p=1\).
The transformed integrand is \(u^{-q}\). Evaluate this power tail directly; its convergence depends on whether \(q>1\).

The critical case p=1

When \(p=1\), the substitution reduces the original integral to \(\int_{\log c}^\infty u^{-q}\,du\). For a finite upper endpoint \(T>\log c\), if \(q\ne 1\), then

\int_{\log c}^{T}u^{-q}\,du = \frac{T^{1-q}-(\log c)^{1-q}}{1-q}.

If \(q>1\), then \(1-q<0\), so \(T^{1-q}\to0\) as \(T\to\infty\). The improper integral has a finite limit. If \(q<1\), then \(1-q>0\), so \(T^{1-q}\to\infty\), and the integral diverges. At the boundary \(q=1\), the antiderivative is \(\log u\), which grows without bound as \(T\to\infty\). Thus the critical power tail converges exactly when \(q>1\).

Worked examples

Worked Example: A convergent tail with a negative q

Evaluate \(\int_e^\infty \frac{(\log x)^3}{x^2}\,dx\). In the theorem's notation, \(p=2\) and \(q=-3\), so \(p>1\) predicts convergence, even though the logarithmic factor is in the numerator.

Use \(u=\log x\). Then \(x=e^u\), \(dx=e^u\,du\), \(x^2=e^{2u}\), and the bounds \(x=e\) and \(x=R\) become \(u=1\) and \(u=\log R\). Therefore

\int_e^R \frac{(\log x)^3}{x^2}\,dx = \int_1^{\log R}u^3e^{-u}\,du.

An antiderivative is \(-e^{-u}(u^3+3u^2+6u+6)\). To check it, let \(P(u)=u^3+3u^2+6u+6\). Then \(P'(u)=3u^2+6u+6\), and

\frac{d}{du}\bigl[-e^{-u}P(u)\bigr] =e^{-u}\bigl(P(u)-P'(u)\bigr) =e^{-u}u^3.

As \(u\to\infty\), \(e^{-u}P(u)\to0\). At \(u=1\), \(P(1)=1+3+6+6=16\). Hence

\int_e^\infty \frac{(\log x)^3}{x^2}\,dx =0-\left(-\frac{16}{e}\right) =\frac{16}{e}.

The result is finite, as predicted. Equivalently, the antiderivative in \(x\) is \(-\frac{(\log x)^3+3(\log x)^2+6\log x+6}{x}\); differentiating it gives \(\frac{(\log x)^3}{x^2}\).

Worked Example: The convergent logarithmic boundary case

Evaluate \(\int_e^\infty \frac{1}{x(\log x)^2}\,dx\). Here \(p=1\) and \(q=2>1\), so the theorem predicts convergence.

For finite \(R>e\), set \(u=\log x\). The bounds become \(1\) and \(\log R\), giving

\int_e^R \frac{1}{x(\log x)^2}\,dx = \int_1^{\log R}u^{-2}\,du = \left[-u^{-1}\right]_1^{\log R} = 1-\frac{1}{\log R}.

Taking \(R\to\infty\) yields the value \(1\). To check in the original variable, an antiderivative is \(-1/\log x\), and

\frac{d}{dx}\left(-\frac{1}{\log x}\right) = \frac{1}{x(\log x)^2}.

The endpoint value \(-1/\log R\) tends to zero, while the lower-end value is \(-1\), confirming the integral is \(1\).

Worked Example: Divergence when p is below 1

Classify \(\int_2^\infty x^{-1/2}\,dx\). This is the case \(p=\frac12<1\) and \(q=0\). For finite \(R>2\),

\int_2^R x^{-1/2}\,dx = \left[2x^{1/2}\right]_2^R = 2\sqrt R-2\sqrt 2.

As \(R\to\infty\), \(2\sqrt R-2\sqrt2\to\infty\), so the improper integral diverges. Differentiating \(2\sqrt{x}\) gives \(x^{-1/2}\), verifying the antiderivative and the calculation.

Worked Example: Divergence at q=1

Consider \(\int_e^\infty \frac{1}{x\log x}\,dx\), where \(p=1\) and \(q=1\). For finite \(R>e\), substitution \(u=\log x\) gives

\int_e^R \frac{1}{x\log x}\,dx = \int_1^{\log R}\frac{1}{u}\,du = \log(\log R).

This tends to infinity, so the integral diverges. The check in the original variable is \(\frac{d}{dx}\log(\log x)=\frac{1}{x\log x}\), and the lower-end value is \(\log(\log e)=0\).

A common mistake: using the logarithmic threshold too broadly

The threshold \(q=1\) applies only when \(p=1\). For example, a mistaken argument might say that \(\int_e^\infty \frac{(\log x)^5}{x^{3/2}}\,dx\) diverges because its \(q=-5\) is less than or equal to \(1\). That applies the critical-case test despite \(p=\frac32\), not \(p=1\). The theorem instead gives convergence because \(p>1\), whatever the fixed real value of \(q\).

The correction is to check \(p\) first. Only if \(p=1\) should you compare \(q\) with \(1\). A logarithm can decide the result at the critical power; it does not replace the power test away from that boundary.

Why the classification matters

Logarithmic factors appear naturally in functions whose growth is close to a power law. The classification gives a precise way to distinguish a genuinely convergent tail from one that merely decreases slowly. In particular, \(1/(x\log x)\) still has an infinite integral, while the additional logarithmic factor in \(1/[x(\log x)^2]\) makes the tail integrable.

When applying this result, first verify that the integrand is positive and defined on the tail in question. Here \(c>1\) guarantees both for all real \(p,q\). Then identify the power \(p\); if it equals \(1\), inspect \(q\). The lower endpoint is finite and the integrand is continuous there under this assumption, so the convergence decision comes from the infinite tail.

Check Your Understanding

Use the classification theorem, and show the key comparison or truncated integral where appropriate.

For which real values of \(q\) does \(\int_3^\infty \frac{1}{x^2(\log x)^q}\,dx\) converge?
Classify \(\int_e^\infty \frac{1}{x(\log x)^{3/4}}\,dx\).
Does \(\int_2^\infty \frac{(\log x)^8}{x^{5/4}}\,dx\) converge or diverge?
Find the value of \(\int_e^\infty \frac{1}{x(\log x)^3}\,dx\).
Why is the test based on comparing \(q\) with \(1\) not the right test when \(p\ne1\)?