The amazing Poisson calculation: a method or a trick?

Denis Bell

April 27, 2017 The Gaussian integral and Poisson’s calculation

The Gaussian function e−x2 plays a central role in probability and statistics. For this reason, it is important to know the value of the integral ∞ Z 2 I = e−x dx. 0 The function e−x2 does not have an elementary antiderivative, so I cannot be evaluated by the FTC. There is the following argument, attributed to Poisson, for calculating I . Consider

∞ ∞ Z 2 Z 2 I 2 = e−x dx · e−y dy. 0 0 Interpret this product as a double integral in the plane and transform to polar coordinates x = r cos θ, y = r sin θ to get

∞ ∞ Z Z 2 2 I 2 = e−(x +y )dxdy 0 0

π/2 ∞ Z Z 2 = e−r r drdθ 0 0 ∞ π Z d 2 π = − e−r dr = . 4 0 dr 4 √ π Thus I = 2 . Can this argument be used to evaluate other seemingly intractable improper integrals?

Consider the integral Z ∞ J = f (x)dx. 0 Proceeding as above, we obtain Z ∞ Z ∞ J2 = f (x)f (y) dxdy. 0 0 In order to take the argument further, there will need to exist functions g and h such that f (x)f (y) = g(x2 + y 2)h(y/x). Transformation to polar coordinates and the substitution u = r 2 will then yield 1 Z ∞ Z π/2 J2 = g(u)du h(tan θ)dθ. (1) 2 0 0 Which functions f satisfy

f (x)f (y) = g(x2 + y 2)h(y/x) (2)

Theorem Suppose f : (0, ∞) 7→ R satisfies (2) and assume f is non-zero on a set of positive Lebesgue measure, and the discontinuity set of f is not dense in (0, ∞). Then f has the form

2 f (x) = Axpecx . (3)

Furthermore, the functions g and h in (2) are given by

p cx g(x) = A1x e ,  x p h(x) = A , 2 1 + x2 2 where A1A2 = A . Assume p > −1 and c < 0 to ensure integrability of f .

Formula (1) with the functions given in the theorem leads to

Z π 2 Γ(p + 1) p J = p+1 sin t dt (−2c) 0 where Γ denotes the gamma function.

Calculation of the integral in a closed form requires that p be an integer. But in this case

∞ Z 2 J = A xpecx dx 0 can be evaluated in terms of the original Gaussian integral I by repeated integration by parts. We conclude that Poisson’s method has essentially no further application as an integration technique! Probabilistic/geometric interpretation of the result

In the case when p is a positive integer and c = −1/2, the normalized form of f in (3)

−1 p −x2/2 f (x) = J x e I{x>0}

q 2 is the pdf of χp+1 distribution, i.e. the distribution of the diameter of a (p + 1)-dimensional box with sides chosen according to independent N(0,1) laws. Discontinuous functions satisfying (2)

1. Let a > 0. It is clear that the function f = I{a} satisfies

f (x)f (y) = g(x2 + y 2)h(y/x) (2)

with g = I{2a2} and h = I{1}.

2. Let m be a multiplicative function, m(x)m(y) = m(xy). Then the functions f = g = m and h(x) = m(x/1 + x2) satisfy (2) since

 y/x  m(x)m(y) = m(xy) = m(x2 + y 2)m . 1 + (y/x)2

If m is continuous, then m has the form xp, but there exist discontinuous multiplicative functions. Such functions are discontinuous everywhere and non-vanishing. Proof of the theorem

Relies on three Lemmas:

Lemma 1. Suppose f satisfies (2) and f is non-zero on a set of positive Lebesgue measure. Then f never vanishes.

An application of Lemma 1 to number theory Let A denote the set of positive algebraic numbers. Using the fact √ that A is closed under the operations +, ×, ÷, , it is easy to show that the function f = IA satisfies (2) (with g = h = IA). The Lemma thus implies that if A has positive Lebesgue measure, then all real numbers are algebraic.

Thus we have a proof using functional equations that that the set of algebraic numbers has zero Lebesgue measure. In fact, we can prove by the same argument the following more general result.

