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Convergence Comparisons · Tutorial 339 of 998

Iterated Logarithms at Critical Thresholds

Learn how each logarithmic layer changes convergence precisely when the preceding power is at its critical exponent.

Intermediate 10 min read

What You'll Learn

  • Classify tails with a power of x and real powers of log x and log log x
  • Apply successive limit comparisons at the critical power exponent
  • Identify the separate thresholds for log x and log log x
  • Use substitutions to expose the relevant power-integral test
  • Avoid assuming that a factor that tends to zero guarantees convergence

Introduction

For many improper integrals, a power of \(x\) determines convergence. At a critical power, however, that power alone no longer decides the result. Factors involving \(\ln x\), and then \(\ln\ln x\), can settle the question. Each new logarithmic layer matters precisely when the previous layer is itself at its threshold.

This tutorial studies tails that have the form of a power of \(x\) multiplied by powers of \(\ln x\) and \(\ln\ln x\). The key is to compare in stages: first inspect the power of \(x\); if it is critical, inspect the power of \(\ln x\); if that too is critical, inspect the power of \(\ln\ln x\).

Scope: We consider integrals from \(e^e\) to infinity. On this interval, \(x\), \(\ln x\), and \(\ln\ln x\) are positive, so their real powers are defined. Changing a finite lower endpoint does not affect convergence if the integrand is continuous between the two lower endpoints.

The Family of Tails and Its Thresholds

For real numbers \(p,q,r\), define the positive function

$$ f(x)=\frac{1}{x^p(\ln x)^q(\ln\ln x)^r}, \qquad x\ge e^e. $$

Negative values of \(q\) or \(r\) are allowed. For example, a negative \(r\) means that \((\ln\ln x)^{-r}\) appears in the numerator. The convergence classification is:

Theorem: The improper integral \(\int_{e^e}^{\infty} f(x)\,dx\) converges if and only if one of the following holds:
  • \(p>1\);
  • \(p=1\) and \(q>1\);
  • \(p=1,\ q=1,\) and \(r>1\).
In every other case it diverges. At each critical exponent, equality moves the decision to the next logarithmic layer.

The theorem gives a hierarchy. If \(p\ne1\), neither logarithmic factor changes the outcome. If \(p=1\), the power \(q\) decides unless \(q=1\). Only when both \(p=1\) and \(q=1\) does the value of \(r\) determine convergence.

Why Logarithms Cannot Overcome a Noncritical Power

We first justify why \(p\) settles the result whenever it differs from \(1\). We use the power-integral test: \(\int_A^\infty x^{-s}\,dx\) converges for \(s>1\) and diverges for \(s\le1\), where \(A>0\).

The comparison works because any fixed powers of logarithms grow more slowly than any positive power of \(x\). More precisely, for \(a>0\) and real \(b,c\),

$$ \lim_{x\to\infty}x^{-a}(\ln x)^b(\ln\ln x)^c=0. $$

To see this, put \(t=\ln x\). The expression becomes \(e^{-at}t^b(\ln t)^c\). For sufficiently large \(t\), \(\ln t\le t\), so the absolute size of the factors \(t^b(\ln t)^c\) is bounded above by some fixed power \(t^M\). Since \(e^{-at}t^M\to0\), the displayed limit follows.

If \(p>1\), choose \(s\) with \(1<s<p\). The ratio of \(f(x)\) to the convergent comparison function \(x^{-s}\) is

$$ \frac{f(x)}{x^{-s}} =x^{s-p}(\ln x)^{-q}(\ln\ln x)^{-r} \longrightarrow 0. $$

The limit comparison theorem, including its zero-limit form, then gives convergence: for sufficiently large \(x\), \(f(x)\le x^{-s}\), and the comparison integral converges.

If \(p<1\), choose \(s\) with \(p<s<1\). Now the same ratio satisfies

$$ \frac{f(x)}{x^{-s}} =x^{s-p}(\ln x)^{-q}(\ln\ln x)^{-r} \longrightarrow \infty. $$

The comparison integral \(\int_{e^e}^{\infty}x^{-s}\,dx\) diverges because \(s<1\). Since \(f(x)\ge x^{-s}\) for all sufficiently large \(x\), the integral of \(f\) diverges as well. These comparisons show that fixed logarithmic powers cannot change the outcome when \(p\ne1\).

