NONELEMENTARY

DANA P WILLIAMS

This is a preliminary version and has not been careful ly proofread Please do not duplicate or

circulate I wil l be happy to provide a nal version on request In the meantime I would be

delighted to get your comments suggestions and especial ly corrections

Decemb er

Abstract I will attempt to give a selfcontained treatmentwhich will make pre

cise and then prove the statement that the antiderivativeofexpx can not b e

written in terms of the usual socalled elementary functions of Calculus

Introduction

For years Ive b egun teaching techniques of integration to rst year college students

bywarning them that integrationmore precisely antidierentiationis harder than

dierentiation In particular I am fond of p ointing out that it is easy to write down

simple expressions suchasexpx that dont haveanantiderivative whichcan

b e expressed in terms of the standard functions of calculus Of course it is a tad

dicult to explain precisely what one means In the previous semester I was quite

happy to p oint out that b y the Fundamental Theorem of Calculus every continuous

had an so

Z

x

expt dt erf x

is certainly an antiderivativeofexpx On the other hand it is not hard to accept

that whatever expressing an antiderivative in elementary terms means erf isnt it

In this article my goal is to show that functions suchas

x

sinx e

x

and hx f xe g x

x x

do not have elementary anti Naturally the rst task is to decide exactly

what an elementary function is

DANA P WILLIAMS

Elementary functions

The approach here is to use the language of eld theoryWe start with the eld

F C z ofcomplex rational functions Of course F is much to o smallweknow

of lots more functions than these so naturally well add new functions by adjoining

elements to F For example wemaywanttoaddf z logz Immediately we run

into the question of domain Here we will content ourselves with restricting attention

to a suitable op en subset of the plane Somewhat more generallywe could for

example let b e part of the usual Riemann surface for log So we assume that

f z

f is an on which satises e z for all z f is called a

branch of log z Then weletK M denote the eld of meromorphic functions

on Then C z K and we can dene F F f to b e the smallest subeld of K

containing the functions

n

g z r z r z f z r z f z

n

for r r r F Naturallyiff f f M then we can iterate the

n n

pro cess to obtain a subeld C z f f ofM that is

n

C z F F F F C z f f

n n

where F F f

k k k

We will b e primarily interested in subelds E of M which are dierential in

 

the sense that E f f f EgE Remarkablymany subelds are automatically

dierential

Lemma Suppose that F is a dierential subeld of M and that f M is



algebraic over F Then f Ff that is F f is a dierential subeld

m

Proof If pX r z r z X is the minimal p olynomial of f over F then

m

m

  

r z r z f z r z f z

n



f z

m

r z r z f z nr z f z

n

Denition Wesay that a dierential subeld of M is an elementary extension

of C z onifE C z f f for functions f M such that either

n k

f is algebraic over F

k k

expf b elongs to F or

k k

f expg for some g F

k k

for k n In case wesaythat f is a of F Similarly

k k

in case wesaythatf is an exp onential of F

k k



One could sp eculate just howmuchweknow ab out functions like f z logz but well give that a pass here

NONELEMENTARY ANTIDERIVATIVES

Remark A subeld of M of the form C z f f with each f satisfying one

n k

of or in Denition is automatically dierential and therefore elementary

To see this it suces to notice that F f is dierential if F is However this

k k k

follows from Lemma in case and is clear in cases and

Finallywe can dene an elementary function

Denition A function f in M is elementary on if f b elongs to some ele

mentary extension of C z on

A few moments reections shows that the denition of an elementary function is

incredibly generous it is dicult to conceive of a function which is not elementary

