The Problem of Integration in Finite Terms
Author(s): Robert H. Risch
Reviewed work(s):
Source: Transactions of the American Mathematical Society, Vol. 139 (May, 1969), pp. 167-189
Published by: American Mathematical Society
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THE PROBLEM OF INTEGRATION IN FINITE TERMS
BY
ROBERT H. RISCH(l)
This paper deals with the problem of telling whether a given elementary function,
in the sense of analysis, has an elementary indefinite integral.
In ?1 of this work, we give a precise definition of the elementary functions and
develop the theory of integration of functions of a single varia' Z. By using functions of a complex, rather than a real variable, we can limit ourselves to exponentiation, taking logs, and algebraic operations in defining the elementary functions,
since sin, tan- 1, etc., can be expressed in terms of these three. Following Ostrowski
[9], we use the concept of a differential field. We strengthen the classical Liouville
theorem and derive a number of consequences.
?2 uses the terminology of mathematical logic to discuss formulations of the
problem of integration in finite terms.
?3 (the major part of this paper) uses the previously developed theory to give an
algorithm for determining the elementary integrability of those elementary functions which can be built up (roughly speaking) using only the rational operations,
exponentiation and taking logarithms; however, if these exponentiations and
logarithms can be replaced by adjoining constants and performing algebraic
operations, the algorithm, as it is presented here, cannot be applied.
The man who established integration in finite terms as a mathematical discipline
was Joseph Liouville (1809-1882), whose work on this subject appeared in the
years 1833-1841. The Russian mathematician D. D. Mordoukhay-Boltovskoy
(1876-1952) wrote much on this and related matters. The present writer received
his introduction to this subject through the book [10] by the American J. F. Ritt
(1893-1951).
The reader need only be familiar with some standard facts from algebra and
complex analysis in order to understand this paper. Some basic results from differential algebra are used, but they are explicitly stated and references are given for
their proofs.
1. Liouville theory of elementary functions. A field 9 with a unary operation
d/dz is said to be a differentialfield iff for any a, b in 9:
(1)
(2)
~~~~~~d
= dd a+d d b
d
dz
dz ~~~~~(a+b)
dz
d
d
d
-(a b) = a d-b+b-a. z
z
ddz
Received by the editors March 28, 1967 and, in revisedform, April 22, 1968.
(1) Presentaddress:T. J. Watson, ResearchCenter, Yorktown Heights, New York.
167
168
R. H. RISCH
[May
We will also write (dldz)a as a'. We will always assume that the characteristic of
our field is 0. A differential field containing 9 is a differentialextension of q. An
isomorphism of the differentialfields 9, and g2 which preserves the differentiation
operation is called a differential isomorphism.K= {a E 9 : (dldz)a=O} is a differential field and is called the constantfield of 9.
Let U be a differential field extension of 9 having the property that for any
finitely generated differential extension Y of 9, there is a differential isomorphism
of ' into U, leaving 9 fixed; then U is a universalextension of B.
It is proved in [4, pp. 768-771], that every differential field has a universal
extension. There a stronger definition of universal extension is used. We will denote
the constant field of U by C.
It may be of help to think of 9 as a field of meromorphic functions, that is
finitely generated over the rational numbers, on some region A of the complex
plane or a Riemann surface, of U as consisting of all functions which are meromorphic on some subregion of A, and of C as the complex numbers.
For 0 E U, 0 and ?f(O) are said to be simple elementary over 9 iff one of the
following conditions holds:
(1) 6 is algebraic over _.
(2) There is an f in q, fo 0, such thatf =fO'.
This situation will be abbreviated as O=logf
(3) There is anf in 9 such that 06= Of'. Here we write 0 = expf
Note that ?q(O)is, in an obvious way, a differential field. Also observe that any
c c C is simple elementary over 9. If we have for 0, y, E U, = logf and ,= log f or
0=expf and f = expf, then 0 = 0 + c, respectively V= cOfor some c E C.
In cases (2) and (3), if the constant field of 9(0) = K and 0 is transcendental over
Q, then 0 is said to be a monomial over ?9. It is easy to show that if 0= logf or
9 then 6 is not a monomial over 9 iff there is a c c gY(O)fl C such
0=expfforfc
that 6 is algebraic over 9(c). A useful example here is 9 = Q(z, e-), where Q =
rationals and z is the identity function on a region A. If 0 and , are two different
logarithms of e2 (say z and z+22ri), then , is not a monomial over g(O).
Let
Oi-1)
Oj.,On), each Oi being simple elementary over -9(01..,
i= 1, .
n. Then 3 and any g E Y is said to be elementary over 9. J and any
g E s are regular elementary over 9 iff each 0i is a monomial or algebraic over
g(019
..
S i-1).
We now turn to a proof and sharpening of the basic Liouville theorem on integration in finite terms. This result is usually proved by an analytic technique
which was introduced by Liouville in [7]. In Mordoukhay-Boltovskoy's book [8],
the technique of partial fractions in monomial extensions was first used to study
elementary integration. That the partial fraction method could be used to prove
Liouville's theorem itself was pointed out by Rosenlicht in [11].
First we study the differentiation of certain types of elements of monomial
extension fields. Let 9 be a differentialfield, 0 a monomial over 9.
1969]
THE PROBLEM OF INTEGRATION
IN FINITE TERMS
169
(a) Let P c g[6], P=Ann +An_16On-1+ +Ao, Ai c q An+O.
Case 1. 0= logf. Here P'= A'n6+ (A'- +nAn(f'/f))6 n-1+.
+(A'+A,(f'/f)).
If A = 0, then A' - 1+ nAn(f '/f)# 0; for if it were, then An 1+ nA,6 = constant and
0 would not be a monomial over _q. Here we use the remark following the definition of simple elementary above. Thus the derivative of a polynomial of degree
n in 0=logf is a polynomial of degree n-I if its leading coefficient is a constant.
It is a polynomial of degree n if the leading coefficient is nonconstant.
Case 2. 0= expf. Then
+(n= (An+nf'An)6n+(A_1
P'
l)f'An1)on-1l+*
+(A +f'A1)6+Ao.
A'+nf'An=#O, for if it were then 0-nh=constantxAn and 0 is not a monomial
over Bq.Thus, the derivative of P is of the same degree as P.
(b) Let P, Q be the elements of g[6], Q monic and irreducible in 0, p = degree P
< q= degree Q
(P/lQn) = -nPQ'/Qn+ 1Ap'/pQn.
Case 1. 6=logf By (a), degree Q'=q-1. Thus, Q{-nPQ', for if it did, it
would have to divide P or Q'.Thus, the partial fraction decomposition of (p/Qn)/
is of the form R/Qn+ 1 + S/Qn, R$O0.
Case 2. 0 = exp f. Suppose QI-nPQ'. Since Q{P we have that QIQ'.Q = 6q +*
so Q'=qfq ' + . so Q'=qf'Q. Therefore, Q=(ef)q= 9q. Since Q is irreducible,
q=1 and Pe-9.
Thus, the partial fraction decomposition of (p/Qn)/, unless Q= 0, is of the form
R/Qn+1+ S/Qn, R$O0.If Q = 0, then it is of the form (P'-nf'P)/Qn, P'-nf'P#O0.
(c) Let P E g[6], P monic, degree P=p.
