Four Ways to Evaluate a Poisson Integral

Transcription

1 290 MATHEMATICS MAGAZINE Four Ways to Evaluate a Poisso Itegral HONGWEI CHEN Christopher Newport Uiversity Newport News, VA hche@pcs.cu.edu I geeral, it is difficult to decide whether or ot a give fuctio ca be itegrated i elemetary ways. I light of this, it is quite surprisig that the value of the Poisso itegral /7 I(x)= l( - 2xcos + x2) do ca be determied precisely. Eve more surprisig is that we ca do so for every value of the parameter x. Usig four differet methods, we will show that 0, if Ixl < 1; )27rllxl, if xl > 1. Our itegral is oe of several kow as the Poisso Itegral; all are related i some way to Poisso's itegral formula, which recovers a aalytic fuctio o the disk from its boudary values, a relatioship we metio below. However, oe of our methods ivolves complex aalysis at all. The first oe uses Riema sums ad relies o a trigoometric idetity. The secod method is based o a fuctioal equatio ad ivolves a sequece of itegral substitutios. The third method uses parametric differetiatio ad the half-agle substitutio. We fiish with a approach based o ifiite series. It is iterestig to see how wide a rage of mathematical topics are exploited. These evaluatios are suitable for a advaced calculus class ad provide a very ice applicatio of Riema sums, fuctioal equatios, parametric differetiatio, ad ifiite series. We begi with three elemetary observatios: 1. I (0) = I(-x)= I(x). 3. I(x) = 2r I I x I + I(l/x), (x g: 0). The reader ca probably supply the proofs for these, but we will demostrate the third. If x :7 0, we have 7r x 2 1 I(x) = l x cos 0 + do = r) = 2lx I(x). =/ y lx2do + I(l/x) = 2rr l I x I + I(l/x). I view of this third observatio, our mai formula follows easily oce we show that I (x) = 0 for I x I < 1. This will be the goal of the ext four sectios. I. Usig Riema sums Sice 1-2xcos > (1 - Ix 1)2, for I x < 1, Mathematical Associatio of America is collaboratig with JSTOR to digitize, preserve, ad exted access to Mathematics Magazie

2 VOL. 75, NO. 4, OCTOBER the itegrad is cotiuous ad itegrable. Partitio the iterval [0, r] ito equal subitervals by the partitio poits Xk = kr/, for 1 < k <. The Riema sum for I(x), R, ca be simplified usig laws of logarithms: JR=Ll (1-2xcos (k)+ x2) =-I (1+ x)2 f 1-2xcos- +2). (1) k=l- To proceed further, let o = exp(ir/). The distict roots of the polyomial x2-1 are ok for - < k <, so -I x2 1 = k=- - ( - wk). Combiig the cojugate factors ad appealig to De Moivre's theorem, we fid -I 2-1 = (2-1) H ( - k)( _ -k) k=l so that = (2 1- ) (1-2x cos (- +x2), k=1 k=l1-2 xos + 2 (2) Substitutig the idetity (2) ito (1), we have ' R= Z -r I l(x x-l + (x2 1)) Sice I x I < 1, x as -- oo. Hece, we obtai jr Ax+l 2 l(x) = lim 7Z = lim - l + ( (2 1) 0. -oo - oo x - Remark This method relied o the trigoometric idetity (2), which is iterestig i its ow right. II. Usig a fuctioal equatio The fuctioal equatio we have i mid is I(x) = I(-x) = I (2). (3) Addig the two itegrals below ad usig laws of logarithms, we obtai I(x) I(x) + I(-x) = = l( l(l -2x2cos20 2x2os2 + ) x4) do. d0. Jo

3 292 MATHEMATICS MAGAZINE Settig a = 20 gives 1 f2't I(x) + I(-x)= - I l(l -2x2cosa + x4)d 2 Jo ()1 =-2 I (2) +- 2 l( -2x2 cosa + x4) da. 2 2 J7, The substitutio a = 27r - t i the last itegral shows that it is exactly the same as the first itegral. Sice the two terms o the left are the same (recallig that I(x) = I(-x)), we obtai (3). Applyig equatio (3) repeatedly, we fid that (x) =- I (X2) = I(X4) = = I(x2) Agai we assume that I x I < 1, so that x2" -+ 0 as - oo ad cosequetly 1 I(x) = lim I (x2) = O. -oo 2 Remark Equatio (3) holds for ay x. I particular, we have that I(0) = 0 ad I (+1) = 0. The latter equatio leads to a added bous: sice, for istace, r/2 rf/2 7 J l(si 0)d = l(cos ) d = - - 2, Jo 2 7r I(1)= l (2-2cos0)d = 27r2+24 l(si0)do, JO JO ad similarly for I (-1). These two itegrals are improper. To show covergece, for example, usig itegratio by parts, we have 7rf/2 l(si 0) do = lim e l(si ) - lim f/2 ocos0 do Jo 0 E-o E-o si0 7r/2 = j-f 0 cot 0 d0. Jo Sice 0 cot 0 is Riema itegrable o [0, r/2], fo/2 I(si 0) do coverges. III. Usig parametric differetiatio Sice I(x) is differetiable for Ix j < 1, we apply the Leibiz rule to I (x) to fid (X) = ( r -2cos 0 + 2x I1-2x cos 0 + X2 do. Clearly, I'(O) = 0. We ow show that I'(x) = 0 for x A 0. First, we prove that r^ 1 - X2 ~1 -. x2,do=,. (4) 1-2x cos 0 + x2 The itegrad i (4) is called Poisso's kerel, which is used to derive solutios of the two-dimesioal Laplace's equatio o uit circle [1, p.135], ad also plays a