Example All the and their inverses are elementaryFor

example

iz iz

e e

sinz

i

and

p

z sin z i log iz

iz

are b oth elementary The rst is contained in C z e The second is the logarithm

of a function algebraic over C z

Example Even complicated comp ositions are easily seen to b e elementary f z

iz i cosz

sin cosz is in C z e e

The ab ove example is no accident and is a general fact

Lemma Al l meromorphic composites of elementary functions are elementary

Sketch of Proof Supp ose that f g and f g are meromorphic on and that f

and g are elementary Wewant to showthatz f g z is elementary Let

f C z f f Since quotients of elementary functions are elementarywecan

n

k

assume that f C z f f f Thus it suces to consider f z r z pz

n n

where r E C z f f and p is either algebraic over E or a logarithm or

n

exp onential of E An induction argumentallows us to assume that z r g z is

elementary and since pro ducts of elementary functions are trivially elementarywe

only have to see that z p g z is elementary

First supp ose that p is algebraic over E Then there are r r E such that

j

j

r z pz r z

j

It is not true that the inverse of an elementary function is elementary This do es not app ear

to b e easy to see and is certainly b eyond the scop e of this article



We only lo ok at comp ositions that are meromorphic This is to avoid worrying ab out what

happ ens near zero for atro cities like q z sinz

DANA P WILLIAMS

for all z By induction we can assume that each function z r g z b elongs

i

 

to some elementary extension E Then z p g z is algebraic over E and is

elementary

If p is a logarithm of E then by denition the function hz exp pz is in E

and induction once again implies that h g is elementary and b elongs to an elementary

 

extension E Then z p g z is a logarithm of E andiselementary

exp hz for some h Ewehave Finallyif p is an exp onential of E say pz



h g in an elementary extension E by induction and z p g z is an exp onential



of E

Example It follows that p erfectly ridiculous functions like lo cal solutions h to

z

tan e sin z cos log z hz hz

are elementary on their domains

Themainresults

In spite of showing that virtually every conceivable function is elementarywe

are going to prove the following theorem which asserts that many common anti

derivatives fail to b e elementaryeven if one is willing to sharply restrict the domain

Theorem Thereisnoopen set on which the functions denedbyexpz

expz z and sinz z have elementary antiderivatives In fact if f C z and

g C z is nonconstant then z f z exp g z has an elementary anti

 

if and only if thereisan a C z such that f a ag

Our pro of of this result is based on two rather dicult results Therstisa

classical result of Liouville suitably generalized by Ostrowski

Theorem Liouville Let F C z or any other dierential eld of mero

morphic functions containing C z considered as a subeld of M for some open

subset Then a function g F has an elementary antiderivative if and only if

thereareconstants c c C and functions u u v F such that

n n

n



X

u

i



c v g

i

u

i

i

The rst step towards the pro of of Theorem is the following lemma whichwill

also play a crucial role in the pro of of the main theorem

g

Lemma Suppose that g is a nonconstant element of C z Then e is

transcendental over C z



Liouville develop ed his results in the p erio d b etween and Actual references maybe

found in the bibliographyof

NONELEMENTARY ANTIDERIVATIVES

This lemma in turn dep ends on the following result which will also have applications

elsewhere

Lemma Suppose that F is a dierential subeld of M containing C z Then

 

if Aut F C z then for al l f F

 

Proof Dene D FF by D f f f It is not hard to see that D is

andD xy D xy xD y and a derivation on F ie D x y D xD y

that D C z f gNowifp is the minimal p olynomial in C z X with p

then





D p p p

   

p p

   

p p



p D



Since p D this proves the claim

Proof of Lemma Supp ose that were not transcendental Then would b e con

tained in a normal algebraic extension F of C z ofdegreen Supp ose that

 

Aut F C z Then Lemma implies that f f for all f M

  

so that g g Since the order of Aut F C z is n and since g is

not constant wehave

Q





X



Q

n g

Q

Put u Note that if p is the minimal monic p olynomial of in C z X

Q

then px X this shows that u C z It follows from that u



Thus u u is in C z and a derivativeof n g in C z The last statement implies



that all the p oles of u u have order On the other hand the fact that u is non



constant implies that u u has p oles and any such p oles must b e of order This

contradiction completes the pro of

Liouvilles original work dep ended on a rather subtle analysis of the structure of

elementary functions This is exploited in but at least I found the language and

style involved there to b e rather dicult to absorb The pro ofs given here will b e

based on the following result of Rosenlicht



This is essentially a pro of that dierentiation is the unique derivation extending dierentiation

