THE GAUSSIAN INTEGRAL KEITH CONRAD Let x J x and x x - PDF document

THEGAUSSIANINTEGRALKEITHCONRADLetI=Z1�1e�1 2x2dx;J=Z10e�x2dx;andK=Z1�1e�x2dx:Thesenumbersarepositive,andJ=I=(2p 2)andK=I=p 2.Theorem.Withnotationasabove,I=p 2,orequivalentlyJ=p =2,orequivalentlyK=1.Wewillgivemultipleproofsofthisresult.(Otherlistsofproofsarein[4]and[9].)Thetheoremissubtlebecausethereisnosimpleantiderivativefore�1 2x2(ore�x2ore�x2).Forcomparison,Z10xe�1 2x2dxcanbecomputedusingtheantiderivative�e�1 2x2:thisintegralis1.1.FirstProof:PolarcoordinatesThemostwidelyknownproof,duetoPoisson[9,p.3],expressesJ2asadoubleintegralandthenusespolarcoordinates:J2=Z10e�x2dxZ10e�y2dy=Z10Z10e�(x2+y2)dxdy:Thisisadoubleintegralovertherstquadrant,whichwewillcomputebyusingpolarcoordinates.Therstquadrantisf(r;):r0and0=2g.Writingx2+y2asr2anddxdyasrdrd,J2=Z=20Z10e�r2rdrd=Z10re�r2dr
Z=20d=�1 2e�r210
 2=1 2
 2= 4:Takingsquareroots,J=p =2.By[1],thisnicemethoddoesn'tapplytoanyotherintegral.2.SecondProof:AnotherchangeofvariablesOurnextproofusesanotherchangeofvariablestocomputeJ2,butthiswillonlyrelyonsingle-variablecalculus.Asbefore,wehaveJ2=Z10Z10e�(x2+y2)dxdy=Z10Z10e�(x2+y2)dxdy;1 2KEITHCONRADbutinsteadofusingpolarcoordinateswemakeachangeofvariablesx=ytontheinsideintegral,withdx=ydt,soJ2=Z10Z10e�y2(t2+1)ydtdy=Z10Z10ye�y2(t2+1)dydt:SinceZ10ye�ay2dy=1 2afora�0,wehaveJ2=Z10dt 2(t2+1)=1 2
 2= 4;soJ=p =2.ThisapproachisduetoLaplace[7,pp.94{96]andhistoricallyprecedesthemorefamiliartechniqueintherstproofabove.WewillseeintheeighthproofthatthiswasnotLaplace'srstmethod.3.ThirdProof:DifferentiatingundertheintegralsignFort�0,setA(t)=Zt0e�x2dx2:TheintegralwewanttocalculateisA(1)=J2andthentakeasquareroot.DierentiatingA(t)withrespecttotandusingtheFundamentalTheoremofCalculus,A0(t)=2Zt0e�x2dx
e�t2=2e�t2Zt0e�x2dx:Letx=ty,soA0(t)=2e�t2Z10te�t2y2dy=Z102te�(1+y2)t2dy:Thefunctionundertheintegralsigniseasilyantidierentiatedwithrespecttot:A0(t)=Z10�@ @te�(1+y2)t2 1+y2dy=�d dtZ10e�(1+y2)t2 1+y2dy:LettingB(t)=Z10e�t2(1+x2) 1+x2dx;wehaveA0(t)=�B0(t)forallt�0,sothereisaconstantCsuchthat(3.1)A(t)=�B(t)+Cforallt�0.TondC,welett!0+in(3.1).TheleftsidetendstoZ00e�x2dx2=0whiletherightsidetendsto�Z10dx=(1+x2)+C=�=4+C.ThusC==4,so(3.1)becomesZt0e�x2dx2= 4�Z10e�t2(1+x2) 1+x2dx:Lettingt!1inthisequation,weobtainJ2==4,soJ=p =2.Acomparisonofthisproofwiththerstproofisin[20]. THEGAUSSIANINTEGRAL34.FourthProof:AnotherdifferentiationundertheintegralsignHereisasecondapproachtondingJbydierentiationundertheintegralsign.IheardaboutitfromMichaelRozman[14],whomodiedanideaonmath.stackexchange[22],andinaslightlylesselegantformitappearedmuchearlierin[18].Fort2R,setF(t)=Z10e�t2(1+x2) 