Case 1. 6=logf Then (log P)'=P'/P where degree P'=p-1.
where N=P'-pf'P.
N=O in
Case 2. 6=expf Then (logP)'=P'/P=N/P+pf'
case P= 0 and otherwise N is a polynomial of degree less than p.
THEOREM OF LIOUVILLE-FIRST STATEMENT. Let ! be a diferential field, Y
elementary over 9?. Suppose ! and Y have the same constantfield K. Let g E t
f E 9 withg'=f. Theng= vo + E1 Ci log vi wherevo,vi are els of 9 andci els of K.
Proof. Y is regular elementary over 9;
O,n)each 6i being a
)=9(01,
monomial or algebraic over g(Ol, . . ., i -1). We use induction on n. Suppose n=O,
then F= 9 and g= vowhere vo E 9. Assume we have proven the theorem for n- 1.
Then g= wo+> 1 di log wi, where wo, wi els 9(01), and di els K.
Case 1. 01 is a monomial over 9. Then
ki
pil
g = P+
L+aj log Q,
(Q1)j
j=1
k2P
+
(Q))+a2
kr
+ E
log Q2
P
:=l~r
rJ i +ar r log Qr +
m
bi log h
R. H. RISCH
170
[May
where P, Pi,j, Qi are in 9[01]; ai, bi els K, hi els -; degree Pij <degree Qi; the
Q, are monic and irreducible.
We use the discussion preceding this theorem to conclude first that each Pi,j= 0,
for otherwise, using part (b) of that discussion, the right-hand side would have,
after differentiation, a term R/(Qi)k,+l. Thus g=P+ >K=1aj log Qj+ im=1 bi loghi.
Now by part (c), each aj=0; for otherwise the partial fraction decomposition off
would have a term S/Qj. Now if 01=expfl, then P must be of degree 0 in 01 and
g=P+:Ei= 1 bi log hi as required. If 01=logf1, then P=c logf1 +v where v E 9 and
c EK. Then g=v+c logf1+ ET 1 bi log hi which is again in our desired form.
Case 2. 01 is algebraic over 9.
f = (WO)'+ E diWj.
i=1
W
Adding conjugates over 9 we get:
k
rn?f= (Trace wo)'+
(Norm
d.i Nomw)wi)'
Norm w,
where m is a positive integer. Thus here too, g is of the desired form and our proof
is complete.
A question that arises is: Are any constants, not in 9, necessary for the integration of elements of Q9? The example
fdz
2
loV1
z-V2
seems to indicate that one must adjoin V/2 to Q(z) in order to perform the integration. We now show this rigorously.
PROPOSITION
1.1. Let F be elementary over K(z), where K is afield of constants
and z is transcendentalover K with z'= 1. Supposef EcF andf'= 1/(Z2 - 2), then
/2 E Proof. Without loss of generality, we may assume that K is the constant field
of -F Thus, by the first statement of Liouville's theorem, we get that f= vo +
1 ci log vi where vo, vi are in K(z) and the ci are in K. We may assume that the vi,
for i>0, are irreducible elements of K[z]. Note that if V2 0 K, then Z2 -2 is
irreducible over K.
We decompose vo into partial fractions over K and then differentiate the expression vo+ k=1 ci log vi. Now reasoning as in the proof of the first statement of
Liouville's theorem, we see that vo must be in K[z], for otherwise 1/(Z2 -2) would
have to have an irreducible factor of degree greater than 1 in its denominator.
Also, the only summand in the second term is c log (Z2 - 2). Furthermore, vo= az
+ b. Thus,f= az + b + c log (Z2 - 2), a, b andc els K. Thenl /(Z2 -2)=a+2cz/(z2-2)
which is impossible, by the uniqueness of a partial fraction decomposition. Thus
2e .
IN FINITE TERMS
THE PROBLEM OF INTEGRATION
1969]
171
It would seem that the new constants needed in integrating an element of 9
should all be algebraic over K. This is indeed the case.
THEOREM OF LIOUVILLE-SECOND
STATEMENT (STRONG LIoUVILLE THEOREM).
Let ! be a differentialfield with constantfield K. Supposef E 9 and there is a g
elementary over ! such that g'=f. Then there are a vo e t, ci e K (K= algebraic
closure of K), vi E KQ9such that f= v' + i1 civ'/vi where every automorphismof
R- over 9 permutes the terms of the sum.
(For the proof, see [3, p. 33].) Let xl, . . ., x, be indeterminatesover 9,
p(x, ..., x.) and q(xl, . . ., xn) are in 9[x1, . . ., xJ]. Suppose that thereare c1, . . ., cn
in C (= constantfield of a universalextension of Q9)such that p(c1, . .., c) =0 and
ql(cl . .* CO)=0. Then there are kl, . . ., kn in K such that p(kl, . .., kn)= 0 while
LEMMA.
q(kl,
.kn)74?
Proof of second statement. By adjoining constants c1,..., cn in C to 9 we can
apply the First Statement to the present situation with Q9(c1,..., cO)replacing B.
Thus we have that
1
9 = po(cl,.
q c
.n+~
c)+>ci
log p(cl,
()
..,Cn)'
where each pi, qi is a polynomial with coefficients in 9.
Pi(C1, .. ., CO)#0 ,
i = 1,.. .,k,
q,(cl, ..., CO):A O,
i = O,... ., k.
By differentiating, we get
op
f=
M
-qp
po~~qo
piqi
=i
where p, and qi are Pt and qi evaluated at (c1,..., cn).
Thus, c1, . . ., Cnsatisfy the polynomial conditions P(x1,...,
xn)=O and
k
Q(Xi,
* * , xn) = q2(Xl,
..
,
Xn)
X7npi(x,
...,
xn)qi(x,
,
xn) #
0,
P being obtained from (*) by a simple transformation. Thus, by the lemma,
P(ki,.. ., kn)=0 and Q(k, ... ., kn)=#0 where (kl, . . ., kn) E Kn.
By backtracking from these conditions we get f= i3'+ :% 1 ki3/lbi where ei are
els of K-9, viz. eh=pi(k1,..., kn)/q,(k1,..., kn), i=, . . ., k.
Taking traces we get If= (Trace eo)'+ =1 Tracekielei, 1 a positive integer.
Taking vo= (1/1) Trace v-, we get our desired result.
OPEN PROBLEM. -In the preceding theorem, if g E F where Y is elementary over
9, can vo, vi, ki, and log vi, i= 1, . . ., k, be chosen so as also to be in F?
172
R. H. RISCH
[May
Let !? be a differentialfield,f E ?t.
(a) Supposef#O0 and 0 = logf is not a monomial over 9?. Then 0 =g + c, g E t,
PROPOSITION 1.2.
c E C.
(b) Suppose 0 = ef is not a monomialover 9?. Then on = dh wheren E Z, d E C, and
h E 9.
Proof. (a) f'/f= 6' where 0 is algebraic over !2'(k),k E C.
Then after adding conjugates, using the argument of the Strong Liouville Theorem
which is based on the lemma and adding conjugates again we get thatf'/f=g' for
g e ?. Thus 6=g+c, cc C.
(b) Here 0'/0=f' with 0 algebraic over t(k), k e C.
Reasoning as in (a) we get that h'/h = nf ', where n E Z and h E ?. Thus on = dh,
h E C.
