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DIFFERENTIATING UNDER THE INTEGRAL SIGN KEITH CONRAD I had learned to do integrals by DIFFERENTIATING UNDER THE INTEGRAL SIGN KEITH CONRAD I had learned to do integrals by

DIFFERENTIATING UNDER THE INTEGRAL SIGN KEITH CONRAD I had learned to do integrals by - PDF document

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DIFFERENTIATING UNDER THE INTEGRAL SIGN KEITH CONRAD I had learned to do integrals by - PPT Presentation

Bader had given me It showed how to di64256erentiate parameters under the integral sign its a certain operation It turns out thats not taught very much in the universities they dont emphasize it But I caught on how to use that method and I used tha ID: 23299

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DIFFERENTIATINGUNDERTHEINTEGRALSIGNKEITHCONRADIhadlearnedtodointegralsbyvariousmethodsshowninabookthatmyhighschoolphysicsteacherMr.Baderhadgivenme.[It]showedhowtodi erentiateparametersundertheintegralsign{it'sacertainoperation.Itturnsoutthat'snottaughtverymuchintheuniversities;theydon'temphasizeit.ButIcaughtonhowtousethatmethod,andIusedthatonedamntoolagainandagain.[If]guysatMITorPrincetonhadtroubledoingacertainintegral,[then]Icomealongandtrydi erentiatingundertheintegralsign,andoftenitworked.SoIgotagreatreputationfordoingintegrals,onlybecausemyboxoftoolswasdi erentfromeverybodyelse's,andtheyhadtriedalltheirtoolsonitbeforegivingtheproblemtome.1RichardFeynman[5,pp.71{72]1.IntroductionThemethodofdi erentiationundertheintegralsign,duetoLeibnizin1697[4],concernsintegralsdependingonaparameter,suchasR10x2e�txdx.Heretistheextraparameter.(Sincexisthevariableofintegration,xisnotaparameter.)Ingeneral,wemightwritesuchanintegralas(1.1)Zbaf(x;t)dx;wheref(x;t)isafunctionoftwovariableslikef(x;t)=x2e�tx.Example1.1.Letf(x;t)=(2x+t3)2.ThenZ10f(x;t)dx=Z10(2x+t3)2dx:Ananti-derivativeof(2x+t3)2withrespecttoxis1 6(2x+t3)3,soZ10(2x+t3)2dx=(2x+t3)3 6 x=1x=0=(2+t3)3�t9 6=4 3+2t3+t6:Thisanswerisafunctionoft,whichmakessensesincetheintegranddependsont.Weintegrateoverxandareleftwithsomethingthatdependsonlyont,notx.AnintegrallikeRbaf(x;t)dxisafunctionoft,sowecanaskaboutitst-derivative,assumingthatf(x;t)isnicelybehaved.Therule,calleddi erentiationundertheintegralsign,isthatthet-derivativeoftheintegraloff(x;t)istheintegralofthet-derivativeoff(x;t):(1.2)d dtZbaf(x;t)dx=Zba@ @tf(x;t)dx: 1Justbeforethisquote,Feynmanwrote\OnethingIneverdidlearnwascontourintegration."Perhapshemeantthatheneverfelthelearneditwell,sincehedidknowit.See[6,Lect.14,15,17,19],[7,p.92],and[8,pp.47{49].Achallengehegavein[5,p.176]suggestshedidn'tlikecontourintegration.1 2KEITHCONRADIfyouareusedtothinkingmostlyaboutfunctionswithonevariable,nottwo,keepinmindthat(1.2)involvesintegralsandderivativeswithrespecttoseparatevariables:integrationwithrespecttoxanddi erentiationwithrespecttot.Example1.2.WesawinExample1.1thatR10(2x+t3)2dx=4=3+2t3+t6,whoset-derivativeis6t2+6t5.Accordingto(1.2),wecanalsocomputethet-derivativeoftheintegrallikethis:d dtZ10(2x+t3)2dx=Z10@ @t(2x+t3)2dx=Z102(2x+t3)(3t2)dx=Z10(12t2x+6t5)dx=6t2x2+6t5x x=1x=0=6t2+6t5:Theansweragreeswithour rst,moredirect,calculation.Wewillapply(1.2)tomanyexamplesofintegrals,inSection12wewilldiscussthejusti cationofthismethodinourexamples,andthenwe'llgivesomemoreexamples.2.Euler'sfactorialintegralinanewlightForintegersn0,Euler'sintegralformulaforn!is(2.1)Z10xne�xdx=n!;whichcanbeobtainedbyrepeatedintegrationbypartsstartingfromtheformula(2.2)Z10e�xdx=1whenn=0.NowwearegoingtoderiveEuler'sformulainanotherway,byrepeateddi erentiationafterintroducingaparametertinto(2.2).Fort�0,letx=tu.Thendx=tduand(2.2)becomesZ10te�tudu=1:Dividingbytandwritinguasx(whyisthisnotaproblem?),weget(2.3)Z10e�txdx=1 t:Thisisaparametricformof(2.2),wherebothsidesarenowfunctionsoft.Weneedt�0inorderthate�txisintegrableovertheregionx0.Nowwebringindi erentiationundertheintegralsign.Di erentiatebothsidesof(2.3)withrespecttot,using(1.2)totreattheleftside.WeobtainZ10�xe�txdx=�1 t2; DIFFERENTIATINGUNDERTHEINTEGRALSIGN3so(2.4)Z10xe�txdx=1 t2:Di erentiatebothsidesof(2.4)withrespecttot,againusing(1.2)tohandletheleftside.WegetZ10�x2e�txdx=�2 t3:Takingoutthesignonbothsides,(2.5)Z10x2e�txdx=2 t3:Ifwecontinuetodi erentiateeachnewequationwithrespecttotafewmoretimes,weobtainZ10x3e�txdx=6 t4;Z10x4e�txdx=24 t5;andZ10x5e�txdx=120 t6:Doyouseethepattern?Itis(2.6)Z10xne�txdx=n! tn+1:Wehaveusedthepresenceoftheextravariablettogettheseequationsbyrepeatedlyapplyingd=dt.Nowspecializetto1in(2.6).WeobtainZ10xne�xdx=n!;whichisouroldfriend(2.1).Voila!Theideathatmadethisworkisintroducingaparametert,usingcalculusont,andthensettingttoaparticularvaluesoitdisappearsfromthe nalformula.Inotherwords,sometimestosolveaproblemitisusefultosolveamoregeneralproblem.Compare(2.1)to(2.6).3.AdampedsineintegralWearegoingtousedi erentiationundertheintegralsigntoproveZ10e�txsinx xdx= 