1+x2dx:ThenF(0)=R10dx=(1+x2)==2andF(1)=0.Dierentiatingundertheintegralsign,F0(t)=Z10�2te�t2(1+x2)dx=�2te�t2Z10e�(tx)2dx:Makethesubstitutiony=tx,withdy=tdx,soF0(t)=�2e�t2Z10e�y2dy=�2Je�t2:Forb�0,integratebothsidesfrom0tobandusetheFundamentalTheoremofCalculus:Zb0F0(t)dt=�2JZb0e�t2dt=)F(b)�F(0)=�2JZb0e�t2dt:Lettingb!1inthelastequation,0� 2=�2J2=)J2= 4=)J=p  2:5.FifthProof:AvolumeintegralOurnextproofisduetoT.P.Jameson[5]anditwasrediscoveredbyA.L.Delgado[3].Revolvethecurvez=e�1 2x2inthexz-planearoundthez-axistoproducethe\bellsurface"z=e�1 2(x2+y2).Seebelow,wherethez-axisisverticalandpassesthroughthetoppoint,thex-axisliesjustunderthesurfacethroughthepoint0infront,andthey-axisliesjustunderthesurfacethroughthepoint0ontheleft.WewillcomputethevolumeVbelowthesurfaceandabovethexy-planeintwoways.FirstwecomputeVbyhorizontalslices,whicharediscs:V=Z10A(z)dzwhereA(z)istheareaofthediscformedbyslicingthesurfaceatheightz.Writingtheradiusofthediscatheightzasr(z),A(z)=r(z)2.Tocomputer(z),thesurfacecutsthexz-planeatapairofpoints(x;e�1 2x2)wheretheheightisz,soe�1 2x2=z.Thusx2=�2lnz.Sincexisthedistanceofthesepointsfromthez-axis,r(z)2=x2=�2lnz,soA(z)=r(z)2=�2lnz.ThereforeV=Z10�2lnzdz=�2(zlnz�z)10=�2(�1�limz!0+zlnz):ByL'Hospital'srule,limz!0+zlnz=0,soV=2.(AcalculationofVbyshellsisin[11].)Nextwecomputethevolumebyverticalslicesinplanesx=constant.Verticalslicesarescaledbellcurves:lookattheblackcontourlinesinthepicture.Theequationofthebellcurvealongthetopoftheverticalslicewithx-coordinatexisz=e�1 2(x2+y2),whereyvariesandxisxed.Then 4KEITHCONRAD V=Z1�1A(x)dx,whereA(x)istheareaofthex-slice:A(x)=Z1�1e�1 2(x2+y2)dy=e�1 2x2Z1�1e�1 2y2dy=e�1 2x2I:ThusV=Z1�1A(x)dx=Z1�1e�1 2x2Idx=IZ1�1e�1 2x2dx=I2.ComparingthetwoformulasforV,wehave2=I2,soI=p 2.6.SixthProof:The�-functionForanyintegern0,wehaven!=Z10tne�tdt.Forx�0wedene�(x)=Z10txe�tdt t;so�(n)=(n�1)!whenn1.Usingintegrationbyparts,�(x+1)=x�(x).Oneofthebasicpropertiesofthe�-function[15,pp.193{194]is(6.1)�(x)�(y) �(x+y)=Z10tx�1(1�t)y�1dt: THEGAUSSIANINTEGRAL5Setx=y=1=2:�1 22=Z10dt p t(1�t):Note�1 2=Z10p te�tdt t=Z10e�t p tdt=Z10e�x2 x2xdx=2Z10e�x2dx=2J;so4J2=R10dt=p t(1�t).Withthesubstitutiont=sin2,4J2=Z=202sincosd sincos=2 2=;soJ=p =2.Equivalently,�(1=2)=p .Anymethodthatproves�(1=2)=p isalsoamethodthatcalculatesZ10e�x2dx.7.SeventhProof:AsymptoticestimatesWewillshowJ=p =2byatechniquewhosestepsarebasedon[16,p.371].Forx0,powerseriesexpansionsshow1+xex1=(1�x).Reciprocatingandreplacingxwithx2,weget(7.1)1�x2e�x21 