Impossibility theorems for elementary integration
Brian Conrad
R
2
Abstract. Liouville proved that certain integrals, most famously e−x dx,
cannot be expressed in elementary terms. We explain how to give precise
meaning to the notion of integration “in elementary terms”, and we formulate Liouville’s theorem that characterizes the possible form of elementary antiderivatives. Using this theorem, we deduce a practical criterion for proving
such impossibility results in special cases.
R
2
This criterion is illustrated for the Gaussian
integral e−x dx from probR
ability theory, the logarithmic integral dt/ log(t) from the study of primes,
and elliptic integrals. Our exposition is aimed at students who are familiar with
calculus and elementary abstract algebra (at the level of polynomial rings F [t]
over a field F ).
1. Introduction
The Central Limit Theorem in probability
R x theory2 assigns a special significance
to the cumulative area function Φ(x) = √12π −∞ e−u /2 du under the Gaussian bell
√
2
curve y = (1/ 2π) · e−u /2 . It is known that Φ(∞) = 1 (i.e., the total area under
the bell curve is 1), as must be the case for applications in probability theory, but
this value is not determined by computing Φ(x) as an “explicit” function of x and
finding its limit as x → ∞. It is a theorem (to be made precise later) that there
is no elementary formula for Φ(x), so the evaluation of Φ(∞) must proceed by a
method different from the calculation of anti-derivatives as in calculus. One uses a
trick from multivariable calculus to express Φ(∞)2 in terms of an integral over the
plane, and this planar integral is computed by a switch to polar coordinates. Note
also that by a change of variable, the study of Φ(x) can be recast as the study of
2
the anti-derivative of e−u (in terms of which the identity Φ(∞) = 1 takes the form
R ∞ −u2
√
e
du = π).
−∞
Let us give another example of an important indefinite integral that lacks an
elementary formula. In number theory, there is much interest in the study of
the step function π(x) = #{1 ≤ n ≤ x | n is prime} of a real variable x that
counts the number of primes up to x. The prime number theorem provides a
1991 Mathematics Subject Classification. Primary 12H05; Secondary 12H20.
Key words and phrases. Elementary integration, differential field.
The author was supported by the NSF and the Sloan Foundation during the writing of this
paper, and many thanks are due to the Clay Mathematics Institute for providing the opportunity
to speak on this charming topic that deserves to be more widely understood.
1
2
BRIAN CONRAD
remarkable asymptotic formula: π(x) ∼ x/ log(x) as x → ∞. That is, as x → ∞,
π(x)/(x/ log(x)) → 1. Asymptotic quantities have relative error that tends to 0,
but the actual difference can explode; for example, x2 ∼ x2 + 3x as x → ∞, even
though the difference 3x explodes in absolute value. In this sense, one can seek
better asymptotic approximations to π(x), and the first in a sequence
of better
Rx
such approximations is given by the logarithmic integral Li(x) = 2 dt/ log(t) for
x > 2. (It is an exercise to show Li(x) ∼ x/ log(x) as x → ∞.) The prime number
theorem was first conjectured by the 14-year-old Gauss in the form π(x) ∼ Li(x)
as x → ∞. As with the Gaussian integral from probability theory, the logarithmic
integral likewise admits no elementary
formula.
R
R Observe also that with the change
of variable u = log t, we have dt/ log(t) = (eu /u)du.
R
R
2
The above examples show the interest in computing e−u du and (eu /u)du,
and in both cases we have said that there is no elementary formula for such antiderivatives. How can one prove such assertions? Of course, to prove impossibility
results of this sort it is necessary to first give a precise definition of “elementary
formula”. Roughly speaking, an elementary formula should be built from the familiar operations and functions in calculus: addition, multiplication, division, rootextraction, trigonometric functions and their inverses, exponential and logarithmic
functions, and arbitrary composition among such functions. For example, the function
(1.1)
πx2 − 3x log x
p
ex − sin(x/(x3 − 7))
should qualify as an elementary function on an open interval where it makes sense.
It is a minor but important technical issue to keep track of the interval on which
we are working; for example, the expression 1/(x2 − 1)1/6 has two possible meanings, depending on whether we work on the interval (−∞, −1) or (1, ∞). We will
generally suppress explicit mention of open intervals of definition, leaving it to the
reader to make such adjustments as are necessary in order that various algebraic
manipulations with functions make sense.