The following easy consequence of 1.2 will be useful to us in formulating the
integration problem.
PROPOSITION 1.3. Let 0 be a monomial over 9?. 0=expf or 0=logf, f ?t.
Suppose b= expf or = logf (i.e., bsatisfies the same definingequationas 0), then
b is also a monomial over 9 and !l'(6) is differentially isomorphic to !t(O) with
6 ---> b.
In the example on page 168 where 0 = z and b= z + 2rTi,?ft(0)= Q(z, ez) while
9(0= Q(27ri, z, ez) and these fields are not isomorphic.
1.4. Let 9 be a differentialfield, f E ?t. Suppose that there is a g,
PROPOSITION
elementaryover ! such that g'-=f Then there is an h, regularelementaryover 9 such
that h'=f.
Proof. By the Strong Liouville Theorem, there is a j in i(o, log v1,..., log Vk),
where a is algebraic over the constant field of 9 and v1, . . ., vk are in 9(a), such that
j = vo +
1 ci log vi, vo E Q, ci E @(a), and ]'=f By 1.2, each log vi differs by an
additive constant from an element of !(o, log vl, . . ., log vi - 1) or else is a monomial over the latter field. Thus, by subtracting a constant from j, we get an
h, regular elementary over q, such that h'=f
2. Formulation of the problem of integration in finite terms. We begin with a
few remarks on recursive functions. For complete details, see [1].
A function on and to the natural numbers is called recursive iff there is an
algorithm for computing f(n) for each n. There are also precise mathematical
definitions of recursive functions which have all been shown to be equivalent to
each other. The hypothesis that they all are equivalent to the definition we have
given is called "Church's thesis."
A set of natural numbers is recursivelyenumerableiff it is the range of a recursive
function. If a set of numbers has a recursive characteristicfunction, then the set is
called recursive. This means that it and its complement are recursivelyenumerable
1969]
THE PROBLEM OF INTEGRATION
173
IN FINITE TERMS
or that there is an algorithm for deciding whether or not a given natural number is
in the set.
Suppose there is a one to one correspondence between a countable set of mutually
distinguishable objects Y and the natural numbers such that given an a in Y
one can compute the associated natural number, and vice versa. Then $- is said
to have been given a Godel numbering. We define recursively enumerable and
recursive subsets of 9Y as sets whose associated Godel numbers form a recursively
enumerable or recursive set of natural numbers.
The Y we will deal with will be our symbols for the elementary functions. We
will describe a language fragment which comes very close to the way a mathematician customarily denotes these functions. Our functions lie in differential
fields of meromorphic functions on a region of the complex plane or a Riemann
surface which are finitely generated over the rational numbers, e.g.,
9
= Q(al, .. ., an, b, z, exp z2, log (z+ exp z2), \/(log (z+ exp z2)))
where a1, . . ., an are constants, algebraically independent over Q, b is algebraic
over Q(a1, . . ., an), z is the identity function, etc. We can represent all members of
9 if we have one symbol for each of the generators of 9 and binary operation
symbols for +, *, -, and
It is easy enough to invent symbols for logarithms and exponentials of functions
already described-just write log or exp in front of the previously defined symbol.
However, algebraic operations cause a problem. Not all of them can be expressed
by means of radical signs. Thus, we have invented "y symbols " to denote them.
We consider sequences of symbols:
e = <Ie, Te, al,
. ..,
aOk, y[(((y1+ePl
ey1
1)+e
(k, m >0, 1> 1, 7Te and the symbol following
+e))],
Ze, 81,
*
m>
ak possibly missing), which we call
elementaryfield descriptions (efd's). Along with e, we will consider the nested
sequence: Qlel QIel Qall . . I Q6m= Qe. An Qais the smallest set of terms containing
Ie, . . ., a and closed under the binary operation symbols +e, -e e * e.
We divide the symbols of e into two sorts-the y symbols and the non-y symbols.
A y symbol is of the form: y[(((yn+eAl*eyn-1)+e.. +eAn))] where n> 1, y'
means (((Y eY)*eeY * )) j times, j > 1, and the Ai are in Q, where a is the symbol
directly preceding, in the sequence e, the y symbol we are discussing. Besides the
y symbol explicitly indicated in the e above, some of the Si may be y symbols.
However, Ie, ITe, al, . . ., ak and Ze are non-y symbols.
Each 3j satisfies one of the following three conditions:
(1) 3j is a y symbol.
(2) There is a 4 inQ6j JQ6o= Qze) such that 3j is the symbol expe C.
(3) There is a 4 in Q6 l, such that 3j is the symbol loge C.
DEFINITION. We call (9A, V) an e-model, where -A is a differential field of
174
R. H. RISCH
[May
meromorphic functions on a region A and Vis a mapping whose domain is a subset
'D(e, V) of Qeand whose range is -9A, iff
(1) V(le)= 1.
(2) V(Te), V(al),..., V(ak) are complex numbers which are algebraically
independent over Q with V(lTe) = T
(3) For the y symbol following ak we have V(y[-]) is a complex number and
V(y[-])+ V(31)V(y[-])1+...
+ V(81)=O and this is the monic irreducible
equation satisfied by V(y[-]) over Q(iT, V(al), .. ., V(ak)).
(4) V(ze)= I where I is the identity function on A.
(5) If j = y[(((+ eAe yn- 1)+e** +eAn))], then V(Sj)n+ V(A1)V(Sj)n-1 ...
+ V(An)= O
(6) If Sj= expe 4 then V(Sj)= eV(?, where e is the exponential function.
= eV(61).
(7) If S = loge 4 then V(W)
(8) If oc,f are in (I(e, V) then so are a+e/3, -ea and aoe/. If furthermore
V(/)#70 then oc . e/3 is also in 'D(e, V) and we have
V(a+ee) = V()+ V(/3)
V(-eea) =
VW
V(oc*e/3)= V() V(),
V( . /)e = V(oc)/V(/).
It is clear that -Q= Q(IT, V(a1), . . ., V(ak), V(y[-]), I, V(81), ..., V(Sm)). The
only reason we use fields of meromorphic functions in defining an e model instead
of abstract differential fields is because of the properties used in Proposition 2.2.
In the meromorphic case expe 4 does not refer to any solution of 0'- 4'o =0 but a
special one. Similarly, loge 4 refers to a member of a certain countable set of
solutions of 4'-6'4=0.
We note that while, on a region A of the complex plane, there is an uncountable
number of elementary functions, we have invented only a countable number of
symbols to denote them. The paradox is explained by the fact that an efd may have
an uncountable number of e models, all on the same region. E.g., <Ie a1, ..., ak, Ze>
has c distinct e models which are obtained by letting a1,..., ak vary over all sets
of k algebraically independent elements of C. All these e models are differentially
isomorphic. For our purposes, we may treat them as identical. For example,
f e'z log (V\rz) dz and f eez log (Vez) dz are either both elementary or both
nonelementary.
DEFINITION. We say that an efd e is regular if there is an e model (PA, V) such
that:
(1) For each Si of the form y[(((yn +eAi*eyn - 1) +e' * +eeA
V(Qi)n+ V(A1)V(Si)n- 1+ . **+ V(An)= 0
is the monic irreducible equation satisfied by V(Si) over Q(7r,V(al),..,
(2) For each Si of the form expe 4 or loge C, V(Si)is a monomial over
Q(tar V(el), ...I, V(S),- ))I),
Thus, 9A
iS
regularelementaryover Q(7T,V(ocl),. . ., I).