When the Power of \(x\) Is Critical

Now let \(p=1\). The integral is

$$ \int_{e^e}^{\infty} \frac{dx}{x(\ln x)^q(\ln\ln x)^r}. $$

Use the substitution \(u=\ln x\), so \(du=dx/x\). The lower endpoint \(x=e^e\) becomes \(u=e\), giving

$$ \int_e^{\infty}\frac{du}{u^q(\ln u)^r}. $$

This substitution reveals the next threshold. When \(q>1\), choose \(s\) with \(1<s<q\). The ratio of \(u^{-q}(\ln u)^{-r}\) to \(u^{-s}\) is \(u^{s-q}(\ln u)^{-r}\), which tends to zero because \(q-s>0\). Since \(\int_e^\infty u^{-s}\,du\) converges, the transformed integral converges.

When \(q<1\), choose \(s\) with \(q<s<1\). The same ratio tends to infinity. The comparison integral \(\int_e^\infty u^{-s}\,du\) diverges, so the transformed integral diverges. Thus, unless \(q=1\), the exponent of \(\ln x\) decides the result.

At \(q=1\), a second substitution is required. Set \(v=\ln u\), so \(dv=du/u\). The remaining integral becomes

$$ \int_1^{\infty}\frac{dv}{v^r}. $$

The power-integral test gives convergence exactly when \(r>1\). If \(r\le1\), this last integral diverges. This proves the full classification: each substitution exposes a new power-integral threshold only when the preceding exponent is exactly critical.

Worked Examples

Worked Example: The Triple Critical Case

Classify

$$ \int_{e^e}^{\infty} \frac{dx}{x\ln x\,\ln\ln x}. $$

Here \(p=1\), \(q=1\), and \(r=1\). Both earlier layers are critical, so we must check the final logarithmic exponent. First set \(u=\ln x\), with \(du=dx/x\):

$$ \int_{e^e}^{\infty}\frac{dx}{x\ln x\,\ln\ln x} = \int_e^{\infty}\frac{du}{u\ln u}. $$

Now set \(v=\ln u\), with \(dv=du/u\). The lower endpoint \(u=e\) becomes \(v=1\):

$$ \int_e^{\infty}\frac{du}{u\ln u} = \int_1^{\infty}\frac{dv}{v} = \lim_{B\to\infty}\left[\ln v\right]_1^B = \lim_{B\to\infty}\ln B = \infty. $$

Therefore the original integral diverges. The check is the transformed power integral: its exponent is \(r=1\), exactly the divergent boundary case.

Worked Example: Convergence Just Beyond the Iterated-Log Threshold

Classify

$$ \int_{e^e}^{\infty} \frac{dx}{x\ln x\,(\ln\ln x)^2}. $$

Here \(p=1\), \(q=1\), and \(r=2\). The first substitution \(u=\ln x\) gives

$$ \int_{e^e}^{\infty} \frac{dx}{x\ln x\,(\ln\ln x)^2} = \int_e^{\infty}\frac{du}{u(\ln u)^2}. $$

Set \(v=\ln u\). Since \(dv=du/u\) and \(v=1\) at \(u=e\),

$$ \int_e^{\infty}\frac{du}{u(\ln u)^2} = \int_1^{\infty}\frac{dv}{v^2} = \lim_{B\to\infty}\left[-\frac{1}{v}\right]_1^B = 1. $$

The integral converges, with value \(1\). Differentiating the antiderivative \(-1/v\) gives \(1/v^2\), confirming the evaluation. This is the threshold result: the final exponent \(r=2\) is greater than \(1\).