Liouville proved that if a function can be integrated in elementary terms, then
such an elementary integral has to have a very special form. For functions of the
2
form f eg with rational functions f and g (e.g., e−u with f = 1 and g = −u2 , or
eu /u with f = 1/u and g = u), Liouville’s theorem gives rise to an “elementary
integrability” criterion in terms of solving a first-order differential equation with
a rational function. This criterion is especially well-suited to the two motivating
integration problems considered above, and in each case we work out Liouville’s
criterion to deduce the asserted impossibility result (i.e., that neither the Gaussian
bell curve integral nor the logarithmic integral are elementary functions). We also
briefly discuss another example, for moreR advanced
readers: the non-elementarity
p
of elliptic integrals and more generally dx/ P (x) for polynomials P (X) with
degree ≥ 3 and no double roots.
The reader who wishes to follow up on further details of the theory described
in these notes is encouraged to read [1]; our proof in §5 of Liouville’s “elementary
integrability” criterion for functions of the form f eg with rational functions f and
g is a variant on an argument in [1] but it is written in a manner that avoids the
abstract formalism of differential fields and so is easier to understand for readers
who do not have extensive experience with abstract algebra.
IMPOSSIBILITY THEOREMS FOR ELEMENTARY INTEGRATION
3
Notation. We write R and C to denote the real and complex numbers respectively. The notation F [X] denotes the polynomial ring in one variable over a field
F , and we write F (X) to denote its fraction field (the “rational functions” in one
variable over F ). Similar notation F [X1 , . . . , Xn ] and F (X1 , . . . , Xn ) is used with
several variables. For example,
we will work with the polynomial ring C(X)[Y ]
P
consisting of polynomials j aj (X)Y j in Y with coefficients that are rational functions of X over C. We may and do view C(X)[Y ] as a subring of the field C(X, Y )
of two-variable rational functions over C. An element of C(X)[Y ] is monic if it is
nonzero and as a polynomial in Y has leading coefficient (in C(X)) equal to 1, such
as Y 3 + ((3X 2 − 2i)/(4X 2 + 7X))Y − (−2 + 7i)/X; any nonzero element of C(X)[Y ]
has a unique C(X)-multiple that is monic (divide by the C(X)-coefficient of the
highest-degree monomial in Y ). We generally write C(X) when we are thinking
algebraically and C(x) when we are thinking function-theoretically.
2. Calculus with C-valued functions
Though it may seem to complicate matters, our work will be much simplified
by systematically using C-valued functions of a real variable x (such as eix =
cos(x) + i sin(x) or (2 + 3i)x3 − 7ix + 12) rather than R-valued functions. As one
indication of the simplification attained in this way, observe that the formulas
eix + e−ix
eix − e−ix
, cos(x) =
2i
2
express trigonometric functions in terms of C-valued exponentials. In general, the
advantage of using C-valued functions is that they encode all trigonometric functions and inverse-trigonometric functions in terms of exponentials and logarithms.
(See Example 2.1.) One also gets algebraic simplifications; for example, the addition formulas for sin(x) and cos(x) may be combined into the single appealing
formula ei(x+y) = eix eiy .
The reader may be concerned that allowing C-valued functions will permit a
more expansive notion of elementary function than one may have wanted to consider
within the framework of R-valued functions, but it is also the case that working
with a more general notion of “elementary function” will strenghten the meaning of
impossibility theorems for elementary integration problems. Since we wish to allow
C-valued functions, we must carry over some notions of calculus to this more general
setting. Briefly put, we carry over definitions using real and imaginary parts. For
example, a C-valued function can be written in the form f (x) = u(x) + iv(x) via
real and imaginary parts, and we say f is continuous when u and v are so, and
likewise for differentiability. In the differentiable case, we make definitions such
as f 0 = u0 + iv 0 , and C-valued integration is likewise defined in terms of ordinary
integration of real and imaginary parts. (The crutch of real and imaginary parts
can be avoided, but we do not dwell on the matter here.)