V(-1)).
1969]
THE PROBLEM OF INTEGRATION
175
IN FINITE TERMS
Let -9 be (regular) elementary over K(z) where K is a subfield
of the complex numberswhichisfinitely generatedover the rationalsand z is a solution
of z'= 1. For any region B of the complex plane, there is a subregionA, afield qA Of
functions meromorphicon A which is differentially isomorphic to Q9,a (regular)
efd e, and a mapping V such that (PA, V) is an e model.
PROPOSITION 2.1.
Proof. This can be easily derived from some well-known results in elementary
complex analysis. The existence of -A is also a consequence of Seidenberg's
theorem that any differential field that is finitely generated over the rationals
is differentially isomorphic to a field of meromorphic functions. See [12].
DEFINITION. Let e be an efd. Then Ne={OaE Qe: 3 an e model (GA, V) and an f,
regular elementary over -A such that a E D(e, V) and V(o)= (d/dz)f}.
Note that if we substitute, in the above definition, "elementary" for "regular
elementary" the new set is the same as the old (Proposition 1.4). Proposition 2.1
tells us that f can be realized as a meromorphic function on some subregion of A.
THE PROBLEM
IN FINITETERMS.IS Xe a recursive subset of Qe?
OF INTEGRATION
This problem can be intuitively stated as: Given an arbitrary symbol for an
elementary function, is there some way of choosing the branches for the logarithmic
and algebraic operations involved so that the resulting function exists and has an
elementary indefinite integral? For example, the answer for f (log ez- z) exp Z2 dz
is yes since we can choose log ez = z. Note that the efd
<le, Ze, expe Ze,
loge
expe Ze,
expe (Ze'eZe)>
is not regular. Cf. the example on page 172.
PROPOSITION
2.2. There is an e such that Xe is not a recursive subset of Qe, i.e.,
for which the above problem is undecidable.
Proof. It follows from [2, p. 430] and [1, p. 103] that there is a polynomial
P(w, x1, ..., Xk, Y1, . . , Yk) with integer coefficients such that the set of natural
numbers S, defined below is not recursive.
s E S " integers
= G(s,
X1,...,
Xl,...
Xk such that P(s, Xl1.,
Xk)
Xk,
.
2
Xk)
= 0.
Let Fj=(1/2Ti)(log exp zi-zi), j= 1,..., k. It is clear that for each s, H(s)
= G(s, F1, . . ., Fk) is a symbol for an elementary function and that f H(s) exp z2 dz
is elementary for some choice of the branches of the logarithms iff s E S.
Thus, for any e such that Qecontains a symbol H(s) exp z2, Xe is not a recursive
subset of Qe
It would be nice to have a proof of the above proposition without having vTplay
its special role.
We will get around the difficulties indicated in Proposition 2.2 by choosing our
symbols so that no matter what branches of functions, which are built up using
176
R. H. RISCH
[May
logarithmic or algebraic operations, are chosen or what region A we use, in
defining our e model, the integrability of a function represented by a symbol
depends only on the symbol. This is achieved by restricting ourselves to regular
efd's. The next proposition which can be derived from 1.3 summarizes the properties of these efd's.
2.3. Let e be a regularefd, (GA, VA) and (SB, VB)e models.Then
PROPOSITION
0(e, VA)-= ((e, VB) and we now write this set as (De. The correspondenceVl(cc)
VA,e)
V2(a), for a E be, is a differential isomorphismof -2A and GJ, Let Ze(PA,
VA,e(a)}.
-2A
such
that
(d/dz)f=
3
elementary
over
f
which
is
regular
E
(De:
an
={la
ThenEe(-A,
VB,e) =EeVA,e) =Ye(-B5
Also note that for a regular efd e, q)e is a recursive subset of Qe and the relation
defined by c_ e/ if V(a) = V(8) for some e model (9A, V) is well defined and
-e,
recursive. Thus we can tell whether or not two elementary functions are equal when
they are both in a common model of a regular efd.
For a regular efd e, ?e is a recursively enumerable subset of (D,. This follows
from the Strong Liouville Theorem.
It should be pointed out that if e is regular and has e models gA and -9,, then
it is not necessarily true that the functions of !9A can be continued analytically
into the corresponding functions of 9B. E.g., log x2 represents two distinct monogenic analytic functions. Thus, the existence of the differential isomorphism of
Proposition 2.3 could not be proved by analytic continuation.
3. Integration of elements of pure monomial extensions. We saw, in ?2, how to
rigorously formulate the problem of integration in finite terms. We intend now to
solve a portion of it with the theorems:
(1) Let 4' be the set of those elementaryfield descriptions e in which each Si is
either of theform expe 4 or loge 71 (i.e., no Si is a y symbol). Thenthe subset of regular
efd's of & is recursive.
(2) Let e be a regular efd of &, then Ye is recursive.
Roughly, this means that we can handle any elementary function which involves
just the rational operations, exponentials and logarithms, providing these latter
two cannot be replaced by adjoining constants and performing algebraic operations.
E.g., we can look at
fexp (zl
z)
log z dz
but not at f exp (1 +4 log (z3 - 1)) dz with our algorithm (as it is given here).
In his book [8], Mordoukhay-Boltovskoy makes a pretense of showing that
" given an integral of a rational transcendental function, then it is always possible,
by means of a finite number of algebraic operations, to express it in a finite number
of terms or assert its inexpressibility in finite terms" [8, pp. 244-245]. What he
means is our theorem (2) above.
Using the concepts and terminology which we have developed, his method is as
1969]
IN FINITE TERMS
THE PROBLEM OF INTEGRATION
177
follows: Suppose we are given a function P(0)/Q(0) where P(0), Q(0) are els of
29[0], Q9=K(z, 01,. .., an-l)) 0=0n, 0i a monomial over K(z, 01,...,
i-1). One
can show that if elementary,
[P(0)_ f
J{(0al)kl
.fQ(0)
P(O)
...(0oan)kn
R1(0
+
cjlog (0-oj)+S,
where one factors the denominator of Q over the algebraic closure of 9, R(0)
E 9[0], ci E K, and S=Edi log Si, di E K, Si E Kg. After differentiating and clearing
denominators, one gets AR'+BR+>im=1 ciCi+S'D=O, where A, B, CT,D are
known polynomials in 0, the ci are unknown constants and S' is an unknown
element of 9. A bound k can be found for the degree of the polynomial R. Thus,
+ ao into the preceding equation, one gets a simulsubstituting R = akOk+ +
taneous system of first order linear differential equations for the ai's. These equations have the parameters ci and S'. One must determine if these equations have a
solution ao, . . ., ak with ai E 9, ci EKR,S' E 9.