2�arctantfort�0.CallthisintegralF(t)andsetf(x;t)=e�tx(sinx)=x,so(@=@t)f(x;t)=�e�txsinx.ThenF0(t)=�Z10e�tx(sinx)dx:Theintegrande�txsinx,asafunctionofx,canbeintegratedbyparts:Zeaxsinxdx=(asinx�cosx) 1+a2eax: 4KEITHCONRADApplyingthiswitha=�tandturningtheinde niteintegralintoade niteintegral,F0(t)=�Z10e�tx(sinx)dx=(tsinx+cosx) 1+t2e�tx x=1x=0:Asx!1,tsinx+cosxoscillatesalot,butinaboundedway(sincesinxandcosxareboundedfunctions),whiletheterme�txdecaysexponentiallyto0sincet�0.Sothevalueatx=1is0.ThereforeF0(t)=�Z10e�tx(sinx)dx=�1 1+t2:Weknowanexplicitantiderivativeof1=(1+t2),namelyarctant.SinceF(t)hasthesamet-derivativeas�arctant,theydi erbyaconstant:forsomenumberC,(3.1)Z10e�txsinx xdx=�arctant+Cfort�0:We'vecomputedtheintegral,uptoanadditiveconstant,without ndinganantiderivativeofe�tx(sinx)=x.TocomputeCin(3.1),lett!1onbothsides.Sincej(sinx)=xj1,theabsolutevalueoftheintegralontheleftisboundedfromabovebyR10e�txdx=1=t,sotheintegralontheleftin(3.1)tendsto0ast!1.Sincearctant!=2ast!1,equation(3.1)ast!1becomes0=� 2+C,soC==2.Feedingthisbackinto(3.1),(3.2)Z10e�txsinx xdx= 2�arctantfort�0:Ifwelett!0+in(3.2),thisequationsuggeststhat(3.3)Z10sinx xdx= 2;whichistrueanditisimportantinsignalprocessingandFourieranalysis.Itisadelicatemattertoderive(3.3)from(3.2)sincetheintegralin(3.3)isnotabsolutelyconvergent.Detailsareprovidedinanappendix.4.TheGaussianintegralTheimproperintegralformula(4.1)Z1�1e�x2=2dx=p 2isfundamentaltoprobabilitytheoryandFourieranalysis.Thefunction1 p 2e�x2=2iscalledaGaussian,and(4.1)saystheintegraloftheGaussianoverthewholereallineis1.ThephysicistLordKelvin(afterwhomtheKelvintemperaturescaleisnamed)oncewrote(4.1)ontheboardinaclassandsaid\Amathematicianisonetowhomthat[pointingattheformula]isasobviousastwicetwomakesfouristoyou."Wewillprove(4.1)usingdi erentiationundertheintegralsign.Themethodwillnotmake(4.1)asobviousas22=4.Ifyoutakefurthercoursesyoumaylearnmorenaturalderivationsof(4.1)sothattheresultreallydoesbecomeobvious.Fornow,justtrytofollowtheargumentherestep-by-step.Wearegoingtoaimnotat(4.1),butatanequivalentformulaovertherangex0:(4.2)Z10e�x2=2dx=p 2 2=r  2: DIFFERENTIATINGUNDERTHEINTEGRALSIGN5CalltheintegralontheleftI.Fort2R,setF(t)=Z10e�t2(1+x2)=2 1+x2dx:ThenF(0)=R10dx=(1+x2)==2andF(1)=0.Di erentiatingundertheintegralsign,F0(t)=Z10�te�t2(1+x2)=2dx=�te�t2=2Z10e�(tx)2=2dx:Makethesubstitutiony=tx,withdy=tdx,soF0(t)=�e�t2=2Z10e�y2=2dy=�Ie�t2=2:Forb�0,integratebothsidesfrom0tobandusetheFundamentalTheoremofCalculus:Zb0F0(t)dt=�IZb0e�t2=2dt=)F(b)�F(0)=�IZb0e�t2=2dt:Lettingb!1,0� 2=�I2=)I2= 2=)I=r  2:IlearnedthisfromMichaelRozman[12],whomodi edanideaonaMathStackexchangequestion[3],andinaslightlylesselegantformitappearedmuchearlierin[15].5.HighermomentsoftheGaussianForeveryintegern0wewanttocomputeaformulafor(5.1)Z1�1xne�x2=2dx:(IntegralsofthetypeRxnf(x)dxforn=0;1;2;:::arecalledthemomentsoff(x),so(5.1)isthen-thmomentoftheGaussian.)Whennisodd,(5.1)vanishessincexne�x2=2isanoddfunction.Whatifn=0;2;4;:::iseven?The rstcase,n=0,istheGaussianintegral(4.1):(5.2)Z1�1e�x2=2dx=p 2:Togetformulasfor(5.1)whenn6=0,wefollowthesamestrategyasourtreatmentofthefactorialintegralinSection2:stickatintotheexponentofe�x2=2andthendi erentiaterepeatedlywithrespecttot.Fort�0,replacingxwithp txin(5.2)gives(5.3)Z1�1e�tx2=2dx=p 2 p t:Di erentiatebothsidesof(5.3)withrespecttot,usingdi erentiationundertheintegralsignontheleft:Z1�1�x2 2e�tx2=2dx=�p 2 2t3=2;so(5.4)Z1�1x2e�tx2=2dx=p 2 t3=2: 6KEITHCONRADDi erentiatebothsidesof(5.4)withrespecttot.Afterremovingacommonfactorof�1=2onbothsides,weget(5.5)Z1�1x4e�tx2=2dx=3p 2 t5=2:Di erentiatingbothsidesof(5.5)withrespecttotafewmoretimes,wegetZ1�1x6e�tx2=2dx=35p 2 t7=2;Z1�1x8e�tx2=2dx=357p 2 t9=2;andZ1�1x10e�tx2=2dx=3579p 2 t11=2:Quitegenerally,whennisevenZ1�1xne�tx2=2dx=135(n�1) t(n+1)=2p 2;wherethenumeratoristheproductofthepositiveoddintegersfrom1ton�1(understoodtobetheemptyproduct1whenn=0).Inparticular,takingt=1wehavecomputed(5.1):Z1�1xne�x2=2dx=135(n�1)p 2:Asanapplicationof(5.4),wenowcompute(1 2)!:=R10x1=2e�xdx,wherethenotation(1 2)!anditsde nitionareinspiredbyEuler'sintegralformula(2.1)forn!whennisanonnegativeinteger.Usingthesubstitutionu=x1=2inR10x1=2e�xdx,wehave1 2!=Z10x1=2e�xdx=Z10ue�u2(2u)du=2Z10u2e�u2du=Z1�1u2e�u2du=p 2 23=2by(5:4)att=2=p  2:6.AcosinetransformoftheGaussianWearegoingtocomputeF(t)=Z10cos(tx)e�x2=2dx DIFFERENTIATINGUNDERTHEINTEGRALSIGN7bylookingatitst-derivative:(6.1)F0(t)=Z10�xsin(tx)e�x2=2dx:Thisisgoodfromtheviewpointofintegrationbypartssince�xe�x2=2isthederivativeofe�x2=2.Soweapplyintegrationbypartsto(6.1):u=sin(tx);dv=�xe�x2dxanddu=tcos(tx)dx;v=e�x2=2:ThenF0(t)=Z10udv=uv 10�Z10vdu=sin(tx) ex2=2 x=1x=0�tZ10cos(tx)e�x2=2dx=sin(tx) ex2=2 x=1x=0�tF(t):Asx!1,ex2=2blowsupwhilesin(tx)staysbounded,sosin(tx)=ex2=2goesto0.ThereforeF0(t)=�tF(t):Weknowthesolutionstothisdi erentialequation:constantmultiplesofe�t2=2.SoZ10cos(tx)e�x2=2dx=Ce�t2=2forsomeconstantC.To ndC,sett=0.TheleftsideisR10e�x2=2dx,whichisp =2by(4.2).TherightsideisC.ThusC=p =2,sowearedone:forallrealt,Z10cos(tx)e�x2=2dx=r  2e�t2=2:Remark6.1.IfwewanttocomputeG(t)=R10sin(tx)e�x2=2dx,withsin(tx)inplaceofcos(tx),theninplaceofF0(t)=�tF(t)wehaveG0(t)=1�tG(t),andG(0)=0.Fromthedi erentialequa-tion,(et2=2G(t))0=et2=2,soG(t)=e�t2=2Rt0ex2=2dx.SowhileR10cos(tx)e�x2=2dx=p  2e�t2=2,theintegralR10sin(tx)e�x2=2dxisimpossibletoexpressintermsofelementaryfunctions.7.TheGaussiantimesalogarithmWewillcomputeZ10(logx)e�x2dx:Integrabilityat1followsfromrapiddecayofe�x2at1,andintegrabilitynearx=0followsfromtheintegrandtherebeingnearlylogx,whichisintegrableon[0;1],sotheintegralmakessense.