Poisson s remarkable calculation - a method or a trick?

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1 Poisso s remarkable calculatio - a method or a trick? Deis Bell 1 Departmet of Mathematics, Uiversity of North Florida 1 UNF Drive, Jacksoville, FL 34, U. S. A. dbell@uf.edu The Gaussia fuctio e x plays a fudametal role i probability ad statistics. For this reaso, it is importat to kow the value of the itegral I = Z e x dx. Sice the itegrad does ot have a elemetary atiderivative, I caot be evaluated directly by the fudametal theorem of calculus. The familiar computatio of the Gaussia itegral is via the followig remarkable trick, attributed to Poisso. Oe forms the square of I, iterprets it as a double itegral i the plae, trasforms to polar coordiates ad the aswer magically pops out. The calculatio is as follows = I = Z π/ Z Z Z re r drdθ = π = πe r 4 e (x +y ) dxdy, Z = π 4. re r dr Hece I = π/. Ca this method be used to compute other seemigly itractable itegrals? I this article we explore this questio. Cosider a improper itegral J Z f(x)dx ad suppose that f satisfies the fuctioal equatio f(x)f(y) = g(x + y )h(y/x), x, y >. (1) 1 Research partially supported by NSF grat DMS

2 Poisso s remarkable calculatio The Poisso s argumet yields the relatio J = 1 Z Z π/ g(x)dx h(ta θ)dθ. () The determiatio of J is thereby reduced to the evaluatio of two, possibly solvable, itegrals. We are therefore led to ask which fuctios f satisfy a equatio of the form (1), where the fuctios g ad h are such that the itegrals i () ca be evaluated? I a previous work [Be], the author characterized the solutios f to equatio (1) uder the somewhat restrictive assumptio that f is asmptotic to a power fuctio at zero. R. Dawso [Da] addressed the problem for the less geeral equatio f(x)f(y) = g(x +y ). I the preset article, we establish the followig geeralizatio of these results. Theorem. Suppose f : (, ) 7 R satisfies equatio (1), f is o-zero o a set of positive Lebesgue measure, ad the discotiuity set of f is ot dese i (, ). The f has the form f(x) = Ax p e cx. Furthermore, the correspodig fuctios g ad h are uique up to multiplicative costats ad are give by g(x) = A 1 x p e cx, (3) x p h(x) = A 1 + x (4) where A 1 A = A. Deote J = Z x p e cx dx ad assume that p > 1 ad c < to esure the existece of the itegral. The decompositio ito (3) ad (4) implied by the Theorem results i the idetity J = Γ(p + 1) ( c) p+1 Z π/ si p tdt where Γ deotes the gamma fuctio. Furthermore, the evaluatio of the above itegral i closed form requires that p be a iteger, thus p {, 1,,... }. But i this case, J ca be reduced to the Gaussia itegral I by a scalig substitutio ad successive itegratio by parts! We coclude that Poisso s argumet has o wider applicability as a itegratio method. This aswers the questio posed i the title of the article. As Dawso observes, it is remarkable that Poisso s method turs out to have essetially oly oe applicatio ad that this sigle applicatio is such a sigificat oe! The proof of the Theorem differs substatially from the argumet i [Be] i focussig o the fuctio g i (1) rather tha o h. The proof will require three prelimiary results.

3 Poisso s remarkable calculatio 3 Lemma 1. Suppose f satisfies (1) ad f is o-zero o a set of positive Lebesgue measure. The f ever vaishes. Proof. Note first that h(1) 6= otherwise takig y = x i ( 1) gives f, cotradictig the hypothesis. We suppose throughout, without loss of geerality, that h(1) = 1. Settig y = x i (1) gives Substitutig for the fuctio g i (1), we obtai f (x) = g(x ). (5) Defie The (6) yields f(x)f(y) = f r x + y h y. (6) x r(x) = f( x), k(x) = h( x). r(x)r(tx) = r x(1 + t) k(t), t, x >. (7) We ow prove the claim: there exists δ > such that k(x) 6= for all x i the iterval (1 δ, 1 + δ). We argue by cotradictio. Sice the map x 7 x is strictly mootoe o (, ), the o-zero set of r has positive Lebesgue measure λ. Hece there exists a iteger N such that a λ(t ) > where T deotes the set {x/ r(x) 6= } [N, N + 1]. Suppose the claim ot hold. The there exists a sequece t 1 such that k(t ) = for all. Equatio (7) yields Defie ad r(x)r(t x) =,, x. (8) T {t x/ x T } V \ [ T k. =1 k= The (8) implies T ad T are disjoit, for all. Hece V T = φ. (9) Furthermore λ(v ) = lim [ λ k= T k lim λ(t ) = lim t λ(t ) = a. (1) If x V, the there exists a subsequece {t k } of {t } ad {x k } T such that x = t k x k. Sice t k 1, this implies x T, i.e. V T.