At this point, Mordoukhay-Boltovskoy waves his hand. (This is not to imply that
he is clear and convincing in getting this far.) He refers to Liouville [6] and Sintzov
[13] for methods of solving linear systems. However, these writers only discuss the
problem when the coefficients are rational functions of z, and the results do not
generalize in any obvious way. Besides, Sintzov's method of solution (?23, beginning on p. 146 of the August issue) is erroneous. We will show that the testing of the
systems, which arise in the integration of elements of Q9(0),for solutions in their
coefficient field can always be reduced to the testing of a succession of single first
order linear equations. Mordoukhay-Boltovskoy does not seem to have been
aware of this important fact. For example, in [8, p. 160] he reduces a system of
two first order equations to a second order equation.
Also very troublesome in this method are the parameters ci and S'-especially
the S'. It is not clear how to handle them. In the trivial examples MordoukhayBoltovskoy works out, no difficulties arise. But there is no certainty that this will
always be so. The problem of handling S' becomes manageable if we first decompose
both our integrand and the functions occurring in the integral into partial fractions
over 9 and then find the conditions on these unknown functions. This idea is found
for the logarithmic case in Ostrowski's paper [9].
We will not be as formal as we were in ?2-e.g., we will usually not make a
distinction between the terms of a language and the functions they represent. Also,
we will not use the language of recursivefunction theory. In ?3, we work exclusively
with fields K(z, 01, . . ., an),K a field of constants which is finitely generated over Q,
z a solution of z' = 1 and each Oia monomial over K(z, 01, . . ., Oi-1). Proposition
2.1 tells us that we have a regular efd for this field and so are able to decide if two
elements of it are equal.
178
R. H. RISCH
[May
We first dispose of the problem of factoring a polynomial in Onover
K(Z,01,.
*, On-A
K= alg. closure of K.
PROBLEM. Let K be finitely generated over Q. Let x1, ..., xn be algebraically
independent over K. To factor a polynomial P(x1, .. ., xn) E K[x1, . . ., xn] into
irreduciblefactors q1, . . ., qkover K. To find an equation, p(b) = 0, which is minimal
for b over K, such that q1, . . ., qk are in K(b)[x1, . . ., xn].
SOLUTION.We refer to results proved on field theoretic algorithms in [5, ?19] and
[14, ?42].
By the Kronecker trick [14, p. 135], we transform P(x1,. ., xn) into P*(t), a
polynomial in a single indeterminate with coefficients in K. Since K is finitely
generated over Q, we can factor a polynomial over it in the manner described in [14].
Thus, we can factor P*(t) over K to see if it has multiple factors or not. If it has,
we replace it by a polynomial without multiple factors, P*(t), having the same
irreduciblefactors as P*(t). Next we find a primitive element b for the splitting field
of P*(t) over K [5, p. 66, Satz 1]. Then we factor P*(t) over K(b) [5, p. 68, Satz 3],
or [14, p. 136]. Then we convert back to a factorization of P(x1, .. ., xn) using the
Kronecker device.
COROLLARY.We may factor a polynomial P(xn) E K(x1,...,
K(Xl, . . ., Xn-1).
xn 1)[xn] over
In our integration algorithm, the polynomial P(On) to be factored over
K(z, 01,
is a factor, irreducible over K(z, 01, .
. .,
-.- -,
On-i)
On-1)
of the denominator of the
M(On)/N(On) E-K(z, 01, -
,
an)
which we wish to integrate. To use the language of ?2, if we start with a regular
efd e=Kle 1,a,,..9
which describes K(z, 01, ..., On)
[k, Y/[], Ze, Si ...
* Xn>
(V(S3)= Os),we now work with the sequence e' = <1ex * * * Xn, Yb1[-]> where
V(Ybl [
]) = b,
is the primitive element obtained according to the preceding discussion for P.
In the course of applying the algorithm to a particular function, we may make
further extensions to our original efd e by y symbols Yb2[-],. .., Yb,[-] in order to
obtain symbols for the factorization of other polynomials which come up.
LEMMA.Let P1 =0, ... XPm= 0 be a set of simultaneouslinear algebraic equations
withcoefficientsin K(z, 01, ..., On).Thenone canfind, in afinite numberof steps, a system Y of simultaneouslinear equationswith coefficientsin K such that a (kl,..., kr)
e Kr satisfies P1 = O, . . ., Pm= O if (k1, . . ., kr) satisfies S.
1969]
THE PROBLEM OF INTEGRATION
179
IN FINITE TERMS
Proof. It suffices to show this for a single equation P = 0.
+ (R,(z)/Sr(z))xr,
For n = 0 our field is K(z). Then P = (Rj(z)/S1(z))xj +
Si,Ri, els K(z). (k1, . .., kr) satisfies P iff (kl, . . ., kr) satisfiesR,(z) Il , 1 Si(z)xl + *
+ R,(z) Fli r Si(z)x,.
By collecting like powers of z we get our desired system Y.
Assume the result is true for n - 1. Then
P = (R1(0n)/S1(0n))x1
+ * * + (Rr(0n)/Sr(0n))xr.
We clear denominators and collect coefficients of like powers of 0n. We get a set
of equations Q = o,. .. Qm=0 with coefficients in K(z, 01, . . 0,On-1) which
are satisfied by a (k, . .., kr) in Kr iff (kl, . . ., kr) satisfies P. By the induction
assumption, we replace each Qiby its equivalent system 9i. Then M= Uim=1 9' is our
desired system.
We can now state the
Y=K(z, 01, .., 0n), K is a field of constants which is
finitely generated over Q, z is transcendentalover K and a solution of z' = 1, each Oia
monomial over K(z, 01, . . ., oi- D.
Y Thenone can determinein afinite numberof steps whetherthere are
(a) Let f E.
evo Jvielsof KF;i= 1, . . ., mandc1,. . . Cmin Ksuch thatf=[vo?+ 2sm=1ci logvi]'.
If they do exist we can find them.
(b) Letf, gi i= 1, .. ., m be els of F. Thenone canfind, in a finite numberof steps,
hl, . . . hr in Y and a simultaneous set Y of linear algebraic equations in m+r
variables, with coefficients in K, such that y' +fy = Jim=1 cigi holdsfor y E Y and ci
els of K if y = jr= 1 y,hi whereyi els K and cl, ... XCm,Y1,..., Yrsatisfy Y.
MAIN THEOREM. Let
Part (b), while not of particular interest in itself, serves to facilitate the proof,
which is by induction on n:
n =O
Y = K(z).
(a) Using the Strong Liouville Theorem, we may write, if f f (i.e., a g such that
g'=f) is elementary:
=
A1,k1
Bk+1zk+l +
+Ao
AkZk+*
+
A1,1
i
l+ p k5
{+k+l
-
}
+B{1
B1kjl+-
+
*+Blz+Bo
B1,O
Pi
skS
-
l
+
+B
Pi
1
J
O
where the Ai's and Bi's are in K: the Ail, Bi,j and pi's are in K[z], the pi's being
monic and irreducible; degree Ai,j and degree Bi,j are both less than degree pi;
Bi, /pi= 11%1 cjq', j/qi,j i= 1,...,
s where pi=JJ_
1 qi,j is the factorization
into monic, irreducible (and therefore linear) factors over K.
of p,
[May
R. H. RISCH
180
We obtain the following conditions on the B's:
Ak =(k+l)Bk+l
Ao
=
B1
0 =Bo
kl - 1B,
kl A
pi
,=i
,)=i
1i
JB
]'
Pi
By [14, p. 88], we can uniquely solve for
giving Al,kl=-(kl-l)piBl,k_l(pl).