(ThisexamplewasbroughttomyattentionbyHaraldHelfgott.) 8KEITHCONRADWealreadyknowR10e�x2dx=p =2,buthowdowe ndtheintegralwhenafactoroflogxisinsertedintotheintegrand?Replacingxwithp xintheintegral,(7.1)Z10(logx)e�x2dx=1 4Z10logx p xe�xdx:Tocomputethislastintegral,thekeyideaisthat(d=dt)(xt)=xtlogx,sowegetafactoroflogxinanintegralafterdi erentiationundertheintegralsigniftheintegrandhasanexponentialparameter:fort��1setF(t)=Z10xte�xdx:(Thisisintegrableforxnear0sinceforsmallx,xte�xxt,whichisintegrablenear0sincet��1.)Di erentiatingbothsideswithrespecttot,F0(t)=Z10xt(logx)e�xdx;so(7.1)tellsusthenumberweareinterestedinisF0(�1=2)=4.ThefunctionF(t)iswell-knownunderadi erentname:fors�0,the�-functionatsisde nedby�(s)=Z10xs�1e�xdx;so�(s)=F(s�1).Therefore�0(s)=F0(s�1),soF0(�1=2)=4=�0(1=2)=4.Fortherestofthissectionweworkoutaformulafor�0(1=2)=4usingpropertiesofthe�-function;thereisnomoredi erentiationundertheintegralsign.Weneedtwostandardidentitiesforthe�-function:(7.2)�(s+1)=s�(s);�(s)�s+1 2=21�2sp �(2s):The rstidentityfollowsfromintegrationbyparts.Since�(1)=R10e�xdx=1,the rstidentityimplies�(n)=(n�1)!foreverypositiveintegern.Thesecondidentity,calledtheduplicationformula,issubtle.Forexample,ats=1=2itsays�(1=2)=p .Aproofoftheduplicationformulacanbefoundinmanycomplexanalysistextbooks.(Theintegralde ning�(s)makessensenotjustforreals�0,butalsoforcomplexswithRe(s)�0,andthe�-functionisusuallyregardedasafunctionofacomplex,ratherthanreal,variable.)Di erentiatingthe rstidentityin(7.2),(7.3)�0(s+1)=s�0(s)+�(s);soats=1=2(7.4)�03 2=1 2�01 2+�1 2=1 2�01 2+p =)�01 2=2�03 2�p :Di erentiatingthesecondidentityin(7.2),(7.5)�(s)�0s+1 2+�0(s)�s+1 2=21�2s(�log4)p �(2s)+21�2sp 2�0(2s):Settings=1hereandusing�(1)=�(2)=1,(7.6)�03 2+�0(1)�3 2=(�log2)p +p �0(2): DIFFERENTIATINGUNDERTHEINTEGRALSIGN9Wecompute�(3=2)bythe rstidentityin(7.2)ats=1=2:�(3=2)=(1=2)�(1=2)=p =2(wealreadycomputedthisattheendofSection5).Wecompute�0(2)by(7.3)ats=1:�0(2)=�0(1)+1.Thus(7.6)says�03 2+�0(1)p  2=(�log2)p +p (�0(1)+1)=)�03 2=p �log2+�0(1) 2+p :Feedingthisformulafor�0(3=2)into(7.4),�01 2=p (�2log2+�0(1)):Itturnsoutthat�0(1)=� ,where :577isEuler'sconstant.Thus,atlast,Z10(logx)e�x2dx=�0(1=2) 4=�p  4(2log2+ ):8.Logsinthedenominator,partIConsiderthefollowingintegralover[0;1],wheret�0:Z10xt�1 logxdx:Since1=logx!0asx!0+,theintegrandvanishesatx=0.Asx!1�,(xt�1)=logx!t.Thereforewhentis xedtheintegrandisacontinuousfunctionofxon[0;1],sotheintegralisnotanimproperintegral.Thet-derivativeofthisintegralisZ10xtlogx logxdx=Z10xtdx=1 t+1;whichwerecognizeasthet-derivativeoflog(t+1).ThereforeZ10xt�1 logxdx=log(t+1)+CforsomeC.To ndC,lett!0+.Ontherightside,log(1+t)tendsto0.Ontheleftside,theintegrandtendsto0:j(xt�1)=logxj=j(etlogx�1)=logxjtbecausejea�1jjajwhena0.Thereforetheintegralonthelefttendsto0ast!0+.SoC=0,whichimplies(8.1)Z10xt�1 logxdx=log(t+1)forallt�0,andit'sobviouslyalsotruefort=0.Anotherwaytocomputethisintegralistowritext=etlogxasapowerseriesandintegratetermbyterm,whichisvalidfor�1t1.Underthechangeofvariablesx=e�y,(8.1)becomes(8.2)Z10e�y�e�(t+1)ydy y=log(t+1): 10KEITHCONRAD9.Logsinthedenominator,partIIWenowconsidertheintegralF(t)=Z12dx xtlogxfort�1.TheintegralconvergesbycomparisonwithR12dx=xt.Weknowthat\att=1"theintegraldivergesto1:Z12dx xlogx=limb!1Zb2dx xlogx=limb!1loglogx b2=limb!1loglogb�loglog2=1:Soweexpectthatast!1+,F(t)shouldblowup.Buthowdoesitblowup?ByanalyzingF0(t)andthenintegratingback,wearegoingtoshowF(t)behavesessentiallylike�log(t�1)ast!1+.Usingdi erentiationundertheintegralsign,fort�1F0(t)=Z12@ @t1 xtlogxdx=Z12x�t(�logx) logxdx=�Z12dx xt=�x�t+1 �t+1 x=1x=2=21�t 1�t:Wewanttoboundthisderivativefromaboveandbelowwhent�1.ThenwewillintegratetogetboundsonthesizeofF(t).Fort�1,thedi erence1�tisnegative,so21�t1.Dividingbothsidesofthisby1�t,whichisnegative,reversesthesenseoftheinequalityandgives21�t 1�t�1 1�t:ThisisalowerboundonF0(t).TogetanupperboundonF0(t),wewanttousealowerboundon21�t.Sinceeaa+1foralla(thegraphofy=exliesonoraboveitstangentlineatx=0,whichisy=x+1),2x=exlog2(log2)x+1forallx.Takingx=1�t,(9.1)21�t(log2)(1�t)+1:Whent�1,1�tisnegative,sodividing(9.1)by1�treversesthesenseoftheinequality:21�t 1�tlog2+1 1�t: DIFFERENTIATINGUNDERTHEINTEGRALSIGN11ThisisanupperboundonF0(t).PuttingtheupperandlowerboundsonF0(t)together,(9.2)1 1�tF0(t)log2+1 1�tforallt�1.WeareconcernedwiththebehaviorofF(t)ast!1+.Let'sintegrate(9.2)fromato2,where1a2:Z2adt 