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The Logarithmic Tail Integral

Use cutoff limits and the antiderivative of one over x to show that its accumulated area grows logarithmically without bound.

Beginner 8 min read

What You'll Learn

  • Define the integral from one to infinity as a limit of finite-interval integrals.
  • Evaluate each truncated integral using the antiderivative ln x.
  • Prove that the truncated values grow without bound.
  • Interpret the constant contribution from each interval whose endpoints double.
  • Avoid treating infinity as a number or mistaking a vanishing integrand for a convergent integral.

Why This Tail Deserves Attention

The function \(1/x\) becomes smaller as \(x\) increases. It is tempting to conclude that its total area must therefore be finite. But the area accumulated over the entire half-line depends not just on the height of the function at distant points, but on how much area continues to accumulate as the endpoint moves farther away.

This example is especially instructive because the area up to any finite endpoint has a simple exact expression: a logarithm. Examining that expression shows that the areas do not settle toward a finite value.

Definition and Finite Cutoffs

An integral with an infinite endpoint is defined by a limit of ordinary definite integrals. In particular, the integral of \(1/x\) from \(1\) to infinity is defined, if the limit exists as a finite real number, by

$$ \int_1^\infty \frac{1}{x}\,dx = \lim_{R\to\infty}\int_1^R \frac{1}{x}\,dx. $$

Here \(R\) is a finite cutoff with \(R\ge 1\). We first calculate the integral for such a finite \(R\), and only then study what happens as \(R\) increases without bound.

Definition: The improper integral \(\int_1^\infty f(x)\,dx\) converges to a finite value if the limit \(\lim_{R\to\infty}\int_1^R f(x)\,dx\) exists as a finite real number. If the truncated integrals grow without bound, the improper integral diverges to infinity.

The natural logarithm \(\ln x\) is defined for \(x>0\), and its derivative is \(\frac{d}{dx}\ln x=1/x\). Thus \(\ln x\) is an antiderivative of \(1/x\) on the positive real numbers. By the Fundamental Theorem of Calculus, for every finite \(R\ge 1\),

$$ \int_1^R \frac{1}{x}\,dx = \left[\ln x\right]_1^R = \ln R-\ln 1 = \ln R. $$

The finite-cutoff answer is therefore \(\ln R\). The remaining question is whether these values approach a finite limit as \(R\) grows.

Worked Example: A Particular Truncated Integral

Worked Example: A Particular Truncated Integral

Evaluate \(\int_1^{e^4}\frac{1}{x}\,dx\), and check the result.

The endpoint \(e^4\) is finite, so the Fundamental Theorem of Calculus applies directly:

$$ \begin{aligned} \int_1^{e^4}\frac{1}{x}\,dx &=\left[\ln x\right]_1^{e^4}\\ &=\ln(e^4)-\ln(1)\\ &=4-0\\ &=4. \end{aligned} $$

To check, differentiate the antiderivative: \(\frac{d}{dx}\ln x=1/x\). Substituting the endpoints into that antiderivative gives \(\ln(e^4)-\ln(1)=4\), as obtained above.

Why the Truncated Values Have No Finite Limit

For \(R\ge 1\), the truncated value is \(A(R)=\ln R\). The logarithm increases as \(R\) increases, but monotonic increase alone would not establish divergence: an increasing function can approach a finite limit. We need to show that the values exceed every fixed bound.

Theorem: The truncated integrals \(\int_1^R\frac{1}{x}\,dx\) grow without bound as \(R\to\infty\). Consequently, \(\int_1^\infty\frac{1}{x}\,dx\) diverges to infinity.

To prove the claim, let \(K\) be any positive real number. Choose the finite cutoff \(R=e^{K+1}\). Since \(e^{K+1}\ge 1\), this is an allowed endpoint. Then

$$ \int_1^{e^{K+1}}\frac{1}{x}\,dx = \ln(e^{K+1}) = K+1 > K. $$

Thus, no matter how large a bound \(K\) is specified, some truncated integral exceeds it. The values cannot approach a finite real number; they grow without bound. By the cutoff-limit definition, the improper integral diverges to infinity.

This argument also explains precisely what the logarithmic antiderivative tells us: the accumulated area at cutoff \(R\) is \(\ln R\), and as the cutoff increases, this quantity eventually exceeds every fixed bound. The notation \(\ln(\infty)\) is not a finite endpoint substitution. The conclusion comes from taking the limit of the finite-cutoff values.

Worked Example: Choosing a Cutoff to Exceed a Target

Worked Example: Choosing a Cutoff to Exceed a Target

Find a finite cutoff \(R\) for which the accumulated area \(\int_1^R\frac{1}{x}\,dx\) is greater than \(12\), and verify it.

