Why Logarithmic Powers Need Their Own Test
A factor of \(1/x\) often appears in integrals over an unbounded interval. When it is multiplied by a power of \(\ln x\), the logarithm changes the rate of decay in a way that is not captured by the usual power-tail test for \(1/x^p\). The deciding factor is the exponent \(q\) on the logarithm.
The substitution \(u=\ln x\) reveals the structure: the factor \(dx/x\) becomes \(du\), leaving a power of \(u\). The resulting integral can then be evaluated using the power rule, with a separate case when that rule would divide by zero.
Definition of the Log-Power Tail
For a real number \(q\) and a fixed lower endpoint \(a>1\), define the log-power tail as the improper integral
provided the limit exists as a finite real number. The condition \(a>1\) ensures \(\ln x>0\) throughout the interval, so the real power \((\ln x)^q\) is defined for every real \(q\). Since the integrand is positive, the limit of the cutoff integrals either is finite or grows without bound.
The lower endpoint affects the value of a convergent integral, but not whether the tail converges. The threshold depends on \(q\). We will first show this for an arbitrary \(a>1\), then use the especially convenient lower endpoint \(e\) in examples.
The Substitution That Reveals the Threshold
Set \(u=\ln x\). Then \(du=dx/x\), and the cutoff endpoints transform as \(x=a\) becomes \(u=\ln a\), while \(x=R\) becomes \(u=\ln R\). Thus each finite cutoff integral is exactly a power integral:
When \(q\ne 1\), the power rule gives the antiderivative \(u^{1-q}/(1-q)\), since its derivative is \(u^{-q}\). When \(q=1\), the integrand is \(1/u\), whose antiderivative is \(\ln u\). These cases lead to different cutoff expressions.
Derivation of the Convergence Test and Value
We can establish both the convergence threshold and the value by evaluating the finite cutoff integral and taking its limit. The argument must consider \(q=1\) separately because the power-rule expression has denominator \(1-q\).
For every finite \(R>a\), \(u=\ln x\) gives \(\int_a^R dx/[x(\ln x)^q]=\int_{\ln a}^{\ln R}u^{-q}\,du\). The substitution is valid on this interval because \(\ln x>0\).
Using the power rule, the cutoff integral equals \(\big((\ln R)^{1-q}-(\ln a)^{1-q}\big)/(1-q)\).
Now \(1-q<0\), so \((\ln R)^{1-q}=1/(\ln R)^{q-1}\to0\) as \(R\to\infty\). The limit is therefore \((\ln a)^{1-q}/(q-1)\), a finite positive number.
Now \(1-q>0\), so \((\ln R)^{1-q}\to\infty\). The cutoff expression tends to \(+\infty\), and the improper integral diverges.
The transformed cutoff integral is \(\int_{\ln a}^{\ln R}du/u=\ln(\ln R)-\ln(\ln a)\). Since \(\ln(\ln R)\to\infty\), this case also diverges.
This proves the theorem. In particular, the borderline \(q=1\) does not converge: the logarithm in the denominator slows the growth of the cutoff integral, but it does not make that growth finite.
Worked Examples
Worked Example: A Convergent Tail with q = 2
Evaluate \(\int_e^\infty dx/[x(\ln x)^2]\). The lower endpoint is \(a=e\), so \(\ln a=1\). With a finite cutoff \(R>e\), substitute \(u=\ln x\), giving
As \(R\to\infty\), \(\ln R\to\infty\), so \(1/\ln R\to0\). Therefore the improper integral equals \(1\).
Check the calculation: the transformed antiderivative is \(-1/u\), and its derivative is \(1/u^2\). The cutoff expression also tends to \(1\), as required by the convergence formula \((\ln e)^{1-2}/(2-1)=1\).
Worked Example: The Boundary Case q = 1
Determine whether \(\int_{e^2}^\infty dx/[x\ln x]\) converges. For a finite \(R>e^2\), the substitution gives lower bound \(u=\ln(e^2)=2\) and upper bound \(u=\ln R\):
As \(R\to\infty\), \(\ln R\to\infty\), and hence \(\ln(\ln R)-\ln 2\to\infty\). The integral diverges. The subtraction of the fixed number \(\ln 2\) changes the cutoff value but cannot prevent it from growing without bound.