4 VOL. VL 75, 7I NO. N 4, OCTOBER OCO importat role i summatio of Fourier series [2, pp ]. Computig the value of this itegral is ofte used to show the usefuless of the residue theorem, a relatively advaced tool [3, p. 303]. We give a more straightforward method usig the half-agle substitutio. Settig t = ta(0/2), we have I li - x2 do _2 dt - 1 2xcos +2 d = 2(1-x2) (1 -)2 + (l + )2t2 293 = 2 arcta ( + t + C I -x = 2arcta ta(0/2)) + C. The fudametal theorem of calculus the gives [ \l- x2 li 2?x N l2xcos?x2 do = lim 2 arcta ta(0/2) = T 1-2xcos9 +x02 -tlr x Usig equatio (4) ad x = 0, we get I'(X) -= 6 I- ( x J 1-12x 1- X2 cos +x2 d = 0. Thus, we have I'(x) = 0 for I x < 1 ad so I(x) = costat. Sice I(0) = 0, we have show that I(x) 0, for I x I < 1. IV. Usig ifiite series We first show that oo l( - 2x ccos 0 + x2) = -2 - cos, (5) where the series coverges uiformly for I x I < 1. Oce we establish this, itegratig (5) with respect to 0 from 0 to 7r, will show that I (x) = 0 oce agai. To prove (5) ad to keep the evaluatio at a elemetary level, istead of usig the Fourier series, we use the relatio 2 cos 0 = eio + e-i1 ad decompose ito partial fractios: 1 - x2 1 - x x cos 0 + x2 (1 - xei)( - xe-) xei 1 - xe-i' The the geometric series expasio leads to, =l I x2 0oo = y x cos 0. (6) 1-2x cos The series (6) coverges uiformly sice x=l I x I coverges for I x < 1. Subtractig 1 from the right-had side of (6) ad the dividig by x, we have =1 x 2cos0-2x 00 cos = 2 2xc 02x x-l cos. (7) 1 - cos 0 + x2 = Sice the series (7) coverges uiformly, itegratig from 0 to x term by term, we have established (5) as desired.

5 294 MATHEMATICS MAGAZINE Remark As a bous, we have aother proof of itegral (4) deduced from series (6). We have see a variety of evaluatios of the Poisso itegral. The iterested reader is ecouraged to ivestigate additioal approaches. Ackowledgmet. I wish to thak Bria Bradie ad the referees for their helpful suggestios. REFERENCES 1. D. Loga, Applied Partial Differetial Equatios, Spriger, New York, G. Tolstov, Fourier Series, Dover Books, New York, J. E. Marsde & M. J. Hoffma, Basic Complex Aalysis, Freema, New York, Cycles i the Geeralized Fiboacci Sequece modulo a Prime DOMINIC VELLA ALFRED VELLA 194 Buckigham Rd. Bletchley, Milto Keyes, UK, MK3 5JB Fiboacci@thevel las.com Sice their ivetio i the thirteeth cetury, Fiboacci sequeces have itrigued mathematicias. As well as modelig the populatio patters of overly eergetic rab- bits, however, they have sparked developmets i more serious mathematics. For ex- ample, geeralized Fiboacci sequeces crop up i all maer of situatios, from fiber optic etworks [3] to computer algorithms [1] to probability theory [2]. I this article, we study geeralized Fiboacci sequeces {G()}, give by the recurrece relatio: G() = ag( - 1) + bg( - 2) for a, b, G(O) ad G(1) itegers. We also study the periods of repetitio i such sequeces whe cosidered modulo p, a prime. For oe particular class of geeralized Fiboacci umbers, we fid a surprisig coectio with Fermat's Last Theorem. Other coectios betwee these two seem- igly urelated subjects have bee discovered i the past [8], but the oe uearthed here allows us to calculate the legth of these repetitios or cycles exactly. Some useful results Whe workig with the geeralized Fiboacci sequeces de- scribed above, we will eed some results to make our lives easier. It is well kow that the usual Fiboacci umbers (that is a = b = 1, G(0) = 0, G(1) = 1) ca be expressed usig Biet's formula [5]: ^ *^Hl)=(liI / - - /5 f. VHG() =(1 2) ( 2) This is easily derived by guessig a solutio to the recurrece of the form G() = A., solvig for A, ad matchig with the iitial coditios. I a similar way, may umber theory texts (see for example, Nive ad Zuckerma [7]) prove that the aalogous Biet formula for our sequece is (A - B)G() = G(1)(A - B) + G(O)(AB - BA), (1)