An endless region with a finite area

The Riemann integral is built for a bounded curve over a bounded interval. Its rectangles have widths that add to the length of the interval and heights that stay below some ceiling, and the whole construction leans on both facts. Take away either — let the interval run off to infinity, or let the curve shoot up to infinity at one end — and the definition simply does not apply.

And yet some of those regions obviously have an area. The region under $1/x^2$ from $1$ outwards is infinitely long, but it thins so fast that it looks like it ought to be finite, and it is: the area is exactly $1$. The region under $1/x$ from $1$ outwards looks almost the same, thins almost as fast, and has infinite area. The difference between those two is the subject here, and it turns out to be a difference with a mirror image in it.

Area out to infinity is a limit of areas #

The way to give an unbounded region an area is to refuse to take it all at once. Cut the region at $x = T$, measure the ordinary area from $1$ to $T$, and let $T$ grow. If those areas approach a number, that number is the area of the whole region; if they grow without bound, the region has infinite area.

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

The right-hand panel of the figure is exactly that limit, taken visibly. Each curve there is the running area as the cut moves outward, and the question of whether the whole region has an area is the question of whether the curve levels off. For $1/x^2$ it does, almost at once: by $T = 10$ the area is already $0.9$, and the remaining infinitely long stretch contributes the last tenth. For $1/x$ it never does. Its running area is $\ln T$, which grows forever, but so slowly that by $T = 10^6$ it has reached only $13.8$ — a curve that is unbounded and looks, on any finite stretch, as though it might be settling.

That is the first thing the definition makes precise and the eye cannot: whether an endless region has a finite area is decided by how fast its height falls, and “fast enough” has a sharp threshold. Against powers of $x$ the threshold sits at $1/x$. Anything that falls faster, even slightly, has finite area; $1/x$ itself and everything slower does not.

The running areas have exact forms, which is what makes the threshold visible. For $p \ne 1$ the antiderivative of $x^{-p}$ is $x^{1-p}/(1-p)$, so

$$\int_1^T \frac{dx}{x^p} = \frac{T^{1-p} - 1}{1 - p}.$$

When $p > 1$ the power $T^{1-p}$ goes to zero and the area approaches $1/(p-1)$. When $p < 1$ it grows without bound. At $p = 1$ the formula breaks — it would divide by zero — and the antiderivative is the logarithm instead, which also grows without bound. The logarithm is the borderline function in a quite literal sense: it is what the power formula turns into at the one exponent where the power formula stops working, and it is the area that names $e$.

Just past the threshold is still out of sight #

The threshold is sharp, and that makes it deceptive near the edge.

At $p = 1.1$ the area is finite: it approaches $1/(1.1 - 1) = 10$. At $p = 0.9$ it is infinite. And on the drawn stretch, out to a million, the three curves are three gently rising lines on a logarithmic plot, the convergent one lowest, none of them visibly flattening. The convergent one reaches $7.5$ by a million and would need $T$ around $10^{20}$ to get within a tenth of its limit.

So the threshold is a fact about the behaviour at infinity, and infinity is the one place no picture reaches. What decides the matter is the exact formula, which says that $T^{-0.1}$ does go to zero, however slowly; the drawing can only show that nothing dramatic happens in the part of the line it covers. This is the same trap the harmonic series sets for sums, and it is no coincidence: the sum of $1/n^p$ and the integral of $1/x^p$ converge for exactly the same exponents, because the rectangles of width one under the curve are the terms of the sum. That comparison is the integral test, and it is the reason condensing a series by doubling and integrating a power arrive at the same boundary from two directions.

The other end, where the curve shoots up #

An improper integral can be improper in a second way: a finite interval, but a curve that goes to infinity at one end of it. The area under $1/\sqrt x$ between $0$ and $1$ is such a region — the curve is unbounded as $x$ approaches $0$ — and it is handled the same way, by cutting at $x = \varepsilon$, measuring from $\varepsilon$ to $1$, and letting $\varepsilon$ shrink.

For the powers the answer comes from the same formula:

$$\int_\varepsilon^1 \frac{dx}{x^p} = \frac{1 - \varepsilon^{1-p}}{1 - p},$$

and now it is $p < 1$ that makes $\varepsilon^{1-p}$ vanish. The area near zero is finite, equal to $1/(1-p)$, exactly when $p < 1$. So $1/\sqrt x$ has finite area near zero — the area is $2$ — and infinite area out to infinity, while $1/x^2$ is the other way round.

The figure lays the two ends side by side and shows a symmetry that the formulas only hint at. The left branch and the right branch are reflections of each other in the line $p = 1$: the area near zero at exponent $1 - t$ equals the area out to infinity at exponent $1 + t$. Every exponent is good at exactly one end, except $1$, which is good at neither.