R and S when they are subjected to the conditions; Rp + Sp' = Al,ki, R, S els K[z],
degree R < degree p', degree S < degree pl. Thus, setting
-(ki- I)Bl,ki -1 = S,
we determine Bl,ki
-I.
into (+) and obtain a new relation:
We substitute Bl,kl-
jePiIjfE
1= 1
,=i
pl JPPi
Pi
Continuing in this manner, we determine Bl,k1 -2, ..., Bi,l. Next A 9** )=Bl,o
which yields
Al,l
p,
) =
E
j=l
C17Lq,j=
ql,j
ClJ
J=l ql,j
and we uniquely determine the cl,j's by breaking up A14* >Ipl into partial
fractions over K.
The other Bi,j's are determined in the same manner. We note that the conditions
put on the B's are always satisfiable so ff is always elementary-as is well known.
Note that no polynomial need be factored over K until the last moment. This is
important in practical considerations(2).
It should be noted that we can use the above conditions to tell whether the integral is in a certain definite form. E.g., to see whether there is a vo E J such that
vI =f, we set Bl,O=0 . . ., B, =0 and obtain the necessary and sufficient conditions A(** ' , . . ., A'** ' .
To determine for a given v, cE whether there is a vo E Y and a c E K such that
(2) Professor George Collins of the University of Wisconsin has pointed out to us that
we can refine the integrationalgorithm so as to avoid factoring into irreduciblesuntil the logs
are to be determined. Using the Euclidean algorithm we can factor any polynomial in K(z)
as follows: P(z)=qlq 2 . qk where the qi are relativelyprime and each is a product of irreducible
polynomials. Everythinggoes through when we replace the irreduciblep,'s used above with the
qi's. It is an interesting open question to determinewhether we can prove the main theorem,
sans the last sentence of (a), with the added stipulation that only the rational operations need
to be appliedto elements of Y in performingthe decision method. Proposition 1.1 indicatesthat
we cannot actually find the vi, i> 0 without factoring over K.
1969]
THE PROBLEM OF INTEGRATION
181
IN FINITE TERMS
(vo+ c log v)' =f, we writev, = a 17= 1pfii,a E K, ji E Z. Then log v1= ji ji f (p/pi).
vi can have no other irreducible factors besides the pi in the denominator off.
j=cjip4, i= 1, .. ., s.
(b) We set
We plug in cjipi for each Bi,o and see if we can satisfy A**
A
p
Pkk
Pi
m
B
Pil
k
j=
C
1
PiL
wk
where pi is a monic, irreducible element of K[z], -aj=ord,,y,
-/i=ordp1f,
=
els
C
A,
B,
-yi Minj ord,, gi;
K[z].
The gi and yi are given to us. We will show that ci > 0 implies that gi > 0 or yi > 0
and find a bound, which is independent of the choice of the cj, for each ci > 0.
?..
The pi-adic expansion of y begins: A1/pi +
The pi-adic expansion of f begins: B61/pi?+?*..
The pi-adic expansion of J.= 1 cjgj begins: CyQ/pyi
where the first nonzero
+
term, in this last expansion, depends on the choice of the cj.
Thus, by plugging these expansions into y' +fy = Cjgjwe get
++a +
= Cyi/py +
-a1p'Aa,/p'?i
B,A/p1i,1pi+.+.
Note that pi a-cip'A,, and pi BBiAa1.Looking at the highest power of l/pi we
must have either
a(+1 _ Yi5
ai+3i
ai+l = aic+ i > Yi.
or
_ Yi,
In the third case - ap'Aa +
=
BO,Aa,=? (pi), -aip'+ BO 0.
=
c*
Thus, we set,
Max (y' 1, yi -/gi, B11/pt if this is an integer). Then y=
a*= YIP where Y and P are in K[z].
Y/p
pk
Substituting this expression for y into y' +fy = T 1 cigi and clearing denominators we find R, S, Ti in K[z], which do not depend on the ci, such that y'+fy
J=1 ciTi. Let
Jt,nt=cigi iff(*): RY'+SY=
y = yaz"+y1z-lZ-1+*
R=
r1oz,+
+Yo,
ro,
***+
S = sYzY+ **+so,
m
E c,T1 = t6z" +
+to,
where r13,sy=#0, and 8 = Max= 1 (degree Tj), ti is a linear homogeneous element of
K[cl, . . ., Cm].
By substituting in (*) we get:
(t)
(r13z1+ .. . )(qy,z - 1 +l . . ) + (syzy+.*)(yza
+*
)
=
.
(t6z6+
When definite values in K are chosen for cl, . . ., cm we must have
a?+/-1
_ 8,
a+y
? 8,
or
- cx+g-1
= a+y
> S.
182
R. H. RISCH
[May
If the third case holds, then ay,rn+ya,sv=0. Let a=r=Max(
-f+l1, 8-y,
- sy/ro if this is an integer). Take hi = z'/P, i= 0, . . ., r. By multiplying (t) out and
equating powers of z, we get our simultaneous system cY which satisfies the conditions of part (b) of this theorem.
Induction step. We assume that both (a) and (b) of our theorem hold for
q=K(z, 01, . . ., On-l) and prove it for J= g(6), 0= On.Besides (a) as stated, we
assume that the simpler variants, which occur when some of the ci and vi are given,
have been established-see the discussion at the end of the proof of (a) for the
K(z) case.
(a) Case 1. 0 =log -. We again apply the Strong Liouville Theorem and the
breaking up off and the vi's into partial fractions to obtain:
Ak6k+
ki
CAs,ks+
+
4Al,
Al,k1
log Dj
Bk+lOk+l?.
+***+B0+2d1j
* ** +Ao0
B
+
B1,k1i1
iki
B
l???
+AS1
+
+Bs,ks-1+
J
Bs,O
.+Bsel+
where the Ai's, Bi's and Dj's are in 9; the Aij's, Bij's and pi's are in g[6], the pi's
being monic and irreducible; degree Aij and degree Bij are both less than the degree
of pi; Bi, /pi=
cijqjlqi ,j i= 1, . . ., s where pi =H 1 q,,j is the factorization of
p, into monic, irreducible factors over Kg; dj els K.
The discussion (page 169) preceding Liouville's Theorem tells us precisely what
terms on the right-hand side will differentiateinto each of the A's. Thus, we obtain
the following relations for the B's:
OB
Ak
=
(k + 1)Bk+ 1,''I
SoBk+leK.
+ Bk.
Thus fAk
=
Bk+(k+
I)Bk+
1 log9 7
By the induction assumption, we can determine whether there is a Bk E 9 and
Bk+1 in K such that the last relation holds. Bk+1 will be uniquely determined since
log - = 6 is a monomial over 9. Bk is determined up to an additive constant. We set
Bk=Bk+bk
where Bk is fixed.
= kBkq'l-q+
Ak-1
Bk-l-
Thus f (Ak-1-kBkq'1/7I)=Bk1+
kbklog19. As before we determine bkeK
Bk-l
up to an additive constant, etc.
Ao = Bl'lb7 +Bo+
di (log Dj)'.
Thus
J(AO
=
Bo+ bi log ?+ Edj log Dj.
