Why certain integrals are ``impossible'.

1 IntroductionElementary Functions and fieldsLiouville s TheoremAn exampleWhy certain integrals are impossible .Pete GoetzDepartment of MathematicsSonoma State UniversityMarch 11, 2009 IntroductionElementary Functions and fieldsLiouville s TheoremAn fields and s Functions and fieldsLiouville s TheoremAn exampleProbabilityCentral Limit theorem (x)=1 2 x e u2/2duFor probability applications, we need ( ) = is not proved by finding a formula for (x) (by findingan explicit antiderivative ofe u2/2) and taking the limit asx .IntroductionElementary Functions and fieldsLiouville s TheoremAn exampleNumber TheoryPrime Number theorem (x) = #{n x|nis prime}Li(x)= x21ln(t)dt (x) Li(x) asx This is not proved by finding an explicit antiderivative of1ln(t).Ifu= ln(t), then 1ln(t)dt= Functions and fieldsLiouville s TheoremAn exampleElementary formulasThe indefinite integrals e u2duand euududo not haveelementary does one prove such claims?

2 First have to give a precise definition of elementary formula .After all e u2du= uae x2dx+Cfor any constantsaandCby Functions and fieldsLiouville s TheoremAn exampleHistoryNewton was perfectly happy to solve an integral by a preferred integration in finite terms and allowedtranscendental functions like Functions and fieldsLiouville s TheoremAn exampleElementary functionAnelementary function(roughly) should be a function ofone variable built out of polynomials, exponentials,logarithms, trigonometric functions, and inverse trigonometricfunctions, by using the operations of addition, multiplication,division, root extraction, and :sin 1(x3 1) lnx+ cos(x/x2+ 1)IntroductionElementary Functions and fieldsLiouville s TheoremAn exampleA simplificationWe will useC-valued functions of therealvariablex, , ourconstants will be complex trigonometric functions and inverse-trigonometricfunctions can be written in terms of complex exponentials (x)=eix e ix2i, cos(x)=eix+e ix2tan 1(x)=12i(ln(x ix+i) i )sin 1(x) = tan 1(x 1 x2), cos 1(x) = tan 1( 1 x2x)IntroductionElementary Functions and fieldsLiouville s TheoremAn exampleMeromorphic functionsAmeromorphic functionis a function defined on an openintervalIof the real numbers whose values are complexnumbers or with the property that sufficiently close to anyx0inIthe function is given by a convergent Laurent series inx functions are meromorphic a meromorphic functionf, bothefand lnfaremeromorphic (one may have to restrict the domain off).

3 IntroductionElementary Functions and fieldsLiouville s TheoremAn exampleFields of meromorphic functionsLetC(x) denote the field of rational functions. Notice thatthis field is closed under elementary function (under our rough definition) shouldbe in some extension ofC(x).IntroductionElementary Functions and fieldsLiouville s TheoremAn exampleFields of meromorphic functionsIff1,..,fnare meromorphic functions, letC(f1,..,fn)denote the set of all meromorphic functionshof the formh=p(f1,..,fn)q(f1,..,fn)for somen-variable polynomialsp,q'= 0 andq(f1,..,fn) isnot identically definition captures the operations of addition,multiplication, and is not hard to show that the setC(f1,..,fn) is a field andthat this field is closed under :K=C(x,sinx,cosx)=C(x,eix).IntroductionE lementary Functions and fieldsLiouville s TheoremAn exampleElementary fieldsA fieldKis anelementary fieldifK=C(x,f1,..,fn) andeachfjisan exponential or logarithm of an element ofKj 1=C(x,f1.)