The reflection is a substitution. Put $x = 1/u$. The interval from $1$ to infinity becomes the interval from $0$ to $1$, the far end becomes the near end, and $dx = -du/u^2$, so

$$\int_1^\infty \frac{dx}{x^p} = \int_0^1 u^{p}\cdot\frac{du}{u^2} = \int_0^1 \frac{du}{u^{2-p}}.$$

The far end of $1/x^p$ is the near end of $1/u^{2-p}$, and $p \mapsto 2 - p$ is the reflection in $p = 1$. The dots in the figure were measured independently on each side, and each near-end area equals the far-end area of the reflected exponent, as the substitution demands.

The consequence is a small theorem that sounds more surprising than it is: no power of $x$ has a finite area over the whole half-line from $0$ to infinity. Whatever the exponent, one end or the other — or, at $p = 1$, both — is infinite. The two conditions pull in opposite directions, and the reflection makes it visible that they cannot both be met.

A horn with finite volume and infinite surface #

Turn the curve $y = 1/x$, from $x = 1$ outwards, about the $x$-axis. The result is an infinitely long trumpet, flaring at one end and narrowing forever — Torricelli’s trumpet, now usually called Gabriel’s horn.

Its volume is a sum of thin discs of radius $1/x$, each of area $\pi/x^2$, so

$$V = \int_1^\infty \frac{\pi}{x^2}\,dx = \pi,$$

finite, by the first figure. Its surface area is a sum of thin bands, each of circumference $2\pi/x$ and slanted width at least $dx$, so

$$S \ge \int_1^\infty \frac{2\pi}{x}\,dx = \infty,$$

infinite, by the same figure’s other curve.

The paradox people draw from this is that the horn can be filled with a finite amount of paint, but its inside cannot be painted — pour in $\pi$ cubic units and the whole inner surface is wet, yet the surface needs infinitely many units of paint to cover. The resolution is that “painting” a surface with a layer of fixed thickness is not what filling does. Far down the horn the tube is narrower than any fixed thickness of paint, so a coat of constant thickness cannot fit inside it, and the filling paint coats the surface with a layer that thins in proportion to the tube. A layer that thin covers infinite area with finite volume without contradiction.

The mathematical content is plainer than the paradox. Volume scales with the square of the radius and surface with the radius itself, so a radius falling like $1/x$ makes the volume integrand $1/x^2$ and the surface integrand $1/x$ — one on each side of the threshold the first figure found. The horn is that threshold turned into a solid.

A finite area with an infinite average #

The same threshold decides a question in probability that looks unrelated, and it produces one of the standard cautionary examples of the subject.

The curve $\frac{1}{\pi(1 + x^2)}$ falls like $1/x^2$ in both directions, so the area under it over the whole line is finite — it is exactly $1$, since the antiderivative is $\arctan x / \pi$ and the arctangent runs from $-\pi/2$ to $\pi/2$. That makes it a legitimate probability density, the Cauchy distribution, and it looks like a slightly heavy-shouldered bell.

Now ask for its average. The mean of a density is the area under $x$ times the density, and $x/(\pi(1 + x^2))$ falls like $1/x$ — exactly on the threshold, at both ends. The area to the right is infinite, the area to the left is minus infinity, and the mean does not exist. Cutting symmetrically at $\pm T$ gives zero every time, which is tempting to report as the answer; cutting at $-T$ and $+2T$ gives $\ln 2/\pi$ instead, and other ways of cutting give anything at all. That is the conditional-convergence hazard of the next section arriving early, and it has a practical consequence: the average of many samples from a Cauchy distribution does not settle down as the sample grows. The average of a thousand samples is exactly as spread out as a single one.

So a density can be integrable while its tail is too heavy for an average, and the dividing line is the same exponent as before, shifted by one because the mean multiplies by $x$. A density falling like $1/x^{p}$ has a finite area when $p > 1$, a finite mean when $p > 2$, a finite variance when $p > 3$. Each extra moment demands one more power of decay, which is why a tail that is not a bell can break the familiar statistics one moment at a time while the area — the total probability — stays exactly one.

A region that has an area only because it cancels #

Everything so far concerned positive curves, where the area either settles or grows. A curve that changes sign allows a third possibility, and it is the most delicate.

The curve $\sin x / x$ oscillates, its lobes alternating above and below the axis and shrinking like $1/x$. The area between it and the axis from $0$ to $T$, counted with sign, is a running total in which each lobe partly undoes the one before.

The signed area converges, and its limit is $\pi/2$ — a fact known as the Dirichlet integral, provable by several routes, none of them short. The unsigned area does not converge. Each lobe, between consecutive multiples of $\pi$, has area about $2/(k\pi)$, and those add like the harmonic series, so the total without signs grows like $\tfrac{2}{\pi}\ln T$.

So the signed integral of $\sin x/x$ exists only because the lobes cancel, exactly as the alternating harmonic series converges only because its terms alternate. Such an integral is called conditionally convergent, and it inherits every hazard of its discrete counterpart. A conditionally convergent series can be rearranged to any total; a conditionally convergent integral can be made to give different answers by taking the limit in different ways — cutting at $T$ that jump from one lobe’s peak to the next, for instance, or cutting two ends at different rates when the region runs to infinity in both directions. The value $\pi/2$ belongs to the specific limit in the definition, and is stable only because that limit is specified.