This last condition is simply that f (Ao- B'/N)
is elementary over 9.
and
THE PROBLEM OF INTEGRATION
1969]
183
IN FINITE TERMS
We determine the B2,j'sas in the K(z) case. Then At** )=Bi,o givingAi*.**)/p
= =_1 ci jqi j/qi,j. The ci,j are uniquely determined, if they exist, from the partial
fraction decomposition of A(i* )/pi.
We note that if in the course of this investigation we find that one of these
conditions is not satisfied, then f does not have an elementary indefinite integral
and we can terminate the proceedings, right there.
Case 2. 0 = exp .
...+AlO
AkOk+
=
+
A
++i
+
|
+B-,O-m+
+ A1,
k
A+.+
***+B_10-1+Bo+:E
+..
+B1,k
+P=
PS
*+B10
+
BkOk
** *+A-10-1+Ao
+A-,O--+
As,l
Ps
P
+
P
+
k-
1
+B1,+
Pi
Pi1
dj log Dj|
Pi
+B+lp PsJ
fBspo
Here, as usual, Bio/pi=j;1=l ci,jq'jlqij.The conditions obtained are
Ak =B+k'Bk
A1 = B' + 4'
B'_1-'B_1
A-,
A-m =
B' m-M4'B-m.
By our induction assumption for (b) of our theorem, we can write Bi =
>j yjiTji,
Tii E 9 where the yji satisfy a linear system Y. This must determineeach B, uniquely
since two solutions would yield that the homogeneous equation B'+ i4'Bi = 0 has
solutions in Bq.This yields a contradiction, as is well known to the reader.
We determine Bl,k1 - 1, .. .., Bl,l; B2,k2- 1, . . ., B2,1;...
; Bsks - 1, *, Bs, in the
same manner as previously. Then we obtain:
E
A
=1
From this, A*.
-+AO
p +B
+(dj
=
Pi
log Dj)'.
i_l_P
-Bi, o(pi). Let degree of qi,j=ni. Then
B+n______
A(*_______
Pi
f
=
Pr
Pi
j=l
qi,j
We determine whether such cij exist, and find them if they do, from consideration
of the partial fraction decomposition of the functions in this equation.
Then AO-;'2 >jnicij (Bo+ 2 dj log Dj)' and our problem is reduced to
telling whether f (AO- ' 2 j nici,j) is elementary over 9.
(b) Here the general method is the same as that for the K(z) case. The problem of
determining bounds is more difficult.
R. H. RISCH
184
[May
Case 1. 0=log r. For y=A/plpx'*.p..c we determine the bounds a*, i-I, .., k
in the same way as in the K(z) case. So we have y= Y/p'*.l p"= Y/P, Y, P els of
$[0]. As before we reduce to the equation:
m
(1)
c1T1.
RY'+SY=
j=l
We set
y=yaa+y
+Yo,
a-la-i+...
R =roO
+ * * + ro,
S=sy0Y+.
.+so,
m
m
?
ci t t 606+
ry i S. :A 0, 8= Max(degree
T).
Ya,
1
* to
a~~~~~~~~~~~~~~~~i
Each y,, rj, sj is in Q9, tj is a linear homogeneous element of 9[c1,
Substitution in (1) yields
(r2 03+ r_ 10, 1+ .)(y'O
+ (y_1? + a(')ya)) O
(SY?sF-1.0Y-1+.
For definite values of cl,...,
:x+ <
when y' =# 0:
a+,y
Cm].
*)
+1?
)
.)(ya?yOl+.
t6G6+.
=
+t0
cm chosen in K, we get
}and when y>
+j
a
,
?-1
8,
aa+y
_ 8+1
or a+P = a+y >
...
< a
or a+?-l-
8+1)
a+y >
When y' # 0 the third condition occurs when
+
= 0 and r#y_ 1
+svya
r#y, syYa
y
+rB_l
+(a(-9'/h)rn+s,- )y,=OLetting ya -1
= vya,
}
J.
-
v E 9 we have that
r,y.v' + (rBya+ syyo)v + rf 1y' + (sy-,1 ? c('/1/7)rfl)ya
v r- 1sj/r +s,.1/r,3+a(q'/'q)
Jrisv-rsr-
=
0,
=0,
v?+a log .
r2
If y
=
0, the third case occurs when re(ya,-1 + a(' /'q)y) + syy
J*sy=
rgi
a-1
Ya
-a
0.
log q
In both instances the induction assumption for part (a) of our Main Theorem
tells us we can determine the constant a so that the integral is of this form. We let
= Max (8 + 1-fi; 8+1 - y; the a's determined in the third case, if they are
integers).
1969]
THE PROBLEM OF INTEGRATION
185
IN FINITE TERMS
We now rewrite (2) in the form:
(rASA+r-1OA
(3)
+* * *)(yA0A +(Y'-
+(SAOA
=
tA+,aO
+SAX-lOA1+
+?,(<R/2) Y )0A1l+*
0A
+.Y
*(Y
**
1+
10A
. * * +to
++
so that at least one of rA, SA, and tA+u is not equalto zero.
We complete the proof of Case 1 by showing, by induction on t, that there is a
simultaneous system of linear algebraic equations, Y, in m+p+q variables, with
coefficients in K, and h1, . . ., hp, jl, ... , j, in g[6], all of which are of degree ? t,
*) satisfies (3) if (y60?+
such that (yO?
)= = dihi+ 2q=leiji where
di, ei are in K and cl,..., Cm,d1, ..., dp, e1,..., eq satisfy Y. The functions in the
statement of (b) of the Main Theorem will be hl/P, .. , jq/P.
In this induction we assume rA or SA#\:AO.If rA=SA=**
= rA + 1=SA
+1=
while rx - I or sA - I =AO, then the system
tA =
0
tA-I-1
=
0
is equivalent, by the lemma, to a linear system in the ci with coefficients in K. Then
we proceed as below with rA_ - SAs
taking the place of rAand SA.
= O. Here (3) takes the form
(rAOA?
+. .)y? + (sAA?+. . .)yo = (tA\A+?.. .
(4)
Looking at the coefficient of OA, we get
rAyO+sAyO = tA
(5)
By the induction assumption for n-I for (b) we get that (5) is equivalent to
els Q, and a linear system Y' with coefficients in K. Plugging this
expression for yo into the remaining Aequations for yo which can be obtained from
(4), yields a simultaneous set of linear algebraic equations in d1, . . ., dp, Ci, . . ., Cm
with coefficients in q, which by the lemma to this theorem is equivalent to a linear
system Y' with coefficients in K, in the same variables. Take Y= ' u -Y' and
ji=0, i= 1,. . ., q and we have our result for yt=0.
Assume the result is true for y - 1. Then the leading coefficient y. satisfies
yo = If= 1 dihi, hi
(6)
rAyU+sNAy
=
tN+g
By the induction assumption for n-I for part (b) we have that (6) is equivalent
to yA= jv= 1 dihi,hi els 9 and a linearalgebraicsystemY" in cl, . . ., cm,di, d.., ip.
Substituting the expression for Y. into (3) we get
(r N0A +
*
*)[Y'
(7)
_10A-1
+ (Y'
_ 2 +(t-1)(
+(SNOA+SA-1OAl+*
=
(UA+u1
A
/Nl)yg
*)[Y_1Gl'-l+y
_1)09
+
0A-2+.