4 Poisso s remarkable calculatio 4 Let U be a arbitrary ope set such that T U. The T V Ū. (11) We coclude from (9) - (11) that λ(u) = λ(ū) λ(t V ) = λ(t ) + λ(v ) a. Thus o a = λ(t ) = if λ(u)/ U ope, T U a. This implies a =, a cotradictio. The claim follows. Now suppose f vaishes, so r(x ) = for some x >. Settig x = x i (7) ad usig the claim, we deduce that r o the iterval x (1 δ/, 1 + δ/). Extrapolatig this property results i the coclusio r, hece f which cotradicts the hypothesis of the Lemma. Before proceedig further, we give some examples of fuctios f satisfyig equatio (1) that do ot have the form specified i the Theorem. Example 1. Let m be a discotiuous fuctio o (, ) with the multiplicative property m(x)m(y) = m(xy), x, y >. (1) (The existece of a large class of such fuctios is well-established, see e.g. [A]). t The equatio (1) is satisfied with f = g = m ad h(t) = m( 1+t ). Note that m is ever zero (otherwise (1) implies m ). Example. Let a > ad defie f = I {a}, where I {a} deotes the idicator fuctio of the sigleto set {a}, g = I {a }, ad h = I {1}. The it is clear that these fuctios satisfy (1). Example 3. Let A deote the set of algebraic umbers i (, ). Defie f = g = h = I A. The (1) follows from the fact that A is closed uder the positivitypreservig arithmetic operatios ad the extractio of square roots. Ideed, this immediately implies that if I A (x)i A (y) = 1, the I A (x + y )I A (y/x) = 1. Coversely, suppose I A (x + y )I A (y/x) = 1. Write x + y = α ad y/x = β where α, β A. Solvig for x ad y, we have r r α α x = 1 + β, y = β 1 + β. Thus x, y A ad so I A (x)i A (y) = 1. This example i cojuctio with Lemma 1, provides a ew proof of the wellkow fact that (assumig at least oe trascedetal umber exists) the set of algebraic umbers has zero Lebesgue measure. I fact, replacig A i this argumet by a arbitrary set, we obtai the followig result. Propositio. Let E be a measurable proper subset of (, ) closed uder additio, multiplicatio, divisio, ad the extractio of square roots. The E has zero Lebesgue measure.

5 Poisso s remarkable calculatio 5 The above examples show that either of the additioal hypotheses i the Theorem is redudat. Lemma. Suppose f satsifes the hypotheses of the Theorem. The the fuctio r is cotiuous everywhere. Proof. I view of Lemma 1, we may write (7) i the form r x(1+t) k(t) r(x) =, t, x >. (13) r(tx) By hypothesis, there exists a iterval (a, b) o which r is cotiuous. Defie c = a+b. Suppose r is discotiuous at some poit x >. Choose t such that c = tx o ad defie x 1 = x+c. Assume r is cotiuous at x 1. The (13) yields lim r(x) = r (x 1 )k(t) = r(x ), x x r(c) i.e. r is cotiuous at x, cotrary to assumptio. We coclude that r is discotiuous at x 1. Iteratio of this argumet shows that r is discotiuous o the sequece of poits {x } defied iductively by x = x 1 + c, 1. Sice x evetually lies i (a, b) this gives a cotradictio, thereby provig the Lemma. Remark. Lemmas 1 ad imply that f ad hece r has costat sig, which we may suppose without loss of geerality, is positive. We deduce from (7) that k is the strictly positive ad everywhere cotiuous. Lemma 3. Suppose f satisfies the hypotheses of the Theorem. The the fuctio log r is itegrable at. Proof. The argumet is a quatitative versio of the iterative step i the proof of Lemma. We make repeated use of (13), which we write as r x(1+t) k(t) r(tx) =, t, x >. (14) r(x) First, choose δ <.15 ad l, L, m, M such that < l, m < 1, M, L > 1 ad l < r(x) < L, x [1, ], m < k(t) < M, t [δ, 1]. Takig x = ad lettig t vary i the rage [δ, 1] i (14), we have l m L < r(x) < L M, x [δ, 1]. (15) l

