Infinite regions with finite area. The p-test, Gabriel’s horn that holds paint but cannot be painted, and the area under the bell curve
A region that stretches out to infinity can still have a finite area, as long as it thins out fast enough. Deciding how fast is fast enough is the business of improper integrals, and the answers range from the predictable to the paradoxical: a horn you can fill with paint but never paint, and a bell curve whose area is the square root of π.
Push R out towards 10⁸ and watch the area under x⁻ᵖ. Then move p across 1, and switch to the spike between 0 and 1, where the verdict flips.
Areas come from the antiderivative: (R^(1 − p) − 1)/(1 − p) and (1 − ε^(1 − p))/(1 − p), or ln R and ln(1/ε) at p = 1. The upper plot shows only the first stretch of the region; the shading fades toward the infinite end, which never fits on any screen.
An integral over an infinite interval is defined as a limit: ∫₁^∞ f(x) dx = lim as R → ∞ of ∫₁ᴿ f(x) dx. If the limit exists the integral converges; otherwise it diverges. Likewise an integrand that blows up at 0 is handled by ∫₀¹ f = lim as ε → 0⁺ of ∫_ε¹ f.
For f(x) = x⁻ᵖ everything is explicit. When p ≠ 1, ∫₁ᴿ x⁻ᵖ dx = (R^(1−p) − 1)/(1 − p), which tends to 1/(p − 1) if p > 1 and to infinity if p < 1. At p = 1 the antiderivative is ln x and the area is ln R, unbounded but very slow. Near 0 the exponents swap roles: ∫₀¹ x⁻ᵖ dx converges exactly when p < 1. No power converges at both ends, and p = 1 diverges at both.
These integrals are the measuring sticks for everything else. By the comparison test, if 0 ≤ f ≤ g and ∫g converges, so does ∫f; so 1/(x² + 1) has a finite tail because it is below x⁻². The same dividing line p = 1 separates the convergent series ∑ 1/nᵖ from the divergent ones, through the integral test.
Lengthen the horn and pour in paint, then switch to painting its outside. The volume never reaches π; the surface area never stops growing.
The horn stands mouth up with its axis vertical, drawn to true scale; drag to orbit and scroll to zoom out along it. The paint level rises at a constant rate of volume, which is why it hurries through the narrow tip; the painted band advances at a constant rate of area. Surface areas are computed numerically.
Spin y = 1/x, x ≥ 1, about the x-axis. By disks, the volume out to L is π∫₁ᴸ x⁻² dx = π(1 − 1/L), which tends to π. The surface area is 2π∫₁ᴸ (1/x)√(1 + 1/x⁴) dx, and since the square root is at least 1 this is at least 2π ln L, which grows without bound. So the infinite horn has finite volume and infinite surface area.
Evangelista Torricelli found the solid in the early 1640s and published it in his Opera geometrica of 1644. A solid of infinite length with finite volume shocked his contemporaries, and philosophers argued about what it meant for decades. The name Gabriel’s horn, after the archangel’s trumpet, came much later.
The paradox in its modern form: you can fill the horn with π units of paint, which surely coats the inside, yet painting the surface takes infinitely much. The resolution is that mathematical paint and real paint behave differently. A coat of real paint has some thickness τ, so it uses about τ times the area, which is infinite; but past x = 1/τ the horn is narrower than the coat, so the paint inside does not form a coat of thickness τ there at all. A coat of mathematical paint has zero thickness and uses no volume, however large the area.
Build the volume under z = e^(−(x² + y²)) from cylindrical shells and watch the total approach π; then rebuild it from slices, which gives the square of the area under the bell curve.
The drawing doubles the heights so the bell is easier to see; every number uses the true surface z = e^(−(x² + y²)). Shells reach out to r = 3.2 and slices to |x| = 3.2, where the surface has fallen below 0.00004; the volume left outside the disc is π·e^(−10.24) ≈ 0.0001. The rest of the gap from π is the midpoint rule’s error, which shrinks like 1/n². Each shell is a cylinder at its midpoint height, so its drawn volume is exactly its term 2πr e^(−r²) Δr in the sum.
Call I = ∫ e^(−x²) dx over the whole line. It converges, since e^(−x²) ≤ e^(−|x|) once |x| ≥ 1, but e^(−x²) has no elementary antiderivative, so the usual route is closed. The trick is to compute I² instead:
I² = ∫e^(−x²)dx · ∫e^(−y²)dy = ∬ e^(−(x² + y²)) dx dy,
the volume under a surface of revolution. In polar coordinates the area element is r dr dθ, and the extra factor r is what makes it work: ∫₀^2π ∫₀^∞ e^(−r²) r dr dθ = 2π·½ = π. So I = √π ≈ 1.77245. Cutting the volume into cylindrical shells is the same calculation done by hand: the shell at radius r has volume 2πr·e^(−r²)·dr, and 2πr·e^(−r²) is the derivative of −π e^(−r²).
Laplace evaluated the integral in the late eighteenth century, and the squaring argument is traditionally credited to Poisson. It is why the normal distribution has the constant 1/√(2π) in front: that factor makes the total area under the bell curve exactly 1.
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