on any algebraic extension of C z

DANA P WILLIAMS

Prop osition Rosenlicht Assume that F is a dierential subeld of M



Suppose that M is transcendental over F and that either For



F Final ly suppose that c c F are linearly independent over Q that

n

v u u F and that

n

n



X

u

i



v c F

i

i

u

i

i



Then v is actual ly in F Furthermore if F then each u must beinF On

i



the other hand if F then for each i n there is an integer such

i

i

F that u

i

We will p ostp one the pro of of Proposition until the very last However our main

result in now a straightforward consequence of Theorem and Proposition

Proof of Theorem Supp ose that z f z exp g z with f g C z f

C If has an elementary antiderivative them we can apply Theorem and g

g

with F C z e Thus

n



X

u

i

g 

fe c v

i

u

i

i

g

with c c C and v u u C z e

n n

Lemma We may assume that the constants c c appearing in arelin

n

early independent over Q

Proof Otherwise wemayaswell assume that c m c m c m with

n n n

m

m m m integers and m For i n let w u d c mand

n i i i

i

W n u Thenifwe put d c wehave d m d m d and

n n n n n n

m

 

mu u

u



i i i

m for i n With these notations w w

m

i

i

u u

i

i

n n

  

X X

w m w w

i

i n i

 

d v v d

i i

w w w

i i n

i i

n

m 

X i

w w

i

n



v d

i

m

i

w w

n

i

i

g

It follows from Lemma that e is transcendental over C z Thus wemay

g g

to conclude that v C z e and apply Proposition with F C z and e

NONELEMENTARY ANTIDERIVATIVES

g 

i

u h e for some functions h C z andintegers In particular u u C z

i i i i i

i

and

g 

fe v C z

P

m

jg

b e with b C z and b Thus On the other hand v

j i m

j

m

X

 jg  

jb g e v b

j

j

j

  

Notice that if b oth j and b are nonzero then ha ving b jb g would imply that g

j j

j

 

had p oles of order a contradiction Thus for each j wehaveb jb g

j

j

g

whenever b However as e is transcendental over C z we certainly have

j

g g mg

f e e e g linearly indep endentover C z Therefore implies that m

g  

and f must b e the co ecientof e in That is f a ag with a C z

This pro ves the last statement in the Theorem as z az exp g z would b e an

 

antiderivativeof in the event f a ag

If z expz then we could apply the ab ovewithf z and g z z

Therefore has an elementary antiderivative if and only if there is a

a such that



a z zaz

However a has a partial fraction decomp osition of the form

m n

r

m

X X

a

r

az pz

r

z z

k

r

k

where p is a p olynomial and a has p oles z z of multiplicities n n with

m m

n for all i Clearly p as otherwise the righthand side of has degree

i

equal to deg p On the other hand if a has a p ole of order s at z then

the righthand side of has a p ole of order s Thisisacontradiction and we

conclude that expz do es not have an elementary antiderivative

If z z exp z so that f z z and g z z This requires that we

solve



a z az

z

However the righthand side of has no p oles of order one Thus cant have

an elementary In fact neither do es z z expw z forany nonzero

constant w C

b

j



must b e nonzero since g is nonconstant and hence has p oles whichmust b e Because g

jb

j of order one

DANA P WILLIAMS

Finally we are left with the question of whether q z sinz z has an elementary

integral Note that if we dene z expiz then q z iz z z

Furthermore wehave just concluded that neither z nor zhave elementary

antiderivatives Unfortunatelyitdoesnotimmediately follow from this that q can

not have an elementary antiderivative

We need to intro duce some notation and conventions for functions in v C z

These functions are rational in C z that is v is a quotient of p olynomials in

z z to denote a function z with co ecients in C z We use the notation v

i

of the form p z q z for p q C z X In particular we can view v as a

function of two indep endentvariables where v z w pw q w keep in mind that

the co ecients of p and q are functions of z Thus we can write D v for the partial

