Root Pi from a Bell Curve, by Leaving the Number Line

Find the area under the bell curve $e^{-x^2}$ across the whole real line.

A $\pi$ appears in the answer to a question with no circle anywhere in it. Where it comes from is the interesting part, and it comes from leaving the line.

The usual method has nothing to grip

The reflex is to find an antiderivative and evaluate it at the two ends. There is no formula in the elementary functions to find. That is a theorem, not a failure of ingenuity: Liouville showed that certain integrands admit no antiderivative built from powers, exponentials, logarithms and trigonometric functions, and $e^{-x^2}$ is the standard example.

Be careful about what is being claimed, because the loose version is false. $\tfrac{\sqrt\pi}{2}\operatorname{erf}(x)$ is an antiderivative, and differentiating it returns $e^{-x^2}$ exactly. What does not exist is an antiderivative written in elementary terms, and $\operatorname{erf}$ is defined by this very integral, so quoting it evaluates nothing.

A concrete slice of Liouville's result is easy to prove and worth doing, because it shows the obstruction rather than asserting it. Suppose $p$ is a polynomial of degree $d$ and $\left(p(x)e^{-x^2}\right)' = e^{-x^2}$. Differentiating the left side gives

The polynomial $p' - 2xp$ has degree $d + 1 \ge 1$, and the constant 1 has degree 0. No $p$ works, at any degree, so the whole family of candidate formulas is empty in one line.

Square the thing you cannot compute

Call the integral $I$. It converges, since $e^{-x^2} \le e^{-|x|}$ once $|x| \ge 1$, so squaring it is legitimate and gives a double integral over the whole plane:

Nothing has been simplified yet. What has changed is that the integrand now depends on $x^2 + y^2$, which is the squared distance from the origin. The function is rotationally symmetric, and a rotationally symmetric function over the plane wants polar coordinates.

The ring that makes it elementary

In polar coordinates the area element is $dA = r\,dr\,d\theta$, and the integrand becomes $e^{-r^2}$ with no angular dependence. So the angular sweep contributes a bare factor of $2\pi$ and everything else is one radial integral:

The extra $r$ is the whole trick. With it, the substitution $u = r^2$ works:

So $I^2 = 2\pi \cdot \tfrac12 = \pi$, and since the integrand is positive, $I = \sqrt\pi$. The $\pi$ came from the angular sweep. It is the $\pi$ of a full turn, and it entered when the problem was moved off the line.

A practical remark about the number itself. The truncated integral is $\int_{-a}^{a} e^{-x^2}dx = \sqrt\pi \operatorname{erf}(a)$, and $\operatorname{erf}(3) = 0.9999779$. Cutting the tails at $\pm 3$ therefore loses 22 parts per million, and cutting at $\pm 6$ loses about $2 \times 10^{-17}$. This is why numerical work with this integrand is so forgiving: the domain is infinite and the useful part of it is a handful of units wide, so a quadrature rule on a finite window can reach machine precision.

The same idea without polar coordinates

If a two-variable change of variables feels like heavy machinery, the trick survives in a lighter form. Let $J = \int_0^{\infty} e^{-x^2}dx = I/2$ and square it. For each fixed $x$, substitute $y = xt$, so $dy = x\,dt$:

The inner integral in $x$ is elementary for the same reason as before, a stray factor of $x$ in front of a Gaussian, and the remaining integral is an arctangent. So $J = \sqrt\pi/2$ and $I = \sqrt\pi$. Here the $\pi$ arrives through $\arctan$ instead of through an angular sweep, which is the same circle wearing a different coat.

Why this cannot be repeated

Squaring an integral to make it easier looks like a technique. It is closer to a coincidence that this particular integrand permits, and the coincidence has two parts.

The integrand must be separable, so that $I^2$ is an integral over the plane of a recognisable function. Exponentials of sums split, so $e^{-x^2}e^{-y^2}$ collapses into $e^{-(x^2+y^2)}$. Then that function of two variables must be rotationally symmetric, so the angular integral is trivial. Both hold here because $x^2 + y^2$ is exactly the squared radius.

Change the exponent and the second condition dies. Squaring $\int e^{-x^4}dx$ gives an integrand $e^{-(x^4+y^4)}$, and $x^4 + y^4$ is not a function of $r$ alone. The level curves are squarish, the angular integral no longer separates, and the whole approach stalls. That integral does have a closed form, $2\Gamma(5/4)$, and it is found by an entirely different route.

What the same argument gives for free

Rescaling $x \mapsto x\sqrt a$ in equation (1) gives $\int e^{-ax^2}dx = \sqrt{\pi/a}$ for every $a > 0$. Setting $a = 1/(2\sigma^2)$ yields $\sigma\sqrt{2\pi}$, which is where the $\sqrt{2\pi}$in the normal density's denominator comes from: it is there to cancel this integral.

In $n$ dimensions the same separation gives $\int_{\mathbb{R}^n} e^{-|x|^2}dx = \pi^{n/2}$, which is the fastest way to reach the volume of a high-dimensional ball. And substituting $t = x^2$ in equation (1) turns it into a statement about the gamma function:

So the factorial of a half is $\sqrt\pi/2$, and the odd appearance of $\pi$ in equation (1) is the same fact seen from a different angle.

Sources and further reading

• The integral and several other derivations of it — Gaussian integral
• The coordinates used in equation (4) — Polar coordinate system
• The theorem behind the dead end — Liouville's theorem (differential algebra)
• The function equation (7) evaluates — Gamma function

Every step was checked numerically as well as symbolically. Simpson quadrature on the line agrees with $\sqrt\pi$ to $6 \times 10^{-14}$; the Cartesian double integral over a 16 by 16 square agrees with $\pi$ to $7 \times 10^{-15}$; the polar route computed independently agrees with the Cartesian one to $1 \times 10^{-14}$, which is the numerical confirmation of the Jacobian factor. And exact rational linear algebra found no polynomial of degree 0 to 12 satisfying equation (2), as the degree argument promises.