1+x2:forallx2R.Foranypositiveintegern,raisethetermsin(7.1)tothenthpowerandintegratefrom0to1:Z10(1�x2)ndxZ10e�nx2dxZ10dx (1+x2)n:Underthechangesofvariablesx=sinontheleft,x=y=p ninthemiddle,andx=tanontheright,(7.2)Z=20(cos)2n+1d1 p nZp n0e�y2dyZ=40(cos)2n�2d:SetIk=R=20(cos)kd,soI0==2,I1=1,and(7.2)implies(7.3)p nI2n+1Zp n0e�y2dyp nI2n�2:Wewillshowthatask!1,kI2k!=2.Thenp nI2n+1=p n p 2n+1p 2n+1I2n+1!1 p 2r  2=p  2andp nI2n�2=p n p 2n�2p 2n�2I2n�2!1 p 2r  2=p  2;soby(7.3)Zp n0e�y2dy!p =2.ThusJ=p =2. 6KEITHCONRADToshowkI2k!=2,rstwecomputeseveralvaluesofIkexplicitlybyarecursion.Usingintegrationbyparts,Ik=Z=20(cos)kd=Z=20(cos)k�1cosd=(k�1)(Ik�2�Ik);so(7.4)Ik=k�1 kIk�2:Using(7.4)andtheinitialvaluesI0==2andI1=1,therstfewvaluesofIkarecomputedandlistedinTable1.k Ik k Ik 0 =2 1 12 (1=2)(=2) 3 2/34 (3=8)(=2) 5 8/156 (15=48)(=2) 7 48/105Table1.FromTable1weseethat(7.5)I2nI2n+1=1 2n+1 2for0n3,andthiscanbeprovedforallnbyinductionusing(7.4).Since0cos1for2[0;=2],wehaveIkIk�1Ik�2=k k�1Ikby(7.4),soIk�1Ikask!1.Therefore(7.5)impliesI22n1 2n 2=)(2n)I22n! 2asn!1.Then(2n+1)I22n+1(2n)I22n! 2asn!1,sokI2k!=2ask!1.ThiscompletesourproofthatJ=p =2.Remark7.1.Thisproofiscloselyrelatedtothefthproofusingthe�-function.Indeed,by(6.1)�(k+1 2)�(1 2) �(k+1 2+1 2)=Z10t(k+1)=2+1(1�t)1=2�1dt;andwiththechangeofvariablest=(cos)2for0=2,theintegralontherightisequalto2R=20(cos)kd=2Ik,so(7.5)isthesameasI2nI2n+1=�(2n+1 2)�(1 2) 2�(2n+2 2)�(2n+2 2)�(1 2) 2�(2n+3 2)=�(2n+1 2)�(1 2)2 4�(2n+1 2+1)=�(2n+1 2)�(1 2)2 42n+1 2�(2n+1 2)=�(1 2)2 2(2n+1): THEGAUSSIANINTEGRAL7By(7.5),=�(1=2)2.Wesawinthefthproofthat�(1=2)=p ifandonlyifJ=p =2.8.EighthProof:Stirling'sFormulaBesidestheintegralformulaZ1�1e�1 2x2dx=p 2thatwehavebeendiscussing,anotherplaceinmathematicswherep 2appearsisinStirling'sformula:n!nn enp 2nasn!1:In1730DeMoivreprovedn!C(nn=en)p nforsomepositivenumberCwithoutbeingabletodetermineC.StirlingsoonthereaftershowedC=p 2andwounduphavingthewholeformulanamedafterhim.WewillshowthatdeterminingthattheconstantCinStirling'sformulaisp 2isequivalenttoshowingthatJ=p =2(or,equivalently,thatI=p 2).Applying(7.4)repeatedly,I2n=2n�1 2nI2n�2=(2n�1)(2n�3) (2n)(2n�2)I2n�4...=(2n�1)(2n�3)(2n�5)


(5)(3)(1) (2n)(2n�2)(2n�4)


(6)(4)(2)I0:Inserting(2n�2)(2n�4)(2n�6)


(6)(4)(2)inthetopandbottom,I2n=(2n�1)(2n�2)(2n�3)(2n�4)(2n�5)


(6)(5)(4)(3)(2)(1) (2n)((2n�2)(2n�4)


(6)(4)(2))2 2=(2n�1)! 2n(2n�1(n�1)!)2 2:ApplyingDeMoivre'sasymptoticformulan!C(n=e)np n,,I2nC((2n�1)=e)2n�1p 2n�1 2n(2n�1C((n�1)=e)n�1p n�1)2 2=(2n�1)2n1 2n�1p 2n�1 2n
