the integrals in gradshteyn and ryzhik. part 7 elementary examples

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1、THEINTEGRALSINGRADSHTEYNANDRYZHIK.PART7:ELEMENTARYEXAMPLESTEWODROSAMDEBERHANANDVICTORH.MOLLAbstract.ThetableofGradshteynandRyzhikcontainssomeelementaryintegrals.Someofthemarederivedhere.1.IntroductionElementarymathematicsleavetheimpressionthatthereismarked

2、differencebetweenthetwobranchesofcalculus.Differentiationisasubjectthatissystematic:everyevaluationisaconsequenceofanumberofrulesandsomebasicexamples.However,integrationisamixtureofartandscience.Thesuccessfulevaluationofanintegraldependsontherightapproach,th

3、erightchangeofvariablesorapatientsearchinatableofintegrals.Infact,thetheoryofindefiniteintegralsofelementaryfunctionsiscomplete[3,4].Risch’salgorithmdetermineswhetheragivenfunctionhasanantiderivativewithinagivenclassoffunctions.However,thetheoryofdefiniteint

4、egralsisfarfromcompleteandthereisnogeneraltheoryavailable.ThelevelofcomplexityintheevaluationofadefiniteintegralishardtopredictascanbeseeninZ∞Z∞√Z∞
−x−x2π−x34(1.1)edx=1,edx=,andedx=Γ.23000Thefirstintegrandhasanelementaryprimitive,thesecondoneistheclassicalG

5、aussianintegralandtheevaluationofthethirdrequiresEuler’sgammafunctiondefinedbyZ∞(1.2)Γ(a)=xa−1e−xdx.0arXiv:0707.2122v1[math.CA]14Jul2007Thetableofintegrals[5]containsalargevarietyofintegrals.Thispapercontin-uestheworkinitiatedin[1,7,8,9,10,11]withtheobjecti

6、veofprovidingproofsandcontextofalltheformulasinthetable[5].Someofthemaretrulyelementary.Inthispaperwepresentaderivationofasmallnumberofthem.2.AsimpleexampleThefirstevaluationconsideredhereisthatof3.249.6:Z1√2(2.1)(1−x)p−1dx=.0p(p+1)2000MathematicsSubjectCla

7、ssification.Primary33.Keywordsandphrases.Elementaryintegrals.Thesecondauthorwishestothepartialsupportofnsf-dms0409968.12TEWODROSAMDEBERHANANDVICTORH.MOLL√Theevaluationiscompletelyelementary.Thechangeofvariablesy=1−xproducesZ1Z1(2.2)I=−2ypdy+2yp−1dy,00andeac

8、hoftheseintegralscanbeevaluateddirectlytoproducetheresult.ThisexamplecanbegeneralizedtoconsiderZ1ap−1(2.3)I(a)=(1−x)dx.0Thechangeofvariablest=xaproducesZ1(2.4)I(a)=a−1t1/a−1(1−t)p−1dt.0Thisintegralappearsas3.