This curve is not an exotic example. It is the shape of a square pulse seen in frequencies, and the overshoot at the start of the signed curve — the running area rises to about $1.85$, well above $\pi/2$, before settling — is the same overshoot that appears as the Gibbs phenomenon when a square wave is built from round ones. The two are the same integral, looked at from its two ends.

Why the Riemann integral could not do this alone #

The definition used throughout — cut the region, measure the bounded part, take a limit — is an addition to the Riemann integral rather than part of it, and the addition carries a cost that is worth naming.

The Riemann integral is defined only for bounded functions on bounded intervals, and it has good properties there: the integral of a sum is the sum of the integrals, a limit of functions that converge uniformly can be integrated term by term, and so on. The improper integral is a limit of Riemann integrals, and limits do not automatically inherit those properties. A sequence of functions each with improper integral zero can converge, at every point, to a function whose integral is one — take a bump of height $1/n$ and width $n$ sliding off to infinity, and the areas stay at one while the heights go to nothing. Interchanging “limit of functions” with “improper integral” is exactly where proofs go wrong, and most of the theorems that license it need a condition that controls the tails.

Lebesgue’s integral, which replaced Riemann’s as the working definition in analysis, handles unbounded regions and unbounded functions within the definition itself, and for positive functions it agrees with everything in the first three figures. It does not agree with the fourth. In Lebesgue’s theory a function is integrable only if the area counted without signs is finite, so $\sin x / x$ over the half-line is not Lebesgue integrable at all, and its value of $\pi/2$ has to be recovered as a limit of integrals exactly as it was here. The conditionally convergent integral is a genuinely different object from an absolutely convergent one, and the more powerful theory is the one that declines to call it an integral.

What the pictures cannot show #

The behaviour at infinity is never on the page. Every figure here stops at some $T$ — a million in the first two, sixty in the fourth — and every conclusion about infinity comes from the exact formulas, not from the drawing. The second figure is the honest warning: the convergent and divergent curves are indistinguishable over a million units, and they would remain indistinguishable over any stretch a figure could show.

The horn’s rings are ellipses standing in for circles seen at an angle. The side view shows a plane section, the shading suggests a solid, and the rings suggest that it is round. The figure establishes nothing about the solid; the volume and surface in the table are computed from the one-dimensional integrals above, and the picture merely helps a reader hold the shape in mind.

Measured and exact agree only up to the truncation. The areas in the second and third figures were computed by strips on a finite range, with the range chosen so that what was left out is smaller than $10^{-20}$. That margin is a property of these curves; for a curve that decays slowly — $1/x^{1.01}$, say — no feasible truncation would come close, and the only reliable value is the formula.

Still open: what the borderline really is #

For powers of $x$ the threshold is exactly $1/x$. But a curve can fall more slowly than every $1/x^{1+\varepsilon}$ and faster than $1/x$, and on that thin sliver the powers say nothing. The answer there is a hierarchy with no end. The area under $1/(x \ln x)$ is infinite, growing like $\ln \ln T$; the area under $1/(x (\ln x)^2)$ is finite. Between those lies $1/(x \ln x \ln\ln x)$, infinite again, growing like $\ln\ln\ln T$, and so on — each borderline function has a borderline beyond it, and no single function separates all convergent integrals from all divergent ones. Du Bois-Reymond proved in 1873 that no such boundary exists: for any sequence of divergent integrands there is a divergent one falling faster than all of them, and for any sequence of convergent ones a convergent one falling slower.

That settles the abstract question and leaves the practical one open in every particular case. There is no universal test that decides, for an arbitrary positive function, whether its improper integral converges; every test compares against some chosen scale, and a function that sits between the steps of the scale has to be handled by hand. The next step from here is to stop asking whether an area is finite and start asking what the area is for functions that are the derivatives of nothing elementary — which is where integration by parts turns out to be a rectangle, and where the factorials and $\pi$ come out of it.

A threshold that has a mirror in it #

The integrals of powers carry the whole structure of the subject in miniature. An infinitely long region has finite area when its height falls faster than $1/x$; a region that shoots up at a point has finite area when it rises slower than $1/x$. The substitution $x \mapsto 1/x$ swaps the two situations, reflects the exponent in $1$, and shows why no power can satisfy both — why $1/x$, sitting exactly on the mirror, fails at both ends at once.

Everything else is that threshold applied. The horn has finite volume and infinite surface because turning a curve about an axis squares its radius for one and not for the other. The integral of $\sin x/x$ is finite and its unsigned version infinite because cancellation can rescue a region whose lobes, taken alone, sit exactly on the threshold. And in each case the figure can show the behaviour on a finite stretch and not the thing that decides the answer, which lives at the one place a figure cannot go.