derivativeof v with resp ect to its rst and D v for the partial with resp ect

to the second

Now backto q Ifq did haveanelementary in tegral u then Theorem would

allow us to assert that there are functions v v v which are rational in C z

r

such that

r

X

uz v z z c log v z z

i i

i

and the ordinary chain rule implies that

z z i z z z D v z D v

iz z

r

X

D v z z D v z z i z

i i

c

i

v z z

i

i

Since is transcendental over C z Lemma must hold with z replaced

byanyvalue in the domain of b oth sides In particular we can replace z by

z forany suciently close to It follows that

z

iz z

P

is the derivativeof u z v z z c log v z z Thiswould imply

i i

i

that

i

u u

is an elementary antiderivativeof z z and this is imp ossible This contradiction

completes the pro of

 k

For example supp ose that q and pX r z r z X r z X ThenD v z w

  k 

k k 

r z r z w r z w while D v z wr z kr z w

 k

  k

NONELEMENTARY ANTIDERIVATIVES

The last bit ab out sinz z is actually a very sp ecial case of a generalization of

Theorem given in I I I x whichsays that assuming g and y are algebraic

i i

g g

n



over C z and no g is a constantmultiple of another g thenw e y e y

i j n

g

i

has an elementary antiderivative only if each e y do es This is b eyond our current

i

scop e

Nowwehavetopay for our fun and provethetwo hard results Theorem and

Proposition on which all the ab ove dep ends

Proof of Theorem By assumption and Remark there is a tower of dierential

elds

F F F F

N



such that g F and for each i N F F f where if f is not

i i i i

N

algebraic over F then f is either a logarithm or exp onential of F

i i i

Notice that if N then the result is trivially true So we can pro ceed by

induction on N Sowe can assume that N and apply the induction hyp othesis

to the tower F F Consequentlywe can assume that

N

r



X

u

i



v g c

i

u

i

i

where c c C as desired but v u u are only known to b e in F Our

r r

task is to showthatwe can mo dify rc c u u suchthat u u F

r r r

For notational convenience we write f so that F F Also the argument

of Lemma implies that w emay assume that the c are linearly indep endentover

i

Q This will b e needed when we try to apply Proposition

We rst concentrate on the case where is transcendental over F If is logarithm

 

of F then a a for some a FWe apply Rosenlich ts Proposition which

implies that each u b elongs to F and that v b elongs to F Then also forces

i



v to b elong to F We will have obtained the result in the presentcaseifwe can show

all this forces v to b e of the form c d with c C and d FBut

m

X

j

v b

j

j

with b FWe can assume that m and that b otherwise were done

j m

Anyway



a

 m   m

b an elementof F of degree m v b mb

m

m m

a

 

Fwe get b sob is a Similarlyif m then we Since v

m

m

  

also have mb a a b which equals mb b Thus mb b

m m m m m

m

DANA P WILLIAMS

is a constant and b elongs to F This contradicts the transcendency of over F and

completes the pro of in the logarithmic case

 

Now supp ose that is an exp onential of F Then a for some a FAgain

i

we can apply Proposition to conclude that each u is of the form a for a F

i i

and an integer while v F Now calculate that

i

   

a a u

i i i



a

i i

u a a

i i i

and plugging into we get

r



X

a



i

g c v a

i i r

a

i

i



Thus wemay assume that each u actually b elongs to F This again forces v to

i

b elong to F andwe only need to see that v F to o But v has the form with

each b in F and b In this case

j m

m

X

  j 

b jb a v

j

j

j

 

If m wehave b mb a This implies that

m

m



m   m m  m 

b b b m b mb

m m m

m m

m  

b mb a

m

m

m

This forces b to b elong to F which is nonsense This disp oses of the cases where

m

is transcendental over F

Finallywe are left only with the case that is algebraic over F Let K b e the

smallest normal extension of F containing F For any in AutK F wehave

r r

 

X X

u u

i

i

 

v v c g c

i i

u u

i i

i i

 