1�tZ2aF0(t)dtZ2alog2+1 1�tdt:UsingtheFundamentalTheoremofCalculus,�log(t�1) 2aF(t) 2a((log2)t�log(t�1)) 2a;solog(a�1)F(2)�F(a)(log2)(2�a)+log(a�1):ManipulatingtogetinequalitiesonF(a),wehave(log2)(a�2)�log(a�1)+F(2)F(a)�log(a�1)+F(2)Sincea�2&#x]TJ/;༷ ;.9;‘ ;&#xTf 1;.51; 0 ;&#xTd [;�1for1a2,(log2)(a�2)isgreaterthan�log2.Thisgivesthebounds�log(a�1)+F(2)�log2F(a)�log(a�1)+F(2)Writingaast,weget�log(t�1)+F(2)�log2F(t)�log(t�1)+F(2);soF(t)isaboundeddistancefrom�log(t�1)when1t2.Inparticular,F(t)!1ast!1+.10.AtrigonometricintegralForpositivenumbersaandb,thearithmetic-geometricmeaninequalitysays(a+b)=2p ab(withequalityifandonlyifa=b).Let'siteratethetwotypesofmeans:fork0,de nefakgandfbkgbya0=a,b0=b,andak=ak�1+bk�1 2;bk=p ak�1bk�1fork1.Example10.1.Ifa0=1andb0=2,thenTable2givesakandbkto16digitsafterthedecimalpoint.Noticehowrapidlytheyaregettingclosetoeachother!k ak bk 0 1 21 1.5 1.41421356237309502 1.4571067811865475 1.45647531512197023 1.4567910481542588 1.45679101393955494 1.4567910310469069 1.4567910310469068Table1.Iterationofarithmeticandgeometricmeans. 12KEITHCONRADGaussshowedthatforeverychoiceofaandb,thesequencesfakgandfbkgconvergeveryrapidlytoacommonlimit,whichhecalledthearithmetic-geometricmeanofxandyandwrotethisasM(a;b).Forexample,M(1;2)1:456791031046906.Gaussdiscoveredanintegralformulaforthereciprocal1=M(a;b):1 M(a;b)=2 Z=20dx p a2cos2x+b2sin2x:Thereisnoelementaryformulaforthisintegral,butifwechangetheexponent1=2inthesquareroottoapositiveintegernthenwecanworkoutalltheintegralsFn(a;b)=2 Z=20dx (a2cos2x+b2sin2x)nusingrepeateddi erentiationundertheintegralsignwithrespecttobothaandb.(Thisexample,withadi erentnormalizationandnocontextforwheretheintegralcomesfrom,isExample4ontheWikipediapagefortheLeibnizintegralrule.)Forn=1wecandoadirectintegration:F1(a;b)=2 Z=20dx a2cos2x+b2sin2x=2 Z=20sec2x a2+b2tan2xdx=2 Z10du a2+b2u2whereu=tanx=2 a2Z10du 1+(b=a)2u2=2 abZ10dv 1+v2wherev=(b=a)u=1 ab:Nowlet'sdi erentiateF1(a;b)withrespecttoaandwithrespecttob,bothbyitsintegralde nitionandbytheformulawejustcomputedforit:@F1 @a=2 Z=20�2acos2x (a2cos2x+b2sin2x)2dx;@F1 @a=�1 a2band@F1 @b=2 Z=20�2bsin2x (a2cos2x+b2sin2x)2dx;@F1 @b=�1 ab2:Sincesin2x+cos2x=1,byalittlealgebrawecangetaformulaforF2(a;b):F2(a;b)=2 Z=20dx (a2cos2x+b2sin2x)2=�1 2a@F1 @a�1 2b@F1 @b=1 2a3b+1 2ab3=a2+b2 2a3b3: DIFFERENTIATINGUNDERTHEINTEGRALSIGN13WecangetarecursionexpressingFn(a;b)intermsof@Fn�1=@aand@Fn�1=@bingeneral:forn2,@Fn�1 @a=2 Z=20�2(n�1)acos2x (a2cos2x+b2sin2x)ndx;@Fn�1 @b=2 Z=20�2(n�1)bsin2x (a2cos2x+b2sin2x)ndx;soFn(a;b)=�1 2(n�1)a@Fn�1 @a�1 2(n�1)b@Fn�1 @b=�1 2(n�1)1 a@Fn�1 @a+1 b@Fn�1 @b:AfewsamplecalculationsofFn(a;b)usingthis,startingfromF1(a;b)=1=(ab),areF2(a;b)=a2+b2 2a3b3;F3(a;b)=3a4+2a2b2+3b4 6a5b5;F4(a;b)=5a6+3a4b2+3a2b4+5b6 12a7b7:11.SmoothlydividingbytLeth(t)beanin nitelydi erentiablefunctionforallrealtsuchthath(0)=0.Theratioh(t)=tmakessensefort6=0,anditalsocanbegivenareasonablemeaningatt=0:fromtheveryde nitionofthederivative,whent!0wehaveh(t) t=h(t)�h(0) t�0!h0(0):Thereforethefunctionr(t)=(h(t)=t;ift6=0;h0(0);ift=0iscontinuousforallt.Wecanseeimmediatelyfromthede nitionofr(t)thatitisbetterthancontinuouswhent6=0:itisin nitelydi erentiablewhent6=0.Thequestionwewanttoaddressisthis:isr(t)in nitelydi erentiableatt=0too?Ifh(t)hasapowerseriesrepresentationaroundt=0,thenitiseasytoshowthatr(t)isin nitelydi erentiableatt=0byworkingwiththeseriesforh(t).Indeed,writeh(t)=c1t+c2t2+c3t3+forallsmallt.Herec1=h0(0),c2=h00(0)=2!andsoon.Forsmallt6=0,wedividebytandget(11.1)r(t)=c1+c2t+c3t3+;whichisapowerseriesrepresentationforr(t)forallsmallt6=0.Thevalueoftherightsideof(11.1)att=0isc1=h0(0),whichisalsothede nedvalueofr(0),so(11.1)isvalidforallsmallx(includingt=0).Thereforer(t)hasapowerseriesrepresentationaround0(it'sjustthepowerseriesforh(t)at0dividedbyt).Sincefunctionswithpowerseriesrepresentationsaroundapointarein nitelydi erentiableatthepoint,r(t)isin nitelydi erentiableatt=0.However,thisisanincompleteanswertoourquestionaboutthein nitedi erentiabilityofr(t)att=0becauseweknowbythekeyexampleofe�1=t2(att=0)thatafunctioncanbein nitelydi erentiableatapointwithouthavingapowerseriesrepresentationatthepoint.Howarewegoingtoshowr(t)=h(t)=tisin nitelydi erentiableatt=0ifwedon'thaveapowerseriestohelpusout?Mightthereactuallybeacounterexample?Thesolutionistowriteh(t)inaverycleverwayusingdi erentiationundertheintegralsign.Startwithh(t)=Zt0h0(u)du: 14KEITHCONRAD(Thisiscorrectsinceh(0)=0.)Fort6=0,introducethechangeofvariablesu=tx,sodu=tdx.Attheboundary,ifu=0thenx=0.Ifu=tthenx=1(wecandividetheequationt=txbytbecauset6=0).Thereforeh(t)=Z10h0(tx)tdx=tZ10h0(tx)dx:Dividingbytwhent6=0,wegetr(t)=h(t) t=Z10h0(tx)dx:Theleftandrightsidesdon'thavetinthedenominator.Aretheyequalatt=0too?Theleftsideatt=0isr(0)=h0(0).TherightsideisR10h0(0)dx=h0(0)too,so(11.2)r(t)=Z10h0(tx)dxforallt,includingt=0.Thisisaformulaforh(t)=twherethereisnolongeratbeingdivided!Nowwe'resettousedi erentiationundertheintegralsign.Thewaywehavesetthingsuphere,wewanttodi