A Vanishing Integrand Can Still Have a Divergent Integral

When a Small Integrand Still Accumulates

A function can become arbitrarily small while its area over an infinite interval continues to grow without bound. The rate at which the function decreases matters: values that are small over a sufficiently long interval can still produce a substantial amount of accumulated area.

The function \(1/x\) provides a precise example. It is positive for \(x\ge 1\), and it approaches zero as \(x\) increases. Nevertheless, the area under its graph from \(1\) to infinity is not finite. We can see exactly how the accumulated area grows by first stopping at a finite cutoff \(R\), then studying the result as \(R\) increases.

Central conclusion: Although \(\lim_{x\to\infty}1/x=0\), the improper integral \(\int_1^\infty 1/x\,dx\) diverges. Its truncated integral up to \(R\) is exactly \(\ln R\), which grows without bound.

Definition Through a Finite Cutoff

An integral with an infinite endpoint is defined by a limit of ordinary integrals over finite intervals. For a function \(f\) defined on \([1,\infty)\), the right-infinite integral is said to converge to a finite value if the limit of its truncated integrals exists and is finite:

\int_1^\infty f(x)\,dx = \lim_{R\to\infty}\int_1^R f(x)\,dx.

If that limit is not finite, the improper integral diverges. Thus, to decide the question for \(f(x)=1/x\), begin with a finite \(R\ge 1\) and calculate

A(R)=\int_1^R\frac{1}{x}\,dx.

The quantity \(A(R)\) is the accumulated area from \(1\) up to the cutoff \(R\). The question is whether these accumulated areas approach a finite number as \(R\) tends to infinity.

The Truncated Integral

For \(x>0\), the natural logarithm \(\ln x\) has derivative \(1/x\). Therefore it is an antiderivative of the integrand on the interval \([1,R]\). By the Fundamental Theorem of Calculus,

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

Here \(\ln 1=0\). This formula gives the exact accumulated area for every finite cutoff \(R\ge 1\), not just an approximation.

Truncated-area formula: For every finite \(R\ge 1\), \(\int_1^R 1/x\,dx=\ln R\). The improper integral converges only if these values have a finite limit as \(R\to\infty\).

Why the Accumulated Area Is Unbounded

The logarithm \(\ln R\) increases without bound as \(R\) increases. One way to make this growth especially clear is to choose cutoffs of the form \(R=e^t\), where \(t\) is positive. Then the truncated-area formula gives

A(e^t)=\ln(e^t)=t.

As \(t\) can be made arbitrarily large, the truncated areas can also be made arbitrarily large. Consequently, \(A(R)\) does not approach a finite limit. By the definition of an improper integral, this proves divergence.

Theorem: The improper integral \(\int_1^\infty 1/x\,dx\) diverges to \(+\infty\). Its cutoff integral is \(\ln R\), so the divergence is logarithmic.

The integrand itself does approach zero. Indeed, given any \(\varepsilon>0\), choose \(X>1/\varepsilon\). Whenever \(x>X\), we have \(x>1/\varepsilon\), and hence \(0<1/x<\varepsilon\). This verifies that \(\lim_{x\to\infty}1/x=0\). The function's limit and the limit of its accumulated integral are different questions: the first is zero, while the second is infinite.

Why Each Larger Scale Adds Area

The exact formula also reveals how the accumulated area changes when the cutoff is multiplied by a fixed factor. If \(c>0\) and both cutoffs are in the domain, then

A(cR)-A(R) = \ln(cR)-\ln R = \ln c.

Thus, multiplying the cutoff by \(c\) adds \(\ln c\) units of area, regardless of the starting cutoff \(R\). For example, each multiplication of the cutoff by \(e\) adds one unit of area, because \(\ln e=1\). The integrand is small at large \(x\), but each successive interval from \(e^k\) to \(e^{k+1}\) still contributes exactly one unit.

Choose a finite cutoff.
Integrate from \(1\) to \(R\) rather than trying to evaluate an integral with an infinite endpoint directly.

Find the accumulated area.
The antiderivative gives \(A(R)=\ln R\).

Take the cutoff limit.
Since \(\ln R\) grows without bound, the improper integral diverges.

Worked Examples

Worked Example: Accumulated Area Up to 10

Find the area under \(1/x\) from \(1\) to \(10\), and verify the result.

First use the truncated-area formula with \(R=10\):

\int_1^{10}\frac{1}{x}\,dx = \left[\ln x\right]_1^{10} = \ln 10-\ln 1 = \ln 10.