A C-valued function f (x) is analytic if its real and imaginary parts u(x) and
v(x) are locally expressible as convergent Taylor series. The property that an Rvalued function is locally expressible as a convergent Taylor series is preserved under
the usual operations on functions (sums, products, quotients, composition, differentiation, integration, inverse function with non-vanishing derivative), so all C-valued
or R-valued functions that we
√ easily “write down” are analytic. For example, (1.1)
is analytic on the interval ( 3 7, ∞), and if f (x) is an R-valued analytic function
then sin(f (x)) and tan−1 (f (x)) are analytic functions. For any positive integer n
(2.1)
sin(x) =
4
BRIAN CONRAD
√
the function n x that is inverse to the function xn is likewise analytic on (0, ∞), so
root extraction of positive functions preserves analyticity. By using the usual definition eu(x)+iv(x) = eu(x) (cos v(x) + i sin v(x)), it likewise follows that if a C-valued
function f (x) is analytic then so is ef (x) . Some elementary algebraic manipulations
with real and imaginary parts show moreover that (ef (x) )0 = f 0 (x)ef (x) , as one
would expect.
0
If a C-valued function f (x) is analytic and non-vanishing, then
R x f 0/f is analytic
as well, so upon choosing a point x0 the integral (log f )(x) = x0 (f (t)/f (t))dt is
an analytic function called a logarithm of f . Such a function depends on the choice
of x0 up to an additive constant
R x (since replacing x0 with another point x1 changes
the function by the constant x01 (f 0 (t)/f (t))dt), but such ambiguity is irrelevant for
our purposes so we will ignore it. Thus, we can equivalently consider a logarithm
of f to be a solution to the differential equation y 0 = f 0 /f . In the special case
x0 = 1 and f (t) = t (on the interval (0, ∞)) this recovers the traditional logarithm
function. If we add a suitable constant to a logarithm of f then we can arrange
that elog f = f , so the terminology is reasonable.
Example 2.1. The formulas (2.1) show that exponentiation recovers trigonometric functions. One checks by differentiation that 2i tan−1 (x) + iπ is a logarithm
of the non-vanishing function (x − i)/(x + i), so by forming logarithms we may
recover the inverse trigonometric function tan−1 . Using identities such as
!
r
1
x
−1
−1
−1
−1
√
− 1 , sin (x) = tan
cos (x) = tan
x2
1 − x2
for |x| < 1, we recover all of the usual inverse trigonometric functions via logarithms
of non-vanishing analytic functions. Thus, we can describe trigonometric functions
and their inverses in terms of exponentials and logarithms of analytic functions.
We manipulate ratios of polynomials (with nonzero denominator) “as if” they
are genuine functions by simply omitting the few points where the denominators
vanish, and we likewise will treat ratios of analytic functions “as if” they are genuine
functions by omitting the isolated points where the denominators vanish. Such
ratios of analytic functions on a fixed non-empty open interval I in R are called
meromorphic functions on I. For example, ex /x is a meromorphic function on the
real line. We will usually not explicitly mention the fixed open interval of definition
being used; the context will always make it clear. The set of meromorphic functions
(on a fixed non-empty open interval) is a field, and on this field we can define the
derivative operator f 7→ f 0 by using the quotient rule in the usual manner. This
derivative operator satisfies the usual properties (e.g., product rule, quotient rule).
The field of meromorphic functions equipped with the derivative operator is the
setting in which Liouville’s theorem takes place.
3. Elementary fields and elementary functions
It is convenient to introduce the following general notation.
Definition 3.1. If f1 , . . . , fn are meromorphic functions then C(f1 , . . . , fn )
denotes the set of meromorphic functions h of the form
P
p(f1 , . . . , fn )
ae ,...,en f1e1 · · · fnen
h=
= P 1
q(f1 , . . . , fn )
bj1 ,...,jn f1j1 · · · fnjn
IMPOSSIBILITY THEOREMS FOR ELEMENTARY INTEGRATION
5
for n-variable polynomials
X
X
p(X1 , . . . , Xn ) =
ae1 ,...,en X1e1 · · · Xnen , q(X1 , . . . , Xn ) =
bj1 ,...,jn X1j1 · · · Xnjn
in C[X1 , . . . , Xn ] with q(f1 , . . . , fn ) 6= 0.
The set C(f1 , . . . , fn ) is clearly a field, via the usual algebraic operations on
functions.