erentiatewithrespecttot;theintegrationvariableontherightisx.Wecanusedi erentiationundertheintegralsignon(11.2)whentheintegrandisdi erentiable.Sincetheintegrandisin nitelydi erentiable,r(t)isin nitelydi erentiable!Explicitly,r0(t)=Z10xh00(tx)dxandr00(t)=Z10x2h000(tx)dxandmoregenerallyr(k)(t)=Z10xkh(k+1)(tx)dx:Inparticular,r(k)(0)=R10xkh(k+1)(0)dx=h(k+1)(0) k+1.12.CounterexamplesandjustificationWehaveseenmanyexampleswheredi erentiationundertheintegralsigncanbecarriedoutwithinterestingresults,butwehavenotactuallystatedconditionsunderwhich(1.2)isvalid.Somethingdoesneedtobechecked.In[14],anincorrectuseofdi erentiationundertheintegralsignduetoCauchyisdiscussed,whereadivergentintegralisevaluatedasa niteexpression.Herearetwootherexampleswheredi erentiationundertheintegralsigndoesnotwork.Example12.1.Itispointedoutin[9,Example6]thattheformulaZ10sinx xdx= 2;whichwediscussedattheendofSection3,leadstoanerroneousinstanceofdi erentiationundertheintegralsign.Rewritetheformulaas(12.1)Z10sin(ty) ydy= 2 DIFFERENTIATINGUNDERTHEINTEGRALSIGN15fort�0bythechangeofvariablesx=ty.Thendi erentiationundertheintegralsignimpliesZ10cos(ty)dy=0;buttheleftsidedoesn'tmakesense.Thenextexampleshowsthatevenifbothsidesof(1.2)makesense,theyneednotbeequal.Example12.2.Forrealnumbersxandt,letf(x;t)=8:xt3 (x2+t2)2;ifx6=0ort6=0;0;ifx=0andt=0:LetF(t)=Z10f(x;t)dx:Forinstance,F(0)=R10f(x;0)dx=R100dx=0.Whent6=0,F(t)=Z10xt3 (x2+t2)2dx=Z1+t2t2t3 2u2du(whereu=x2+t2)=�t3 2u u=1+t2u=t2=�t3 2(1+t2)+t3 2t2=t 2(1+t2):Thisformulaalsoworksatt=0,soF(t)=t=(2(1+t2))forallt.ThereforeF(t)isdi erentiableandF0(t)=1�t2 2(1+t2)2forallt.Inparticular,F0(0)=1 2.Nowwecompute@ @tf(x;t)andthenR10@ @tf(x;t)dx.Sincef(0;t)=0forallt,f(0;t)isdi eren-tiableintand@ @tf(0;t)=0.Forx6=0,f(x;t)isdi erentiableintand@ @tf(x;t)=(x2+t2)2(3xt2)�xt32(x2+t2)2t (x2+t2)4=xt2(x2+t2)(3(x2+t2)�4t2) (x2+t2)4=xt2(3x2�t2) (x2+t2)3:Combiningbothcases(x=0andx6=0),(12.2)@ @tf(x;t)=(xt2(3x2�t2) (x2+t2)3;ifx6=0;0;ifx=0: 16KEITHCONRADInparticular@ @t t=0f(x;t)=0.Thereforeatt=0theleftsideofthe\formula"d dtZ10f(x;t)dx=Z10@ @tf(x;t)dx:isF0(0)=1=2andtherightsideisR10@ @t t=0f(x;t)dx=0.Thetwosidesareunequal!Theprobleminthisexampleisthat@ @tf(x;t)isnotacontinuousfunctionof(x;t).Indeed,thedenominatorintheformulain(12.2)is(x2+t2)3,whichhasaproblemnear(0;0).Speci cally,whilethisderivativevanishesat(0;0),ifwelet(x;t)!(0;0)alongthelinex=t,thenonthisline@ @tf(x;t)hasthevalue1=(4x),whichdoesnottendto0as(x;t)!(0;0).Theorem12.3.Theequationd dtZbaf(x;t)dx=Zba@ @tf(x;t)dx;whereacouldbe�1andbcouldbe1,isvalidatarealnumbert=t0inthesensethatbothsidesexistandareequal,providedthefollowingtwoconditionshold:f(x;t)and@ @tf(x;t)arecontinuousfunctionsoftwovariableswhenxisintherangeofintegrationandtisinsomeintervalaroundt0,fortinsomeintervalaroundt0thereareupperboundsjf(x;t)jA(x)andj@ @tf(x;t)jB(x),bothboundsbeingindependentoft,suchthatRbaA(x)dxandRbaB(x)dxexist.Proof.See[10,pp.337{339].Iftheintervalofintegrationisin nite,RbaA(x)dxandRbaB(x)dxareimproper.InTable2weincludechoicesforA(x)andB(x)forthefunctionswehavetreated.Sincethecalculationofaderivativeatapointonlydependsonanintervalaroundthepoint,wehavereplacedat-rangesuchast�0withtc�0insomecasestoobtainchoicesforA(x)andB(x).Section f(x;t) xrange trange twewant A(x) B(x) 2 xne�tx [0;1) tc�0 1 xne�cx xn+1e�cx3 e�txsinx x (0;1) tc�0 0 e�cx e�cx4 1 1+x2e�t2(1+x2)=2 [0;1) tc�0 allt0 1 1+x2 1 p ee�c2x2=25 xne�tx2 R tc�0 1 xne�cx2 xn+2e�cx26 cos(tx)e�x2=2 [0;1) R allt e�x2=2 jxje�x2=27 xt�1e�x (0;1) 0tc 1/2,1,3/2,2 xc�1e�x xc�1jlogxje�x8 xt�1 logx (0;1] 0tc 1 1�xc logx 19 1 xtlogx [2;1) tc�1 t�1 1 x2logx 1 xc11 xkh(k+1)(tx) [0;1] jtjc 0 maxjyjcjh(k+1)(y)j maxjyjcjh(k+2)(y)jTable2.SummaryWedidnotputthefunctionfromSection10inthetablesinceitwouldmakethewidthtoolonganditdependsontwoparameters.Puttingtheparameterintothecoecientofcos2xinSection10, DIFFERENTIATINGUNDERTHEINTEGRALSIGN17wecantakef(x;t)=1=(t2cos2x+b2sin2x)nforx2[0;=2],0ctc0(thatis,keeptboundedawayfrom0and1),A(x)=1=(c2cos2x+b2sin2x)nandB(x)=2c0n=(c2cos2x+b2sin2x)n+1.Corollary12.4.Ifa(t)andb(t)arebothdi erentiable,thend dtZb(t)a(t)f(x;t)dx=Zb(t)a(t)@ @tf(x;t)dx+f(b(t);t)b0(t)�f(a(t);t)a0(t)ifthefollowingconditionsaresatis ed:thereare andc1c2suchthatf(x;t)and@ @tf(x;t)arecontinuouson[ ; ](c1;c2),wehavea(t)2[ ; ]andb(t)2[ ; ]forallt2(c1;c2),thereareupperboundsjf(x;t)jA(x)andj@ @tf(x;t)jB(x)for(x;t)2[ ; ](c1;c2)suchthatR A(x)dxandR B(x)dxexist.Proof.ThisisaconsequenceofTheorem12.3andthechainruleformultivariablefunctions.SetafunctionofthreevariablesI(t;a;b)=Zbaf(x;t)dxfor(t;a;b)2(c1;c2)[ ; ][ ; ].