Proposition. Let E be a measurable proper subset of (0, ∞) closed under addition, multiplication, division, and the extraction of square roots. Then E has zero Lebesgue measure. Lemma 2. Suppose f satisfies the hypotheses of the theorem. Then f is continuous everywhere. √ √ Proof. Define r(x) = f ( x) and k(x) = h( x). In view of Lemma (1), we may write (2) in the form

2 x(1+t)  r 2 k(t) r(x) = , t, x > 0. (4) r(tx)

By hypothesis, there exists an interval (a, b) on which r is a+b continuous. Define c = 2 . Suppose f (and hence r) is discontinuous at some point x0 > 0. Choose t such that c = txo x0+c and define x1 = 2 . Suppose r is continuous at x1. Let x → x0 x(1+t) x0(1+t) in (4). Then 2 → 2 = x1 and tx → tx1 = c and r is continuous at both points. Hence 2 x0(1+t) r ( 2 )k(t) r(x) → = r(x0), r(tx0) i.e. r is continuous at x0, contradicting the assumption. We conclude that r is discontinuous at x1.

Assuming r is discontinuous at x0, we have shown that r is discontinuous at x + c x = 0 1 2 Iterating this argument shows that r is discontinuous on the sequence of points defined inductively by x + c x = n−1 , n ≥ 1. n 2

This gives a contradiction since xn eventually lies in (a, b).

We conclude that r is continuous everywhere. Lemma 3. Suppose f satisfies the hypotheses of the theorem. Then the function log r is integrable at 0.

The result is non-trivial since solutions to (2) can blow up as x → 0+, e.g. xp with p < 0. Proof of the theorem. Define 1 Z x G(x) = log r(x) − log r(u) du x 0 Z 1 = log r(x) − log r(xu) du, x > 0. 0 Note that by the Lemmas, all these expressions exist. Taking logs in (4), we obtain

x(1 + t) log r(x) + log r(tx) − 2 log r = log k(t), t, x > 0. 2 Thus x(1 + t) G(x) + G(tx) − 2G 2 Z 1 xu(1 + t) = log k(t) −  log r(xu) + log r(txu) − 2 log r du 0 2 Z 1 = log k(t) − k(t) dt = 0. 0 Setting y = tx, we have x + y  G(x) + G(y) = 2G . (5) 2 This is a variant of Cauchy’s functional equation. It is easy to show that the only continuous functions satisfying (5) are linear functions. We thus have 1 Z x cx + p log r(x) − log r(u) du = x 0 2 for constants c and p.

Multiplying by x and differentiating yields xr 0(x) p = cx + . r(x) 2 Solving for r gives r(x) = Axp/2ecx . Hence f (x) = r(x2) = Axpecx2 as claimed.

Using (4), we can solve for k (and thus h), then obtain g by taking y = x in Eq. (2). A concluding remark

The argument used to prove the theorem can be applied to the more general functional equation

f (x)g(y) = p(x2 + y 2)q(y/x), (6)

with a similar conclusion. This answers the question: when does a product function of cartesian coordinates transform to a product of polar coordinates? Some background to the problem and references

The subject of functional equations, although not a major area of study in recent years, has a long and distinguished history going back to Abel and Cauchy. The leader of the subject in modern times is J. Aczel:

J. Aczel, Functional Equations and Their Applications, 1966.

A very similar equation to (6) was previously studied by J.A. Baker using different techniques, ”On the functional equation f (x)g(y) = p(x + y)q(x/y)”, Aequationes Math., 1976. Application to the Poisson technique

The functonal equation f (x)f (y) = g(x2 + y 2)h(y/x) was first studied by the speaker (Math. Mag., 1993) in this context, assuming an additional asymptotic condition: f (x) ∼ xp as x → 0+.

The problem was treated by R. Dawson (Am. Math. Monthly, 2005) without the asymptotic condition, but for the more restricted equation f (x)f (y) = g(x2 + y 2).

The present result was proved by B., Elem. Math, 2010. Thank you for your attention!