Check the transformed antiderivative: the derivative of \(\ln u\) is \(1/u\) for \(u>0\), and the lower endpoint contributes \(-\ln 2\), exactly as shown.
Worked Example: A Divergent Tail with q = 1/2
Consider \(\int_{e^3}^\infty dx/[x(\ln x)^{1/2}]\). Here the lower bound after substitution is \(\ln(e^3)=3\). For \(R>e^3\),
Because \(\sqrt{\ln R}\to\infty\), the cutoff integral tends to \(+\infty\); the improper integral diverges. This agrees with the test because \(q=1/2<1\).
Check: differentiating \(2\sqrt{u}\) gives \(2\cdot\frac12u^{-1/2}=u^{-1/2}\). Evaluating at \(u=\ln R\) and \(u=3\) gives the displayed cutoff expression, whose limit is infinite.
Worked Example: A Convergent Tail from a Different Lower Endpoint
Evaluate \(\int_{e^2}^\infty dx/[x(\ln x)^{3/2}]\). Since \(q=3/2>1\), the tail converges. To calculate its value directly, take \(R>e^2\) and set \(u=\ln x\):
As \(R\to\infty\), the second term tends to zero, so the integral equals \(2/\sqrt2=\sqrt2\). Check against the general formula: \((\ln(e^2))^{1-3/2}/(3/2-1)=2^{-1/2}/(1/2)=2/\sqrt2=\sqrt2\). The derivative of \(-2u^{-1/2}\) is \(u^{-3/2}\), confirming the antiderivative.
Common Mistake: Losing the Factor \(dx/x\)
A frequent error is to substitute \(u=\ln x\) but forget that \(dx/x=du\). For example, one might incorrectly claim that
This is not a valid transformed integral: after changing variables to \(u\), the differential must also be expressed in the new variable. Since \(du=dx/x\), the correct equation is
Another error is to treat the \(q=1\) case as though the formula \(u^{1-q}/(1-q)\) still applied. At \(q=1\), its denominator is zero, so it is not an antiderivative formula for that case. The correct antiderivative is \(\ln u\), and its cutoff limit shows divergence.
Why the Threshold Matters
The test provides a quick way to decide whether a slowly decaying positive function has finite total area on an infinite interval. The numerator here is constant, and the factor \(1/x\) alone sits at the boundary between convergence and divergence for power tails. Dividing by a sufficiently large power of \(\ln x\) changes the outcome: the integral converges precisely when that logarithmic exponent exceeds \(1\).
The comparison across values of \(q\) is summarized below. Each statement concerns the tail beginning at any fixed \(a>1\).
| Exponent | Cutoff behavior | Conclusion |
|---|---|---|
| \(q>1\) | The cutoff integral approaches \((\ln a)^{1-q}/(q-1)\). | Converges |
| \(q=1\) | The cutoff integral grows as \(\ln(\ln R)\), up to a constant. | Diverges |
| \(q<1\) | The cutoff integral contains a term proportional to \((\ln R)^{1-q}\), which grows without bound. | Diverges |
For \(q<1\), the divergence is not caused by a negative contribution or cancellation: the integrand is positive, and the finite cutoff integrals increase as the cutoff moves to the right. The computed limit is infinite. For \(q>1\), the same limit procedure produces a finite exact value, not merely a statement that the tail is small.
Check Your Understanding
Use a finite cutoff before taking an infinite limit. In each question, state the convergence conclusion and show the key transformed integral.
- For \(a>1\), what is the transformed finite integral for \(\int_a^R dx/[x(\ln x)^q]\), including its bounds?
- Determine whether \(\int_e^\infty dx/[x(\ln x)^{5/4}]\) converges, and if it does, evaluate it.
- For \(q=1\), write the cutoff integral from \(a>1\) to \(R\) and explain why its limit is infinite.
- Evaluate \(\int_{e^4}^\infty dx/[x(\ln x)^2]\).
- What is wrong with applying the antiderivative \(u^{1-q}/(1-q)\) when \(q=1\)?