(Hereaandbarenotfunctionsoft.)Then(12.3)@I @t(t;a;b)=Zba@ @tf(x;t)dx;@I @a(t;a;b)=�f(a;t);@I @b(t;a;b)=f(b;t);wherethe rstformulafollowsfromTheorem12.3(itshypothesesaresatis edforeachaandbin[ ; ])andthesecondandthirdformulasaretheFundamentalTheoremofCalculus.Fordi erentiablefunctionsa(t)andb(t)withvaluesin[ ; ]forc1tc2,bythechainruled dtZb(t)a(t)f(x;t)dx=d dtI(t;a(t);b(t))=@I @t(t;a(t);b(t))dt dt+@I @a(t;a(t);b(t))da dt+@I @b(t;a(t);b(t))db dt=Zb(t)a(t)@f @t(x;t)dx�f(a(t);t)a0(t)+f(b(t);t)b0(t)by(12:3):Aversionofdi erentiationundertheintegralsignfortacomplexvariableisin[11,pp.392{393].Example12.5.ForaparametricintegralRtaf(x;t)dx,whereais xed,Corollary12.4tellsusthat(12.4)d dtZtaf(x;t)dx=Zta@ @tf(x;t)dx+f(t;t)providedthehypothesesofthecorollaryaresatis ed:(i)thereare andc1c2suchthatfand@f=@tarecontinuousfor(x;t)2[ ; ](c1;c2),(ii) a and(c1;c2)[ ; ],and(iii)thereareboundsjf(x;t)jA(x)andj@ @tf(x;t)jB(x)for(x;t)2[ ; ](c1;c2)suchthattheintegralsR A(x)dxandR B(x)dxbothexist.Let'sapplythistotheintegralF(t)=Zt0log(1+tx) 1+x2dx 18KEITHCONRADwheret�0.Heref(x;t)=log(1+tx)=(1+x2)and@ @tf(x;t)=x (1+tx)(1+x2).Fixc�0anduse =c1=0and =c2=c.Thenwecanjustify(12.4)for(x;t)2[0;c](0;c)usingA(x)=log(1+c2)=(1+x2)andB(x)=c=(1+x2),soF0(t)=Zt0x (1+tx)(1+x2)dx+log(1+t2) 1+t2=Zt01 1+t2�t 1+tx+t+x 1+x2dx+log(1+t2) 1+t2:Splittinguptheintegralintothreeterms,F0(t)=�1 1+t2log(1+tx)+t 1+t2arctan(x)+log(1+x2) 2(1+t2) t0+log(1+t2) 1+t2=�log(1+t2) 1+t2+tarctan(t) 1+t2+log(1+t2) 2(1+t2)+log(1+t2) 1+t2=tarctan(t) 1+t2+log(1+t2) 2(1+t2):ClearlyF(0)=0,sobytheFundamentalTheoremofCalculusF(t)=Zt0F0(y)dy=Zt0yarctan(y) 1+y2+log(1+y2) 2(1+y2)dy:Usingintegrationbypartsonthe rstintegrandwithu=arctan(y)anddv=y 1+y2dy,F(t)=uv t0�Zt0vdu+Zt0log(1+y2) 2(1+y2)dy=arctan(y)log(1+y2) 2 t0�Zt0log(1+y2) 2(1+y2)dy+Zt0log(1+y2) 2(1+y2)dy=1 2arctan(t)log(1+t2):Thusfor0tcwehaveZt0log(1+tx) 1+x2dx=1 2arctan(t)log(1+t2):Sincecwasarbitrary,thisequationholdsforallt�0(andtriviallyatt=0too).Settingt=1,Z10log(1+x) 1+x2dx=1 2arctan(1)log2=log2 8:13.TheFundamentalTheoremofAlgebraBydi erentiatingundertheintegralsignwewilldeducethefundamentaltheoremofalgebra:anonconstantpolynomialp(z)withcoecientsinChasarootinC.TheproofisduetoSchep[13].Arguingbycontradiction,assumep(z)6=0forallz2C.Forr0,considerthefollowingintegralaroundacircleofradiusrcenteredattheorigin:I(r)=Z20d p(rei):Thisintegralmakessensesincethedenominatorisnever0,so1=p(z)iscontinuousonC.Letf(;r)=1=p(rei),soI(r)=R20f(;r)d. DIFFERENTIATINGUNDERTHEINTEGRALSIGN19WewillprovethreepropertiesofI(r):(1)Theorem12.3canbeappliedtoI(r)forr�0,(2)I(r)!0asr!1,(3)I(r)!I(0)asr!0+(continuityatr=0).Takingtheseforgranted,let'sseehowacontradictionoccurs.Forr�0,I0(r)=Z20@ @rf(;r)d=Z20�p0(rei)ei p(rei)2d:Since@ @f(;r)=�p0(rei) p(rei)2irei=ir@ @rf(;r);forr�0wehaveI0(r)=Z20@ @rf(;r)d=Z201 ir@ @f(;r)d=1 irf(;r) =2=0=1 ir1 p(r)�1 p(r)=0:ThusI(r)isconstantforr�0.SinceI(r)!0asr!1,theconstantiszero:I(r)=0forr�0.SinceI(r)!I(0)asr!0+wegetI(0)=0,whichisfalsesinceI(0)=2=p(0)6=0.ItremainstoprovethethreepropertiesofI(r).(1)Theorem12.3canbeappliedtoI(r)forr�0 :Sincep(z)andp0(z)arebothcontinuousonC,thefunctionsf(;r)and(@=@r)f(;r)arecontinuousfor2[0;2]andallr0.Thiscon rmsthe rstconditioninTheorem12.3.Foreachr0�0thesetf(;r):2[0;2];r2[0;2r0]isclosedandbounded,sothefunctionsf(;r)and(@=@r)f(;r)arebothboundedabovebyaconstant(independentofrand)onthisset.Therangeofintegration[0;2]is nite,sothesecondconditioninTheorem12.3issatis edusingconstantsforA()andB().(2)I(r)!0asr!1 :Letp(z)haveleadingtermczd,withd=degp(z)1.Asr!1,jp(rei)j=jreijd!jcj�0,soforalllargerwehavejp(rei)jjcjrd=2.Forsuchlarger,jI(r)jZ20d jp(rei)jZ20d jcjrd=2=4 jcjrd;andtheupperboundtendsto0asr!1sinced�0,soI(r)!0asr!1.(3)I(r)!I(0)asr!0+ :Forr�0,(13.1)I(r)�I(0)=Z201 p(rei)�1 p(0)d=)jI(r)�I(0)jZ20 1 p(rei)�1 p(0) d:Since1=p(z)iscontinuousat0,for"�0thereis�0suchthatjzj)j1=p(z)�1=p(0)j".Thereforeif0r,(13.1)impliesjI(r)�I(0)jR20"d=2".14.AnexampleneedingachangeofvariablesOurnextexampleistakenfrom[1,pp.78,84].Forallt2R,wewillshowbydi erentiationundertheintegralsignthat(14.1)ZRcos(tx) 1+x2dx=e�jtj:Forexample,takingt=1,ZRcosx 1+x2dx= e: 20KEITHCONRADIn(14.1),setf(x;t)=cos(tx)=(1+x2).Sincef(x;t)iscontinuousandjf(x;t)j1=(1+x2),theintegralin(14.1)existsforallt.Thefunctione�jtjisnotdi erentiableatt=0,soweshouldn'texpecttobeabletoprove(14.1)att=0usingdi erentiationundertheintegralsign;thisspecialcasecanbetreatedwithelementarycalculus:ZRdx 1+x2=arctanx 1�1=:Theintegralin(14.1)isanevenfunctionoft,sotocomputeitfort6=0itsucestotreatthecaset�0.2LetF(t)=ZRcos(tx) 1+x2dx:IfwetrytocomputeF0(t)fort�0usingdi erentiationundertheintegralsign,weget(14.2)F0(t)?=ZR@ @tcos(tx) 1+x2dx=�ZRxsin(tx) 1+x2dx:Unfortunately,thereisnoupperboundj@ @tf(x;t)jB(x)thatjusti esdi erentiatingF(t)undertheintegralsign(orevenjusti esthatF(t)isdi erentiable).Indeed,whenxisnearalargeoddmultipleof(=2)=t,theintegrandin(14.2)hasvaluesthatareapproximatelyx=(1+x2)1=x,whichisnotintegrableforlargex.Thatdoesnotmean(14.2)isactuallyfalse,althoughifweweren'talreadytoldtheanswerontherightsideof(14.1)thenwemightbesuspiciousaboutwhethertheintegralisdi erentiableforallt�0;afterall,youcan'teasilytellfromtheintegralthatitisnotdi erentiableatt=0.Havingalreadyraisedsuspicionsabout(14.2),wecangetsomethingreallycrazyifwedi erentiateundertheintegralsignasecondtime:F00(t)?