Worked Example: A Logarithmic Factor Cannot Rescue a Subcritical Power

Classify

$$ \int_{e^e}^{\infty} \frac{dx}{x^{0.8}\ln x\,(\ln\ln x)^2}. $$

The power exponent is \(p=0.8<1\), so we compare with \(x^{-0.9}\), whose integral diverges. The ratio is

$$ \frac{x^{-0.8}(\ln x)^{-1}(\ln\ln x)^{-2}}{x^{-0.9}} = \frac{x^{0.1}}{(\ln x)(\ln\ln x)^2}. $$

This ratio tends to infinity: its reciprocal \(x^{-0.1}(\ln x)(\ln\ln x)^2\) tends to zero by the logarithmic-growth limit above. Thus the original integrand is at least \(x^{-0.9}\) for all sufficiently large \(x\). Since

$$ \int_{e^e}^{\infty}x^{-0.9}\,dx = \lim_{R\to\infty}\left[10x^{0.1}\right]_{e^e}^{R} = \infty, $$

the original integral diverges by direct comparison. This verifies the classification even though the logarithmic factors appear in the denominator: they cannot overcome the subcritical power.

Worked Example: Convergence When the Logarithmic Exponent Exceeds Its Threshold

Classify

$$ \int_{e^e}^{\infty}\frac{dx}{x(\ln x)^{3/2}}. $$

Here \(p=1\), \(q=3/2>1\), and \(r=0\). Set \(u=\ln x\), so \(du=dx/x\), and the lower limit becomes \(e\):

$$ \int_{e^e}^{\infty}\frac{dx}{x(\ln x)^{3/2}} = \int_e^{\infty}u^{-3/2}\,du = \lim_{B\to\infty}\left[-2u^{-1/2}\right]_e^B = \frac{2}{\sqrt e}. $$

The value is finite, so the integral converges. Differentiating \(-2u^{-1/2}\) gives \(u^{-3/2}\), confirming the transformed evaluation. The classification also checks directly: after \(p=1\), the exponent \(q=3/2\) exceeds \(1\).

A Common Mistake at the Final Threshold

A frequent error is to stop after recognizing that both \(p\) and \(q\) are \(1\), then assume the integral converges because the denominator contains two logarithmic factors. The presence of logarithms alone does not decide the result; their powers matter.

For instance, the calculation

$$ \int\frac{dx}{x\ln x\,\ln\ln x} \overset{\text{wrong}}{=} \int\frac{du}{u} $$

omits a factor. With \(u=\ln x\), we have \(du=dx/x\), but \(\ln\ln x=\ln u\) remains in the denominator. The correct calculation is

$$ \int\frac{dx}{x\ln x\,\ln\ln x} = \int\frac{du}{u\ln u} = \int\frac{dv}{v}, \qquad v=\ln u. $$

The final integral diverges. Keeping the remaining logarithmic factor at every substitution is essential: each layer is tested only after the preceding substitution has exposed it.

Using the Thresholds Efficiently

For this family, a reliable classification procedure is to inspect the exponents in order. Do not compare every factor at once. If \(p\ne1\), stop: the power of \(x\) decides. If \(p=1\), inspect \(q\). If \(q\ne1\), stop: the power of \(\ln x\) decides. Only if \(p=q=1\) should you inspect \(r\).

1
Check the power of \(x\).
If \(p>1\), the tail converges; if \(p<1\), it diverges.
2
If \(p=1\), check the power of \(\ln x\).
If \(q>1\), the tail converges; if \(q<1\), it diverges.
3
If \(p=q=1\), check the power of \(\ln\ln x\).
The tail converges exactly when \(r>1\).

This method is useful because it makes clear which factor is doing the decisive work. It also prevents a common overgeneralization: a factor tending to zero does not automatically make an integral converge. Its rate must be compared with the critical scale. The successive substitutions and comparisons identify that scale precisely.

Check Your Understanding

For each integral, decide whether it converges or diverges. Identify the exponent that determines the result, and use substitutions or comparisons to justify your classification.

  1. \(\int_{e^e}^{\infty}\frac{dx}{x(\ln x)^{1/2}(\ln\ln x)^3}\)
  2. \(\int_{e^e}^{\infty}\frac{dx}{x\ln x(\ln\ln x)^{3/2}}\)
  3. \(\int_{e^e}^{\infty}\frac{dx}{x^{3/2}(\ln x)^{-2}}\)
  4. \(\int_{e^e}^{\infty}\frac{dx}{x(\ln x)(\ln\ln x)}\)
  5. State which exponent must equal \(1\) before the value of \(r\) can affect convergence.

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