6 Poisso s remarkable calculatio 6 Now settig x = 4δ i (14) ad usig (15) yields l 5 m 3 L M < r(x) < L5 M 3 l m, x [4δ, δ]. (16) Settig x = 8δ ad usig (16) i (14) we have l 1 m 8 L 7 M 5 < r(x) < L1 M 8 l 7 m 5, x [8δ3, 4δ ] Note that the powers of l, m, L, M i these estimates are icreasig (roughly) by a factor of 3 each time. Iteratig this process, we see that there exist costats D < 1 ad E > 1 such that where = δ. Let D 4 < r(x) < E 4, < x < 1 (17) q = 4 1 log ad ote that q > 1/e by choice of δ. Substitutig t = i (17) gives log r(x) < q log t max( log D, log E), x [t, t 1 ]. Sice Z 1 q log t dt = this implies ad we are doe. Proof of the Theorem. Defie Z 1 Z (eq) x dx < log r(x) dx < G(x) = log r(x) 1 x = log r(x) Z 1 Z x log r(u)du (18) log r(xu)du, x >. (Note that Lemma 3 implies that the itegrals exist.) Takig logarithms i (13), we have x(1 + t) log r(x) + log r(tx) log r = log k(t), t, x >. Thus = log k(t) Z 1 x(1 + t) G(x) + G(tx) G x(1 + t)u o log r(xu) + log r(txu) log r du

7 Poisso s remarkable calculatio 7 Settig y = tx gives = log k(t) Z 1 log k(t)du =. G(x) + G(y) = G x + y, x, y >. (19) Equatio (19) is a variat of the Cauchy fuctioal equatio. It is well-kow (ad easy to show) that the oly cotiuous fuctios G satisfyig (19) are liear fuctios x 7 ax + b. We ca therefore write log r(x) 1 x Z x log r(u)du = cx + p for costats c ad p. Multiplyig by x ad differetiatig yields Solvig this ODE for r gives Hece xr (x) r(x) = cx + p. r(x) = Ax p/ e cx. f(x) = Ax p e cx as claimed. Substitutig this expressio ito (5) ad (6), we obtai the fuctios g ad h. We coclude with the followig remarks. The methods of this paper ca be used to treat the more geeral fuctioal equatio f(x)g(y) = F (x + y )G(y/x), x, y >. Assumig the discotiuity sets of f ad g are o-dese ad f ad g are ozero o sets of positive Lebesgue measure, we ca show that (up to multiplicative costats) the fuctios ecessarily have the form where p 1 + p = p. f(x) = x p1 e cx, g(x) = x p e cx, F (x) = x p e cx, G(x) = x p (1 + x ) p, The subject of fuctioal equatios, which origiated with Cauchy ad Abel, has spawed a extesive body of advaced techiques (see, e.g. [A]). These techiques have bee used to prove far more geeral results tha those preseted here (cf. [Ba], [L] ad [M]). The advatage of the preset approach is that it provides a complete aalysis of equatio (1) by direct ad elemetary meas.

8 Poisso s remarkable calculatio 8 Refereces [A] J. Aczel, Lectures o Fuctioal Equatios ad Their Applicatios, Academic Press, New York, [Ba] J. A. Baker, O the fuctioal equatio f(x)g(y) = p(x + y)q(x/y), Aequatioes Math, 14 (1976), [Be] D. Bell. O the limitatios of a well-kow itegratio techique, Math. Mag. 66 (1993) [Da] R. Dawso. O a sigular itegratio techique of Poisso, Amer. Math. Mothly 11 (5), o. 3, 7 7. [L] K. Lajko, Remark to a paper by J. A. Baker, Aequatioes Math. 19 (1979), [M] F. Meszaros, Numbers, fuctios, equatios 8, De La Motte Castle, Noszvaj, Hugary, 8.