4 ,fj 1)orfjisalgebraicoverKj 1, that isfjis a solution to anequationgltl+ +g1t+g0= 0 whereg0,g1,..,gl Kj 1An elementary field is built from the the field of rationalfunctions in finitely many steps by adjoining an exponential, alogarithm, or a solution to a is captured by adjoining exponentials orlogarithms. Root extraction is captured by the adjunction ofalgebraic fields are closed under Functions and fieldsLiouville s TheoremAn exampleElementary functionsA meromorphic functionfis anelementary functionif it liesin some elementary :f(x)=3 lnx+ cos(xx2+i) is an elementaryfunctionC(x) C(x,lnx) C(x,lnx,ei(xx2+i)) C(x,lnx,ei(xx2+i),f)IntroductionElementa ry Functions and fieldsLiouville s TheoremAn exampleElementary integrationA meromorphic functionfcan beintegrated in elementarytermsiff=g for some elementary an elementary field is closed under differentiation so iffcan be integrated in elementary terms, then necessarilyfisalso Functions and fieldsLiouville s TheoremAn exampleDifferential Galois theoryWe can rephrase our problem: Given an elementary functionf, when does the differential equationdydx f= 0 have anelementary solution?

5 The answer is in the affirmative precisely when we can find atower of fields with special the analogy with ordinary Galois Functions and fieldsLiouville s TheoremAn exampleLiouville s ThereomTheorem (Liouville, 1835):Letfbe an elementary functionand letKbe any elementary field containingf. Iffcan beintegrated in elementary terms then there exist nonzeroc1,..,cn C, nonzerog1,..,gn K, and an elementh Ksuch thatf= cjg jgj+h .Iff= cjg jgj+h , theng= cjln(gj)+his an elementaryantiderivative theorem is proved by induction on the length of a towerof fields constructingK(g) wheregis an antiderivative Functions and fieldsLiouville s TheoremAn exampleAn important corollaryCorollary:Letfandgbe inC(x) withf'= 0 andgnonconstant. Iff(x)eg(x)can be integrated in elementaryterms then there is a functionR(x) inC(x) such thatR (x)+g (x)R(x)=f(x).IfR(x) C(x) satisfiesR (x)+g (x)R(x)=f(x), thenR(x)eg(x) is an antiderivative off(x)eg(x).

6 We can apply this corollary to show thate x2andex/xhaveno elementary Functions and fieldsLiouville s TheoremAn exampleProof fore x2 Takingf= 1 andg= x2in the Corollary, we must show thedifferential equationR (x) 2xR(x) = 1 ( )has no solution forR(x) C(x).ODE s shows the general solution of ( ) isR(x)=ex2( e x2dx+c) for anyc but this doesn thelp!IntroductionElementary Functions and fieldsLiouville s TheoremAn exampleProof fore x2 Suppose thatR(x) C(x) is a solution to ( ).Rcannot be a constant or a polynomial inx(by degreeconsiderations).WriteR(x)=p(x)q(x) for some nonzero relatively primepolynomialsp(x),q(x) withq(x) Cbe a root ofq(x) of multiplicity 1. Thenp(z0)'= 0 andp(x)/q(x)=h(x)/(x z0) withh(x) C(x)having numerator and denominator that are non-vanishing Functions and fieldsLiouville s TheoremAn exampleProof fore x2 The quotient rule yields(p(x)q(x)) = h(x) (x z0) +1+h (x)(x z0) Asz z0inCthe absolute value of (p(x))/q(x)) |x=zblowsup likeA/|z z0| +1withA=|h(z0)/ |'= 0.

7 | 2z (p(z)/q(z))|has growth bounded by a constantmultiple of 1/|z z0| asz |((p(x)q(x)) 2x (p(x)q(x)))|x=z| A|z z0| +1asz contradicts the identityR (x) 2xR(x) = Functions and fieldsLiouville s TheoremAn exampleReferences1B. Conrad, Impossibility theorems for elementary integration, conrad/ Kasper, Integration in Finite Terms: The Liouville Theory,Mathematics Magazine, Vol. 53, No. 4 (Sep., 1980), A. Marchisotto and G. Zakeri, An Invitation to Integrationin Finite Terms,The College Mathematics Journal, Vol. 25,No. 4 (Sep., 1994), pp. Rosenlicht, Integration in Finite Terms,The AmericanMathematical Monthly, Vol. 79, No. 9 (Nov., 1972).