]
]
+1 .
where the ui's are linear homogeneous elements of K[cl, . ., cm
., dp].
[May
R. H. RISCH
186
By our induction assumption for , -1, we have that there are jl'... , jq in 9[4],
)
9=1 eiji and a
with degreeji < - 1 such that (7) is equivalent to (y, 160 +
linear algebraic system Y"'in cl, .. ., cm,dl, . . . Xdp,el,.. ., eq.We take Y = Y'u JY"
and by writing (y,u+ ? =- )
dihi6g+19=1 eiji we have found conditions
equivalent to (3). This completes our induction on ,.
Case 2. 0 = exp 4. Here, in distinction to the K(z) case and Case 1, degree pi
=degree pi so things are slightly different. We set
B
A
Pil
f
a'
al..
*Pkk
pil
..
m
z.g
Pok
Pk
__
c
..Pk
. .pk
Pg -
and show that ci > 0 implies that either gi > 0, yi > 0, or pi = 6 and we find a bound
for each ai.
Looking at the pi-adic expansion of the terms involved in y' +fy = 2m 1 cjgj,
we get if pi {P':
- aripi'A
i +1+
al/p
Since pi -aip'A
... =
W+ **
flA ai/p~i+i
.* +B
yi
,, we have that either
ai +3i
(Xi+1 _ Yi,
or
<yL,
ai+l
> Yi.
= ai+/i
In the latter case
O(pi),
-aipi'A,,,+BO,A,,
- aipi+BO
-
aipi + B
0 (pi),
-ainil'pi
=
where ni = degree Pi.
So we set here
if this is an integer).
as* Max (yi-1, y - li,
B,0,/(p'-ni`'pi)
pi Ip' iffpi==
(see page 169). Thus, here, the pi-adic expansion runs
+
i+
Aoci
pioci
B
3
~~Pi+i
=
+
yi
Yi
Then
aj _ Yi,
as+/
? Yi,
or
ai
= ai+/i
> Ys.
In the third case (A',/Aai) - ci' + Bo= 0 and by our induction hypothesis (for (a))
we determine if there is an a E K and an Aa, in 9 such that f Bo= ai;-log Aa,.
We take here
ca*
=
= Max (yi, yi - gi, the ci determined in the third case if it is an integer).
Setting y= Y/paL .pat = Y/P as before, we obtain the equation R Y'+ S Y
2g= 1 cjTj. Equation (2) of Case 1 is replaced by
(r6+0
* **)[(ya+ac'ya)Oa+
]+(sy0y+
*
*)(ya06+*)
=
t50+***
+to.
THE PROBLEM OF INTEGRATION
1969]
187
IN FINITE TERMS
For definite values of c, . . ., cm chosen in K, we get,
a+: - 8,
or
a+y _ 8,
a+g=
a+y >
In the third case roy' + (a.r'rf+ sy)ya,= 0. Then f sy/r6= - a-log ya. By our
induction hypothesis (for (a)) we determine if there is an a E K and a y,, in 9 such
that the above holds.
Thus, let M=Max (S- g; 8-y; the a determined in the third case, if it is an
integer).
With this bound for the degree of Y determined, we proceed in the same manner
as in Case 1 to find the desired functions and linear system.
This completes the proof of the Main Theorem.
From this result it is easy to see how to complete the proof of the two theorems
given at the beginning of ?3. For (1) we must show how to decide whether a given
efd in &, e=<le, i, ..., v,..., n> has an e model (Q, V) in which V(Si)is a
monomial over Q(V(aj),..., V(S,- )). By Proposition 1.2, it suffices to be able to
Q is f (f'/f) in 9 and is ff'= (1/m) log h for some m EZ
decide for a given f 9,
and h E 9 These are simple variants of (a) of the Main Theorem.
E.g., for the problem of telling if for a givenf E K(z, 01, . . ., 0.), where On= exp ;,
is ff'= (1/nm)log h, h E K(z, 01, . .., On),m e Z, we write h = Okpki . .pkrD, pi a
monic and irreducible member of K(z, 01,..., On-1)[On],degree Pi = ni,
D e K(z,
01,...,
On-i).
Then f' =Ao+All/pl+
+Ar/pr=[qO +(1/m) log
where qi= kilm.
For i > O, qip'= Ai (Pi) determines each qi.
AO =
q0
'-
Thus, the problem is reduced to "is
oE Q?",
- +
m D
E
J(Ao->X=1
D+qj
logp1+
+qrlogPr]'
n
qiniC')=qoC+(1/n) log D where
etc.
The proof of the second theorem given at the beginning of ?3 is clear from (a) of
the Main Theorem.
We now turn to some simple examples of the algorithm.
Investigate. f exp z2. We show first that <Ie, ze, expeze eze> is regular. 2z
=(1/n) log D, for D E Q(z), is clearly impossible. Thus exp Z2 is a monomial over
$
Q(Z).
Z2 = B1 exp Z2+BO,
Texp
Bo = 0
B1,Boels Q(z).
so Bo = constant.
B +2zB1 = 1,
a* = Max (1,
B1 = ylz+yo.
-
B1 must be in K[z] and thus B1 = yyz +
1, third case impossible).
+yo.
188
R. H. RISCH
y1+2z(ylz+yo)
01
o
2y=
y=
Y,=
1
=
[May
1.
This system can not be satisfied so the integral is not
elementary.
Investigate.
j'[2z exp Z2 log
z+e
z
+(2/z) log z+ I/z+ 1]
[iogz)-2
First we must show that log z is a monomial over Q(z, exp Z2). If not, then f (l/z)
must be in Q(z, exp Z2). Then f (l1z) =AO, Ao e Q(z). But this is clearly impossible,
as we see from looking at the partial fraction decomposition of AO.
Our integral must equal
B2(og
Z)2
+ B1 log z +Bo +:dt log Dt +
B 1_ + +
(log Z)2+ Z
log [(log
Z)2 + Z],
B' = 0 so B2 is a constant. (2/z)B2+ B' = 2z exp Z2. Thus,
B1 = (j2z exp z2)
2B2 log1z = exp z2+b1-2B2 log z,
b, a constant. B2 = 0 since B1 must be in Q(z, exp Z2). Thus, B1 = exp Z2 +
bl.
z
exp z2++ b +~.og
/-exp
exp_ b1 + B'o+ (Z di log Di)' =exz
z
z
z
Bo + dt log Dt + b1 log z
constant
=
so b1 can be taken to be 0.
2
-B1, (log
log z-2
z+ )
((log
We must solve A [(2/z) log z +1] + B[(log Z)2 +
B=O.
By the Euclidean
algorithm,
Z]
=
Z)2 + Z).
log z -2, degree A < 2, degree
Thus B,1 =-log
A = log z, B=-2/z.
z. Sub-
stituting this into
log z-2
(2/z) log z+1/z+ I
B,1_
[(log z)2+z]2
(log z)2+z
(logz) +z
+ C log [logZ)2 + Z]
we get c1((2/z) log z + 1) = (2/z) log z + 1. Therefore, c1= 1. So our integral is
exp Z2 log Z
+loglog
zZ)2~+log [(log
Z)2+z]Z
+ Z].
1969]
THE PROBLEM OF INTEGRATION IN FINITE TERMS
189
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