=�ZRx2cos(tx) 1+x2dx:Ifthismadesensethen(14.3)F00(t)�F(t)=�ZR(x2+1)cos(tx) 1+x2dx=�ZRcos(tx)dx=???:Allisnotlost!Let'smakeachangeofvariables.Fixingt�0,sety=tx,sody=tdxandF(t)=ZRcosy 1+y2=t2dy t=ZRtcosy t2+y2dy:Thisnewintegralwillbeaccessibletodi erentiationundertheintegralsign.(AlthoughthenewintegralisanoddfunctionoftwhileF(t)isanevenfunctionoft,thereisnocontradictionbecausethisnewintegralwasderivedonlyfort�0.)Fixc0�c�0.Fort2(c;c0),theintegrandinZRtcosy t2+y2dyisboundedaboveinabsolutevaluebyt=(t2+y2)c0=(c2+y2),whichisindependentoftandintegrableoverR.Thet-partialderivativeoftheintegrandis(y2�t2)(cosy)=(t2+y2)2,whichisboundedaboveinabsolutevalueby(y2+t2)=(t2+y2)2=1=(t2+y2)1=(c2+y2),whichis 2Areaderwhoknowscomplexanalysiscanderive(14.1)fort�0bytheresiduetheorem,viewingcos(tx)astherealpartofeitx. DIFFERENTIATINGUNDERTHEINTEGRALSIGN21independentoftandintegrableoverR.Thisjusti estheusedi erentiationundertheintegralsignaccordingtoTheorem12.3:forctc0,andhenceforallt&#x-278;0sinceweneverspeci edcorc0,F0(t)=ZR@ @ttcosy t2+y2dy=ZRy2�t2 (t2+y2)2cosydy:WewanttocomputeF00(t)usingdi erentiationundertheintegralsign.For0ctc0,thet-partialderivativeoftheintegrandforF0(t)isboundedaboveinabsolutevaluebyafunctionofythatisindependentoftandintegrableoverR(exercise),soforallt&#x-287;0wehaveF00(t)=ZR@2 @t2tcosy t2+y2dy=ZR@2 @t2t t2+y2cosydy:Itturnsoutthat(@2=@t2)(t=(t2+y2))=�(@2=@y2)(t=(t2+y2)),soF00(t)=�ZR@2 @y2t t2+y2cosydy:UsingintegrationbypartsonthisformulaforF00(t)twice(startingwithu=�cosyanddv=(@2=@y2)(t=(t2+y2)),weobtainF00(t)=�ZR@ @yt t2+y2sinydy=ZRt t2+y2cosydy=F(t):TheequationF00(t)=F(t)isasecondorderlinearODEwhosegeneralsolutionisaet+be�t,so(14.4)ZRcos(tx) 1+x2dx=aet+be�tforallt�0andsomerealconstantsaandb.Todetermineaandbwelookatthebehavioroftheintegralin(14.4)ast!0+andast!1.Ast!0+,theintegrandin(14.4)tendspointwiseto1=(1+x2),soweexpecttheintegraltendstoRRdx=(1+x2)=ast!0+.Tojustifythis,wewillboundtheabsolutevalueofthedi erence ZRcos(tx) 1+x2dx�ZRdx 1+x2 ZRjcos(tx)�1j 1+x2dxbyanexpressionthatisarbitrarilysmallast!0+.ForN�0,breakuptheintegraloverRintotheregionsjxjNandjxjN.WehaveZRjcos(tx)�1j 1+x2dxZjxjNjcos(tx)�1j 1+x2dx+ZjxjN2 1+x2dxZjxjNtjxj 1+x2dx+ZjxjN2 1+x2dx=tZjxjNjxj 1+x2dx+4 2�arctanN:TakingNsucientlylarge,wecanmake=2�arctanNassmallaswewish,andafterdoingthatwecanmakethe rsttermassmallaswewishbytakingtsucientlysmall.Returningto(14.4),lettingt!0+weobtain=a+b,soforallt�0,(14.5)ZRcos(tx) 1+x2dx=aet+(�a)e�t: 22KEITHCONRADNowlett!1in(14.5).Theintegraltendsto0bytheRiemann{LebesguelemmafromFourieranalysis,althoughwecanexplainthisconcretelyinourspecialcase:usingintegrationbypartswithu=1=(1+x2)anddv=cos(tx)dx,wegetZRcos(tx) 1+x2dx=1 tZR2xsin(tx) (1+x2)2dx:Theabsolutevalueofthetermontherightisboundedabovebyaconstantdividedbyt,whichtendsto0ast!1.Thereforeaet+(�a)e�t!0ast!1.Thisforcesa=0,whichcompletestheproofthatF(t)=e�tfort�0.15.Exercises1.Fort�0,showbycalculusthatZ10dx x2+t2= 2tandthenprovebydi erentiationundertheintegralsignthatZ10dx (x2+t2)2= 4t3,Z10dx (x2+t2)3=3 16t5,andZ10dx (x2+t2)n=2n�2n�1 (2t)2n�1foralln1.2.StartingfromtheformulaZRcos(tx) 1+x2dx= etin(14.1)fort�0,makeachangeofvariablesandthendi erentiateundertheintegralsigntoproveZRcosx (x2+t2)2dx=(t+1) 2t3etift�0.3.FromtheformulaZ10e�txsinx xdx= 2�arctantfort�0,inSection3,useachangeofvariablestoobtainaformulaforZ10e�axsin(bx) xdxwhenaandbarepositive.Thenusedif-ferentiationundertheintegralsignwithrespecttobto ndaformulaforZ10e�axcos(bx)dxwhenaandbarepositive.(Di erentiationundertheintegralsignwithrespecttoawillproduceaformulaforZ10e�axsin(bx)dx,butthatwouldbecircularinourapproachsinceweusedthatintegralinourderivationoftheformulaforZ10e�txsinx xdxinSection3.)4.BytheformulaZ10e�txsinx xdx= 2�arctantfort�0,letx=ayfora�0toseeZ10e�taysin(ay) ydy= 2�arctant;sotheintegralontheleftisindependentofaandthushasa-derivative0.Di erentiationundertheintegralsign,withrespecttoa,impliesZ10e�tay(cos(ay)�tsin(ay))dy=0:Verifythatthisapplicationofdi erentiationundertheintegralsignisvalidwhena�0andt�0.Whathappensift=0?5.ShowZ10sin(tx) x(x2+1)dx= 2(1�e�t)fort�0byjustifyingdi erentiationundertheintegralsignandusing(14.1). DIFFERENTIATINGUNDERTHEINTEGRALSIGN236.ProveZ10e�txcosx�1 xdx=logt p 1+t2fort�0.Whathappenstotheintegralast!0+?7.ProveZ10log(1+t2x2) 1+x2dx=log(1+t)fort�0(itisobviousfort=0).Thendeduce,fora�0andb�0,Z10log(1+a2x2) b2+x2dx=log(1+ab) b:8.ProveZ10(e�x�e�tx)dx x=logtfort�0byjustifyingdi erentiationundertheintegralsign.Thisis(8.2)fort��1.DeducethatZ10(e�ax�e�bx)dx x=log(b=a)fora�0andb�0.9.ProveZ10e�1 2x2�t2 2x2dx=r  2e�jtjforalltbyjustifyingdi erentiationundertheintegralsignfort�0.3(AsinSection14,theintegralisnotdi erentiableatt=0.)DeducethatZ10e�ax2�b=x2dx=p  2p ae�2p abfora�0andb�0.