Wesaw in the pro of of Lemma that f f for all f K Therefore

Q



r r



X X X X X



u

u

i

i



Q

K F g v v c

i

u u

i i

i i

Q Q

Since b oth u and v are invariant under AutK F they are in F Thus

i

we can divide b oth sides byK F to get the desired result

Before I tackle Rosenlichts fundamental result Ill need a little lemma

NONELEMENTARY ANTIDERIVATIVES

Lemma Suppose that p p are distinct irreducibles in K Also suppose

N

Nn

m

that f a g arerelatively prime arepolynomials in K such that p and a

mk m mn

m

mk

for al l m N Then

N n

m

X X

a

mk

k

p

m

m

k

is not in K

Proof Let M max n If were in K then we could multiply by

m m

M



p p p to conclude that is a subset of f p g q q suchthat

N m N



N

X

a

m

q

m

m



is in K with N N and each a relatively prime to each q But equals

m m

Q



N

q q b a

m

m

Q



N

q

m

m

for some b K Thus if is in K then q divides the numerator of and

Q

M

therefore q divides a q This is a contradiction and completes the pro of

m

m

Proof of Proposition We let K b e a nite normal algebraic extension of F in

which the functions v u u split into linear factors That is we assume that

n

N

Y

ij

u g i n z

i i j

j

N M

X X

v h z an elementof K

j j

j

where h and z are elementsof K each g is a nonzero elementof K and is an

j j i ij

integer Note that and h will often b e zero By an argument only slightly more

ij j

complicated than that in Remark we see that K is dierential and therefore so

is K Moreover K is closed under dierentiationaswell

With these notations our basic assumption is that

n

 



X X X

z

g



j

i

h z c K c

j j i ij i

g z

i j

ij j i

We could replace K byanyUFDandK by its fraction eld

m

For each f K consider the miminal p olynomial p FX off IfpX r z r z X

 m

for functions r F then f z is given by the formula Thus f K i

DANA P WILLIAMS

Iintend to show that unless z all the and h are zero This will b e shown

j ij j

to suce

  

Wehave either a or a for some a FWe will consider that cast a

rst Then



z a z

j j

z z

j j

Next I claim that is in lowest terms Recall that pq is in lowest terms if p and

q are relatively prime in K Since the denominator is irreducible this amounts to

  

showing that a z Butifa z then a z z for all AutK F

j j

j j

P





see the pro of of Lemma Therefore K F a z and a b for some

j



b F This implies that b so that t b is a constantand FThisis

nonsense and establishes the claim



In the case a wehave



a z z

j j

z z

j j

and again I claim that the quotientisinlowest terms provided that z Here

j

 

that means that z az If not then wewould have z z a and once again

j j

j j



z z a for all AutK F Then

j j

Q





X

z

z

j

j

Q

K F a

z z

j j



Letting D K F wehave Da b b for some nonzero b in F Thus

D   

b

D

D

b

D  D

which implies that b This in turn implies that b F whichis



a contradiction Thus wehave shown that in all cases but one the faction



z z isinlowest terms The exceptional case o ccurs when F and

j j

 

z Then z z F

j j j

Now consider



  

z z h z h z h

j j j j j

j j

As ab ove except in the exceptional case or when h is of the form K

j

plus a fraction which in lowest terms had a linear numerator and a denominator of



degree In the exceptional case where z and a Fwehave

j





h a h h z

j j j j

NONELEMENTARY ANTIDERIVATIVES



I claim that if h thenalso h h a If the claim were false then we

j j

j





have h h aand h h a for all AutK F Once again

j j j

j

Q





X

h

h

j

j

Q

K F a

h h

j j

N  N

so that Na bb with N K F and b F But then

 N  N

Na b b and b This implies that b F whichisa

contradiction So in the exceptional case has a fractional part which in lowest

terms has a denominator of degree It follows from Lemma that if

any h then can not reduce to a p olynomial in K In particular v

j

K F F

Thus we are left with

 

X

z

j

K c

i ij

z

j

ij

z 

j

P

Using Lemma again it follows that c whenever z Since the c are

i ij j i

i



indep endentover Q by assumption wehave unless z and F In

ij j



any case there are integers all zero in the event F and equal to

n ij



i i

KF F KButthenu when z and Fsuchthat u

i j i

This completes the pro of of Rosenlichts result and completes our investigation

References

Joseph F Ritt Integration in nite terms Columbia University Press New York

Maxwell Rosenlicht Liouvil les Theorem on functions with elementary Pacic J Math



Department of Mathematics Bradley Hall Dartmouth College Hanover NH

USA

Email address danawilliamsdartmouthedu