22(n�1)Ce(n�1)2n1 (n�1)2(n�1) 2asn!1.Foranya2R,(1+a=n)n!eaasn!1,so(n+a)neann.Substitutingthisintotheaboveformulawitha=�1andnreplacedby2n,(8.1)I2ne�1(2n)2n1 p 2n 2n
22(n�1)Ce(e�1nn)21 n2n 2= Cp 2n:SinceIk�1Ik,theoutertermsin(7.3)arebothasymptotictop nI2n=(Cp 2)by(8.1).ThereforeZp n0e�y2dy! Cp 2asn!1,soJ==(Cp 2).ThereforeC=p 2ifandonlyifJ=p =2. 8KEITHCONRAD9.NinthProof:TheoriginalproofTheoriginalproofthatJ=p =2isduetoLaplace[8]in1774.(AnEnglishtranslationofLaplace'sarticleismentionedinthebibliographiccitationfor[8],withpreliminarycommentsonthatarticlein[17].)Hewantedtocompute(9.1)Z10dx p �logx:Settingy=p �logx,thisintegralis2R10e�y2dy=2J,soweexpect(9.1)tobep .Laplace'sstartingpointforevaluating(9.1)wasaformulaofEuler:(9.2)Z10xrdx p 1�x2sZ10xs+rdx p 1�x2s=1 s(r+1) 2forpositiverands.(Laplacehimselfsaidthisformulaheld\whateverbe"rors,butifs0thenthenumberunderthesquarerootisnegative.)Accepting(9.2),letr!0inittoget(9.3)Z10dx p 1�x2sZ10xsdx p 1�x2s=1 s 2:Nowlets!0in(9.3).Then1�x2s�2slogxbyL'Hopital'srule,so(9.3)becomesZ10dx p �logx2=:Thus(9.1)isp .Euler'sformula(9.2)looksmysterious,butwehavemetitbefore.Intheformulaletxs=coswith0=2.Thenx=(cos)1=s,andaftersomecalculations(9.2)turnsinto(9.4)Z=20(cos)(r+1)=s�1dZ=20(cos)(r+1)=sd=1 (r+1)=s 2:WeusedtheintegralIk=R=20(cos)kdbeforewhenkisanonnegativeinteger.Thisnotationmakessensewhenkisanypositiverealnumber,andthen(9.4)assumestheformII+1=1 +1 2for=(r+1)=s�1,whichis(7.5)withapossiblynonintegralindex.Lettingr=0ands=1=(2n+1)in(9.4)recovers(7.5).Lettings!0in(9.3)correspondstolettingn!1in(7.5),sothe7thproofisinessenceamoredetailedversionofLaplace's1774argument.10.TenthProof:ResiduetheoremWewillcalculateZ1�1e�x2=2dxusingcontourintegralsandtheresiduetheorem.However,wecan'tjustintegratee�z2=2,asthisfunctionhasnopoles.Foralongtimenobodyknewhowtohandlethisintegralusingcontourintegration.Forinstance,in1914Watson[19,p.79]wrote\Cauchy'stheoremcannotbeemployedtoevaluatealldeniteintegrals;thusZ10e�x2dxhasnotbeenevaluatedexceptbyothermethods."Inthe1940sseveralcontourintegralsolutionswerepublishedusingawkwardcontourssuchasparallelograms[10],[12,Sect.5](see[2,Exer.9,p.113]forarecentappearance).OurapproachwillfollowKneser[6,p.121](seealso[13,pp.413{414]or[21]),usingarectangularcontourandthefunctione�z2=2 1�e�p (1+i)z: THEGAUSSIANINTEGRAL9Thisfunctioncomesoutofnowhere,soourrsttaskistomotivatetheintroductionofthisfunction.Weseekameromorphicfunctionf(z)tointegratearoundtherectangularcontour