The amazing Poisson calculation: a method or atrick?

Denis Bell

April 27, 2017

The Gaussian integral and Poisson’s calculation

The Gaussian function e−x2

plays a central role in probability andstatistics. For this reason, it is important to know the value of theintegral

I =

∫ ∞0

e−x2dx .

The function e−x2

does not have an elementary antiderivative, so Icannot be evaluated by the FTC.

There is the following argument, attributed to Poisson, forcalculating I . Consider

I 2 =

∫ ∞0

e−x2dx ·

∫ ∞0

e−y2dy .

Interpret this product as a double integral in the plane andtransform to polar coordinates x = r cos θ, y = r sin θ to get

I 2 =

∫ ∞0

∫ ∞0

e−(x2+y2)dxdy

=

∫ π/2

0

∫ ∞0

e−r2r drdθ

= −π4

∫ ∞0

d

dre−r

2dr =

π

4.

Thus I =√π2 .

Can this argument be used to evaluate other seemingly intractableimproper integrals?

Consider the integral

J =

∫ ∞0

f (x)dx .

Proceeding as above, we obtain

J2 =

∫ ∞0

∫ ∞0

f (x)f (y) dxdy .

In order to take the argument further, there will need to existfunctions g and h such that

f (x)f (y) = g(x2 + y2)h(y/x).

Transformation to polar coordinates and the substitution u = r2

will then yield

J2 =1

2

∫ ∞0

g(u)du

∫ π/2

0h(tan θ)dθ. (1)

Which functions f satisfy

f (x)f (y) = g(x2 + y2)h(y/x) (2)

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

f (x) = Axpecx2. (3)

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

g(x) = A1xpecx ,

h(x) = A2

( x

1 + x2

)p,

where A1A2 = A2.

Assume p > −1 and c < 0 to ensure integrability of f .

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

J2 =Γ(p + 1)

(−2c)p+1

∫ π

0sinp t dt

where Γ denotes the gamma function.

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

J = A

∫ ∞0

xpecx2dx

can be evaluated in terms of the original Gaussian integral I byrepeated integration by parts.

We conclude that Poisson’s method has essentially no furtherapplication as an integration technique!

Probabilistic/geometric interpretation of the result

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

f (x) = J−1xpe−x2/2I{x>0}

is the pdf of√χ2p+1 distribution, i.e. the distribution of the

diameter of a (p + 1)-dimensional box with sides chosen accordingto 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 + y2)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). Thenthe functions f = g = m and h(x) = m(x/1 + x2) satisfy (2) since

m(x)m(y) = m(xy) = m(x2 + y2)m( y/x

1 + (y/x)2

).

If m is continuous, then m has the form xp, but there existdiscontinuous multiplicative functions. Such functions arediscontinuous 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 ofpositive Lebesgue measure. Then f never vanishes.

An application of Lemma 1 to number theoryLet A denote the set of positive algebraic numbers. Using the factthat A is closed under the operations +,×,÷,√, it is easy to showthat the function f = IA satisfies (2) (with g = h = IA). TheLemma thus implies that if A has positive Lebesgue measure, thenall real numbers are algebraic.

Thus we have a proof using functional equations that that the setof algebraic numbers has zero Lebesgue measure.

In fact, we can prove by the same argument the following moregeneral result.

Proposition. Let E be a measurable proper subset of (0,∞)closed under addition, multiplication, division, and the extractionof 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

r(x) =r2(x(1+t)

2

)k(t)

r(tx), t, x > 0. (4)

By hypothesis, there exists an interval (a, b) on which r iscontinuous. Define c = a+b

2 . Suppose f (and hence r) isdiscontinuous at some point x0 > 0. Choose t such that c = txoand define x1 = x0+c

2 . Suppose r is continuous at x1. Let x → x0

in (4). Then x(1+t)2 → x0(1+t)

2 = x1 and tx → tx1 = c and r iscontinuous at both points.

Hence

r(x)→r2( x0(1+t)

2 )k(t)

r(tx0)= r(x0),

i.e. r is continuous at x0, contradicting the assumption. Weconclude that r is discontinuous at x1.

Assuming r is discontinuous at x0, we have shown that r isdiscontinuous at

x1 =x0 + c

2

Iterating this argument shows that r is discontinuous on thesequence of points defined inductively by

xn =xn−1 + c

2, n ≥ 1.

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 asx → 0+, e.g. xp with p < 0.

Proof of the theorem. Define

G (x) = log r(x)− 1

x

∫ x

0log r(u) du

= log r(x)−∫ 1

0log r(xu) du, x > 0.

Note that by the Lemmas, all these expressions exist. Taking logsin (4), we obtain

log r(x) + log r(tx)− 2 log r(x(1 + t)

2

)= log k(t), t, x > 0.

Thus

G (x) + G (tx)− 2G(x(1 + t)

2

)= log k(t)−

∫ 1

0

{log r(xu) + log r(txu)− 2 log r

(xu(1 + t)

2

)}du

= log k(t)−∫ 1

0k(t) dt = 0.

Setting y = tx , we have

G (x) + G (y) = 2G(x + y

2

). (5)

This is a variant of Cauchy’s functional equation. It is easy toshow that the only continuous functions satisfying (5) are linearfunctions. We thus have

log r(x)− 1

x

∫ x

0log r(u) du =

cx + p

2

for constants c and p.

Multiplying by x and differentiating yields

xr ′(x)

r(x)= cx +

p

2.

Solving for r givesr(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 takingy = x in Eq. (2).

A concluding remark

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

f (x)g(y) = p(x2 + y2)q(y/x), (6)

with a similar conclusion. This answers the question: when does aproduct function of cartesian coordinates transform to a product ofpolar coordinates?

Some background to the problem and references

The subject of functional equations, although not a major area ofstudy in recent years, has a long and distinguished history goingback to Abel and Cauchy. The leader of the subject in moderntimes is J. Aczel:

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

A very similar equation to (6) was previously studied by J.A. Bakerusing different techniques, ”On the functional equationf (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 + y2)h(y/x) was firststudied by the speaker (Math. Mag., 1993) in this context,assuming an additional asymptotic condition: f (x) ∼ xp asx → 0+.

The problem was treated by R. Dawson (Am. Math. Monthly,2005) without the asymptotic condition, but for the morerestricted equation f (x)f (y) = g(x2 + y2).

The present result was proved by B., Elem. Math, 2010.

Thank you for your attention!