The cutoff formula gives \(\int_1^R\frac{1}{x}\,dx=\ln R\). To make this exceed \(12\), choose \(R=e^{13}\). Then

$$ \begin{aligned} \int_1^{e^{13}}\frac{1}{x}\,dx &=\left[\ln x\right]_1^{e^{13}}\\ &=\ln(e^{13})-\ln(1)\\ &=13-0\\ &=13>12. \end{aligned} $$

For the check, \(\ln(e^{13})=13\) and \(\ln(1)=0\), so evaluating the antiderivative at the two endpoints gives \(13\), which is indeed greater than \(12\). The particular cutoff is not unique; any \(R>e^{12}\) also gives a value greater than \(12\).

A Constant Contribution on Each Doubling Interval

The logarithmic growth can also be seen by dividing the half-line into intervals whose endpoints double. For any positive integer \(k\), the area over the interval from \(2^k\) to \(2^{k+1}\) is

$$ \int_{2^k}^{2^{k+1}}\frac{1}{x}\,dx = \ln(2^{k+1})-\ln(2^k) = (k+1)\ln 2-k\ln 2 = \ln 2. $$

Every such interval contributes the same positive amount, \(\ln 2\). This does not mean that the integrand has constant height: \(1/x\) decreases across each interval. Rather, the width of each interval also increases, and the total area on the interval remains \(\ln 2\).

Worked Example: Accumulating Area Across Doubling Intervals

For a positive integer \(n\), calculate the area from \(1\) to \(2^n\) by adding the contributions from the intervals with endpoints \(1,2,4,\ldots,2^n\). Check the sum using the antiderivative.

There are \(n\) intervals, each contributing \(\ln 2\). Therefore their total area is

$$ \underbrace{\ln 2+\ln 2+\cdots+\ln 2}_{n\text{ terms}} = n\ln 2. $$

The direct calculation gives the same result:

$$ \begin{aligned} \int_1^{2^n}\frac{1}{x}\,dx &=\left[\ln x\right]_1^{2^n}\\ &=\ln(2^n)-\ln(1)\\ &=n\ln 2-0\\ &=n\ln 2. \end{aligned} $$

Since \(\ln 2>0\), the values \(n\ln 2\) grow without bound as the positive integer \(n\) increases. The antiderivative check confirms the accumulated value, while the interval calculation shows how that growth is built up from repeated positive contributions.

A Common Mistake: Confusing Small Heights with Finite Total Area

A common incorrect argument is: “Since \(1/x\) approaches \(0\) as \(x\) grows, its total area from \(1\) to infinity must be finite.” The conclusion does not follow from the stated fact. The height of the curve tends to zero, but the relevant improper integral is determined by the limit of the accumulated areas.

For this particular integrand, a related error is to write

$$ \int_1^\infty\frac{1}{x}\,dx = \left[\ln x\right]_1^\infty = \ln(\infty)-\ln(1) $$

and then treat \(\ln(\infty)\) as if it were a finite number. Infinity is not a real endpoint at which an antiderivative can be evaluated. The correct calculation keeps a finite cutoff until after the integration:

$$ \int_1^\infty\frac{1}{x}\,dx = \lim_{R\to\infty}\left(\ln R-\ln 1\right) = \lim_{R\to\infty}\ln R = +\infty. $$

The final expression means that the finite-cutoff values grow without bound; it does not assign a real-number value to the improper integral.

Why the Logarithmic Tail Matters

This example is a useful reference point for studying improper integrals. The integrand is positive on the entire interval, its finite-cutoff integrals are straightforward to compute, and the divergence is visible in the exact expression \(\ln R\). It shows that an integrand can decrease toward zero while still accumulating unbounded area.

The central procedure is to keep the cutoff explicit: compute the ordinary definite integral for finite \(R\), simplify the result, and then determine whether its limit is finite. Here the cutoff values are \(\ln R\), and the argument using \(R=e^{K+1}\) proves that they exceed any prescribed bound \(K\). Therefore the logarithmic tail does not converge to a finite area.

Check Your Understanding

Use the cutoff definition and the logarithmic antiderivative to answer these questions.

  1. For a finite \(R\ge 1\), what is the value of \(\int_1^R\frac{1}{x}\,dx\)?
  2. How can you choose a finite cutoff \(R\) to make this truncated integral exceed a given positive number \(K\)?
  3. Why does proving that the truncated values exceed every fixed bound establish divergence to infinity?
  4. Calculate \(\int_4^8\frac{1}{x}\,dx\) using the logarithmic antiderivative.
  5. Why is \(\ln(\infty)\) not a valid finite endpoint substitution?

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