(Hint:LetF(t)betheintegral.Usedi erentiationundertheintegralsignandachangeofvariablestoshowF0(t)=�F(t)ift�0.)10.Incalculustextbooks,formulasfortheinde niteintegralsZxnsinxdxandZxncosxdxarederivedrecursivelyusingintegrationbyparts.Findformulasfortheseintegralswhenn=1;2;3;4usingdi erentiationundertheintegralsignstartingwiththeformulasZcos(tx)dx=sin(tx) t;Zsin(tx)dx=�cos(tx) tfort�0.11.Ifyouarefamiliarwithintegrationofcomplex-valuedfunctions,showfory2RthatZ1�1e�(x+iy)2dx=p 2:Inotherwords,showtheintegralontheleftsideisindependentofy.(Hint:Usedi eren-tiationundertheintegralsigntocomputethey-derivativeoftheleftside.) 3ThisexamplewasbroughttomyattentionbyGregoryMarkowsky.TheearliestreferenceforitthatIknowisacalculustextbook[2,p.106{107]from1888. 24KEITHCONRADAppendixA.JustifyingpassagetothelimitinasineintegralInSection3wederivedtheequation(A.1)Z10e�txsinx xdx= 2�arctantfort�0;whichbynaivepassagetothelimitast!0+suggeststhat(A.2)Z10sinx xdx= 2:Toprove(A.2)iscorrect,wewillshowR10sinx xdxexistsandthenshowthedi erence(A.3)Z10sinx xdx�Z10e�txsinx xdx=Z10(1�e�tx)sinx xdxtendsto0ast!0+.Thekeyinbothcasesisalternatingseries.Ontheinterval[k;(k+1)],wherekisaninteger,wecanwritesinx=(�1)kjsinxj,soconvergenceofR10sinx xdx=limb!1Rb0sinx xdxisequivalenttoconvergenceoftheseriesXk0Z(k+1)ksinx xdx=Xk0(�1)kZ(k+1)kjsinxj xdx:Thisisanalternatingseriesinwhichthetermsak=R(k+1)kjsinxj xdxaremonotonicallydecreasing:ak+1=Z(k+2)(k+1)jsinxj xdx=Z(k+1)kjsin(x+)j x+dx=Z(k+1)kjsinxj x+dxak:On[k;(k+1)]wehave01=jxj1=(k)andtheintervalhaslength,soak=(k)=1=kfork1.Thusak!0,soR10sinx xdx=Pk0(�1)kakconverges.Toshowtherightsideof(A.3)tendsto0ast!0+,wewriteitasanalternatingseries.Breakinguptheintervalofintegration[0;1)intoaunionofintervals[k;(k+1)]fork0,(A.4)Z10(1�e�tx)sinx xdx=Xk0(�1)kIk(t);whereIk(t)=Z(k+1)k(1�e�tx)jsinxj xdx:Since1�e�tx�0fort�0andx�0,theseriesPk0(�1)kIk(t)isalternating.Theupperbound1�e�tx1tellsusIk(t)1=kfork1bythesamereasoningweusedonakabove,soIk(t)!0ask!1.ToshowthetermsIk(t)aremonotonicallydecreasingwithk,setthisupastheinequality(A.5)Ik(t)�Ik+1(t)&#x]TJ/;༕ ;.9;‘ ;&#xTf 1;.92; 0 ;&#xTd [;0fort&#x]TJ/;༕ ;.9;‘ ;&#xTf 1;.92; 0 ;&#xTd [;0:EachIk(t)isafunctionoftforallt,notjustt&#x]TJ/;༕ ;.9;‘ ;&#xTf 1;.92; 0 ;&#xTd [;0(noteIk(t)onlyinvolvesintegrationonaboundedinterval).Thedi erenceIk(t)�Ik+1(t)vanisheswhent=0(infactbothtermsarethen0),andI0k(t)=R(k+1)ke�txjsinxjdxforalltbydi erentiationundertheintegralsign,so(A.5)wouldfollowfromthederivativeinequalityI0k(t)�I0k+1(t)&#x]TJ/;༕ ;.9;‘ ;&#xTf 1;.92; 0 ;&#xTd [;0fort&#x]TJ/;༕ ;.9;‘ ;&#xTf 1;.92; 0 ;&#xTd [;0.Byachangeofvariablesy=x�intheintegralforI0k+1(t),I0k+1(t)=Z(k+1)ke�t(y+)jsin(y+)jdy=e�tZ(k+1)ke�tyjsinyjdyI0k(t):Thiscompletestheproofthattheseriesin(A.4)fort&#x-277;0satis esthealternatingseriestest. DIFFERENTIATINGUNDERTHEINTEGRALSIGN25IfwetruncatetheseriesPk0(�1)kIk(t)aftertheNthterm,themagnitudeoftheerrorisnogreaterthantheabsolutevalueofthenextterm:Xk0(�1)kIk(t)=NXk=0(�1)kIk(t)+rN;wherejrNjjIN+1(t)j1 N+1:Since01�e�yyfory0, NXk=0(�1)kIk(t) Z(N+1)0(1�e�tx)jsinxj xdx=Z(N+1)0tdx=t(N+1):Thus Z10(1�e�tx)sinx xdx = Xk0(�1)kIk(t)  NXk=0(�1)kIk(t) +jrNjt(N+1)+1 N+1:For"�0wecanmakethesecondtermatmost"=2byasuitablechoiceofN.Thenthe rsttermisatmost"=2forallsmallenought(dependingonN),andthatshows(A.3)tendsto0ast!0+.References[1]W.Appel,MathematicsforPhysicsandPhysicists,PrincetonUniv.Press,Princeton,2007.[2]W.Byerly,ElementsoftheIntegralCalculus,2nded.,Ginn&Co.,1888.Onlineathttps://archive.org/details/cu31924004779447/page/n5.[3]S.Cao,IntegratingR10e�x2dxusingFeynman'sparametrizationtrick,URL(version:2017-09-08):http://math.stackexchange.com/questions/390850/.[4]Conifold,answertoWhyisdi erentiationundertheintegralsignnamedtheLeibnizrule?,URL(version:2019-01-06):https://hsm.stackexchange.com/questions/8132.[5]R.P.Feynman,SurelyYou'reJoking,Mr.Feynman!,Bantam,NewYork,1985.[6]R.P.Feynman,MathematicalMethods,1946CornellUniversitynotesbyJamesC.Keck,http://james-keck-memorial-collection.unibs.it/JCKeck-papers/JCKeck-class-notes-from-Feynman's-course-on-MathematicalMethods-at-Cornell-1946.pdf[7]R.P.Feynman,HughesLecturesVol.3,notesbyJohnT.Neer,http://www.thehugheslectures.info/wp-content/uploads/lectures/FeynmanHughesLectures Vol3.pdf.[8]R.P.Feynman,HughesLecturesVol.5,notesbyJohnT.Neer,http://www.thehugheslectures.info/wp-content/uploads/lectures/FeynmanHughesLectures Vol5.pdf.[9]S.K.GoelandA.J.Zajta,\ParametricIntegrationTechniques,"Math.Mag.62(1989),318{322.[10]S.Lang,UndergraduateAnalysis,2nded.,Springer-Verlag,NewYork,1997.[11]S.Lang,ComplexAnalysis,3rded.,Springer-Verlag,NewYork,1993.[12]M.Rozman,\EvaluateGaussianintegralusingdi erentiationundertheintegralsign,"CoursenotesforPhysics2400(UConn),Spring2017.Onlineathttp://www.phys.uconn.edu/phys2400/downloads/gaussian-integral.pdf.[13]A.R.Schep,\ASimpleComplexAnalysisandanAdvancedCalculusProofoftheFundamentalTheoremofAlgebra,"Amer.Math.Monthly116(2009),67{68.[14]E.Talvila,\SomeDivergentTrigonometricIntegrals,"Amer.Math.Monthly108(2001),432{436.[15]J.vanYzeren,\Moivre'sandFresnel'sIntegralsbySimpleIntegration,"Amer.Math.Monthly86(1979),690{693.