Numerically, \(\ln 10\approx 2.302585\), so the area rounded to three decimal places is \(2.303\). To check the evaluation, differentiate the antiderivative: \(\frac{d}{dx}\ln x=1/x\). Evaluating it at the endpoints gives \(\ln 10-\ln 1=\ln 10\), as required.

Worked Example: Area Up to 1000

Find the accumulated area from \(1\) to \(1000\), and compare it with the area up to \(10\).

The exact area is

\int_1^{1000}\frac{1}{x}\,dx = \ln 1000 = \ln(10^3) = 3\ln 10.

Using \(\ln 10\approx 2.302585093\), the value is \(3(2.302585093)=6.907755279\), or \(6.908\) to three decimal places. The area from \(1\) to \(10\) was \(\ln 10\), so increasing the cutoff from \(10\) to \(1000\) added \(2\ln 10\), not an amount proportional to the change in cutoff. The check is exact: \(\ln 1000=3\ln 10\), and the derivative of \(\ln x\) is the original integrand.

Worked Example: Equal Area on Successive Exponential Intervals

For a positive integer \(n\), calculate the area from \(e^n\) to \(e^{n+1}\). Then find the total area from \(1\) to \(e^3\) by adding these intervals.

On the interval from \(e^n\) to \(e^{n+1}\), evaluate directly:

\int_{e^n}^{e^{n+1}}\frac{1}{x}\,dx = \ln(e^{n+1})-\ln(e^n) = (n+1)-n = 1.

The three intervals for \(n=0,1,2\) are \([1,e]\), \([e,e^2]\), and \([e^2,e^3]\). Each contributes \(1\), so their total area is \(1+1+1=3\). Checking against the single truncated integral gives

\int_1^{e^3}\frac{1}{x}\,dx = \ln(e^3)-\ln 1 = 3-0 = 3.

The two calculations agree. More generally, every additional interval from \(e^n\) to \(e^{n+1}\) adds another unit of area, so an arbitrarily large number of such intervals gives an arbitrarily large total.

Worked Example: Choosing a Cutoff for a Target Area

How large must the cutoff \(R\) be for the accumulated area from \(1\) to \(R\) to equal \(5\)?

Set the exact formula \(A(R)=\ln R\) equal to \(5\):

\ln R=5 \qquad\Longrightarrow\qquad R=e^5.

Thus \(R=e^5\approx 148.413\). The check is substitution into the exact area formula: \(A(e^5)=\ln(e^5)=5\). The decimal cutoff is an approximation to the exact cutoff \(e^5\); the accumulated area at the exact cutoff is exactly \(5\).

A Common Mistake

A tempting but incorrect argument is: “Since \(\lim_{x\to\infty}1/x=0\), the integral converges, or even equals zero.” The first step is correct, but it does not establish either conclusion. A limit of function values describes what happens at large individual inputs; it does not calculate the total accumulated area over an unbounded interval.

For this example, the correction is explicit:

\lim_{x\to\infty}\frac{1}{x}=0, \qquad\text{but}\qquad \int_1^R\frac{1}{x}\,dx=\ln R\longrightarrow+\infty.

The integral must be judged by the limit of its truncated integrals, not by the limit of the integrand alone. The example shows that tending to zero is not, by itself, enough to guarantee finite accumulated area.

Why This Example Matters

When studying an improper integral, it is useful to keep two separate quantities in view: the value of the integrand near the far end of the interval, and the area accumulated as the cutoff moves outward. The first quantity describes the height of the graph; the second depends on both height and the length of the interval being added.

For \(1/x\), the height decreases as \(x\) grows, but the exact cutoff formula shows that the added areas never stop accumulating: every interval from \(e^n\) to \(e^{n+1}\) contributes \(1\). This explains the divergence without suggesting that the integrand stays large. It becomes arbitrarily small, while its total accumulated area remains unbounded.

The important conclusion is limited and precise: this example disproves the claim that an integrand approaching zero must have a convergent improper integral. It does not say that every function approaching zero has a divergent integral. To decide a particular case, evaluate or otherwise analyze the cutoff integral and determine whether it has a finite limit.

Check Your Understanding

Use the cutoff integral to answer each question. Keep the limit of the integrand separate from the limit of the accumulated area.

What is the exact value of \(\int_1^R 1/x\,dx\) for finite \(R\ge 1\)?
Why does this formula imply that \(\int_1^\infty 1/x\,dx\) diverges?
What area is added when a cutoff \(R\) is multiplied by \(e\)?
How much area lies between \(e^4\) and \(e^6\)?
Explain why \(\lim_{x\to\infty}1/x=0\) does not prove that the improper integral converges.