Rinthegurebelow,withverticesat�R,R,R+ib,and�R+ib,wherebwillbexedandweletR!1. Supposef(z)!0alongtherightandleftsidesof RuniformlyasR!1.ThenbyapplyingtheresiduetheoremandlettingR!1,wewouldobtain(iftheintegralsconverge)Z1�1f(x)dx+Z�11f(x+ib)dx=2iXaResz=af(z);wherethesumisoverpolesoff(z)withimaginarypartbetween0andb.ThisisequivalenttoZ1�1(f(x)�f(x+ib))dx=2iXaResz=af(z):Thereforewewantf(z)tosatisfy(10.1)f(z)�f(z+ib)=e�z2=2;wheref(z)andbneedtobedetermined.Let'stryf(z)=e�z2=2=d(z),foranunknowndenominatord(z)whosezerosarepolesoff(z).Wewantf(z)tosatisfy(10.2)f(z)�f(z+)=e�z2=2forsome(whichwillnotbepurelyimaginary,so(10.1)doesn'tquitework,but(10.1)isonlymotivation).Substitutinge�z2=2=d(z)forf(z)in(10.2)givesus(10.3)e�z2=2 1 d(z)�e�z�2=2 d(z+)!=e�z2=2:Supposed(z+)=d(z).Then(10.3)impliesd(z)=1�e�z�2=2;andwiththisdenitionofd(z),f(z)satises(10.2)ifandonlyife2=1,orequivalently222iZ.Thesimplestnonzerosolutionis=p (1+i).Fromnowonthisisthevalueof,soe�2=2=e�i=�1andthenf(z)=e�z2=2 d(z)=e�z2=2 1+e�z;whichisKneser'sfunctionmentionedearlier.Thisfunctionsatises(10.2)andwehenceforthignorethemotivation(10.1).Polesoff(z)areatoddintegralmultiplesof=2.Wewillintegratethisf(z)aroundtherectangularcontour Rbelow,whoseheightisIm(). 10KEITHCONRAD Thepolesoff(z)nearesttheoriginareplottedinthegure;theyliealongtheliney=x.Theonlypoleoff(z)inside R(forR�p =2)isat=2,sobytheresiduetheoremZ Rf(z)dz=2iResz==2f(z)=2ie�2=8 (�)e�2=2=2ie32=8 �p (1+i)=p 2:AsR!1,thevalueofjf(z)jtendsto0uniformlyalongtheleftandrightsidesof R,sop 2=Z1�1f(x)dx+Z�1+ip 1+ip f(z)dz=Z1�1f(x)dx�Z1�1f(x+ip )dx:Inthesecondintegral,writeip as�anduse(real)translationinvarianceofdxtoobtainp 2=Z1�1f(x)dx�Z1�1f(x+)dx=Z1�1(f(x)�f(x+))dx=Z1�1e�x2=2dxby(10:2):11.EleventhProof:FouriertransformsForacontinuousfunctionf:R!Cthatisrapidlydecreasingat1,itsFouriertransformisthefunctionFf:R!Cdenedby(Ff)(y)=Z1�1f(x)e�ixydx:Forexample,(Ff)(0)=R1�1f(x)dx.HerearethreepropertiesoftheFouriertransform.Iffisdierentiable,thenafterusingdierentiationundertheintegralsignontheFouriertransformoffweobtain(Ff)0(y)=Z1�1�ixf(x)e�ixydx=�i(F(xf(x)))(y): THEGAUSSIANINTEGRAL11UsingintegrationbypartsontheFouriertransformoff,withu=f(x)anddv=e�ixydx,weobtainF(f0)(y)=iy(Ff)(y):IfweapplytheFouriertransformtwicethenwerecovertheoriginalfunctionuptointeriorandexteriorscaling:(11.1)(F2f)(x)=2f(�x):Let'sshowtheappearanceof2in(11.1)isequivalenttotheevaluationofIasp 2.Fixinga�0,setf(x)=e�ax2,sof0(x)=�2axf(x):ApplyingtheFouriertransformtobothsidesofthisequationimpliesiy(Ff)(y)=�2a1 �i(Ff)0(y),whichsimpliesto(Ff)0(y)=�1 2ay(Ff)(y).Thegeneralsolutionofg0(y)=�1 2ayg(y)isg(y)=Ce�y2=(4a),so(Ff)(y)=Ce�y2=(4a)forsomeconstantC.Lettinga=1 2,sof(x)=e�x2=2,weobtain(Ff)(y)=Ce�y2=2=Cf(y):Settingy=0,theleftsideis(Ff)(0)=R1�1e�x2=2dx=I,soI=Cf(0)=C.ApplyingtheFouriertransformtobothsidesoftheequation(Ff)(y)=Cf(y),weget2f(�x)=C(Ff)(x)=C2f(x).Atx=0thisbecomes2=C2,soI=C=p 2.SinceI�0,thenumberIisp 2.Ifwedidn'tknowtheconstantontherightsideof(11.1)were2,whateveritsvalueiswouldwindupbeingC2,sosaying2appearsontherightsideof(11.1)isequivalenttosayingI=p 2.References[1]D.Bell,\Poisson'sremarkablecalculation{amethodoratrick?"Elem.Math.65(2010),29{36.[2]C.A.BerensteinandR.Gay,ComplexVariables,Springer-Verlag,NewYork,1991.[3]A.L.Delgado,\ACalculationofZ10e�x2dx,"TheCollegeMath.J.34(2003),321{323.[4]H.Iwasawa,\GaussianIntegralPuzzle,"Math.Intelligencer31(2009),38{41.[5]T.P.Jameson,\TheProbabilityIntegralbyVolumeofRevolution,"MathematicalGazette78(1994),339{340.[6]H.Kneser,Funktionentheorie,VandenhoeckandRuprecht,1958.[7]P.S.Laplace,TheorieAnalytiquedesProbabilites,Courcier,1812.[8]P.S.Laplace,\Memoiresurlaprobabilitedescausesparlesevenemens,"OeuvresCompletes8,27{65.(Englishtrans.byS.Stigleras\MemoirontheProbabilityofCausesofEvents,"StatisticalScience1(1986),364{378.)[9]P.M.Lee,http://www.york.ac.uk/depts/maths/histstat/normal_history.pdf.[10]L.Mirsky,TheProbabilityIntegral,Math.Gazette33(1949),279.Onlineathttp://www.jstor.org/stable/3611303.[11]C.P.NicholasandR.C.Yates,\TheProbabilityIntegral,"Amer.Math.Monthly57(1950),412{413.[12]G.Polya,\RemarksonComputingtheProbabilityIntegralinOneandTwoDimensions,"pp.63{78inBerkeleySymp.onMath.Statist.andProb.,Univ.CaliforniaPress,1949.[13]R.Remmert,TheoryofComplexFunctions,Springer-Verlag,1991.[14]M.Rozman,\EvaluateGaussianintegralusingdierentiationundertheintegralsign,"CoursenotesforPhysics2400(UConn),Spring2016.[15]W.Rudin,PrinciplesofMathematicalAnalysis,3rded.,McGraw-Hill,1976.[16]M.Spivak,Calculus,W.A.Benjamin,1967.[17]S.Stigler,\Laplace's1774MemoironInverseProbability,"StatisticalScience1(1986),359{363.[18]J.vanYzeren,\Moivre'sandFresnel'sIntegralsbySimpleIntegration,"Amer.Math.Monthly86(1979),690{693. 12KEITHCONRAD[19]G.N.Watson,ComplexIntegrationandCauchy'sTheorem,CambridgeUniv.Press,Cambridge,1914.[20]http://gowers.wordpress.com/2007/10/04/when-are-two-proofs-essentially-the-same/#comment-239.[21]http://math.stackexchange.com/questions/34767/int-infty-infty-e-x2-dx-with-complex-analysis.[22]http://math.stackexchange.com/questions/390850/integrating-int-infty-0-e-x2-dx-using-feynmans-parametrization-trick