Example 3.2. The field K = C(x, sin(x), cos(x)) is the set of ratios
p(x, sin(x), cos(x))
q(x, sin(x), cos(x))
for polynomials p, q ∈ C[X, Y, Z] such that q(x, sin(x), cos(x)) 6= 0. For example,
we cannot use q = Y 2 + Z 2 − 1 since sin(x)2 + cos(x)2 − 1 = 0. By (2.1) we have
K = C(x, eix ), so elements of K can also be written in the form g(x, eix )/h(x, eix )
with g, h ∈ C[X, Y ] and h 6= 0. There is no restriction imposed on the nonzero
polynomial h by the requirement that the function h(x, eix ) not vanish identically,
due to Lemma 5.1 below.
Definition 3.3. A field K of meromorphic functions is an elementary field if
K = C(x, f1 , . . . , fn ) with each fj either an exponential or logarithm of an element
of Kj−1 = C(x, f1 , . . . , fj−1 ) or else algebraic over Kj−1 in the sense that P (fj ) = 0
for some P (T ) = T m + am−1 T m−1 + · · · + a0 ∈ Kj−1 [T ] with all ak ∈ Kj−1 . A
meromorphic function f is an elementary function if it lies in an elementary field
of meromorphic functions.
Example 3.4. Consider the function f given by (1.1). An elementary field
containing f is
p
3
K = C(x, log x, ex , eix/(x −7) , ex − sin(x/(x3 − 7))).
p
Example 3.5. Root-extractions such as 3 sin(x) − 7x are examples of the algebraic case in Definition 3.3 (this solves the cubic equation T 3 − (sin(x) − 7x) = 0
with coefficients in the elementary field C(x, eix )). A more
example of
√
√ elaborate
such an “algebraic function” is K = C(x, f (x)) with f = x + 3 x: we may take
P (T ) = T 6 − 3xT 4 − (x3 − x2 )T 3 − 3x2 T 2 − 6x2 T − (x3 − x2 )
in C(x)[T ] as a polynomial√satisfied
to this
√ by f over C(x). A simpler
√ approach
√
√
latter example is to view√ x + 3 x as lying in √
the field C(x, x, 3 x), with x
algebraic over C(x) and 3 x algebraic over C(x, x) (and even over C(x)).
The notion of elementary function as just defined includes all functions that one
might ever want to consider to be elementary. It is not obvious is how to prove that
there exist non-elementary (meromorphic) functions, and Liouville’s results will
give a method to prove that specific meromorphic functions are not elementary.
Theorem 3.6. If K is an elementary field, then it is closed under the operation
of differentiation.
Proof. We write K = C(x, f1 , . . . , fn ) as in Definition 3.3 and we induct on n.
The case n = 0 is the case K = C(x). It follows from the usual formulas for derivatives of sums, products, and ratios that C(x) is closed under differentiation. For the
general case, by induction K0 = C(x, f1 , . . . , fn−1 ) is closed under differentiation,
6
BRIAN CONRAD
and we have K = K0 (fn ) with fn either algebraic over K0 or a logarithm or exponential of an element of K0 . Let us now check that itP
suffices to prove fn0 ∈ K0 (fn ).
Under this assumption, for any polynomial P (T ) = j≥0 aj T j ∈ K0 [T ] we have
X
P (fn )0 = a00 +
(a0j fnj + jaj−1 fnj−1 fn0 ) ∈ K0 (fn )
j≥1
a0j
since
∈ K0 for all j (as K0 is closed under differentiation). Thus, if P, Q ∈ K0 [T ]
are polynomials over K0 and Q(fn ) 6= 0 then
0
P (fn )
Q(fn )P (fn )0 − P (fn )Q(fn )0
=
∈ K0 (fn ) = K
Q(fn )
Q(fn )2
since the numerator and denominator lie in K0 (fn ).
It remains to check that the function fn that is either algebraic over K0 or is an
exponential or logarithm of an element of K0 has derivative fn0 that lies in K0 (fn ).
If fn = eg for some g ∈ K0 then fn0 = g 0 fn . Thus, fn0 ∈ K0 (fn ) since g 0 ∈ K0
(as K0 is closed under differentiation). If fn is a logarithm of some g ∈ K0 then
fn0 = g 0 /g ∈ K0 ⊆ K0 (fn ). Finally, we treat the algebraic case. Suppose P (fn ) = 0
for a polynomial P = T m + am−1 (x)T m−1 + · · · + a0 (x) ∈ K0 [T ]. Take P with
minimal degree, so P 0 (T ) := mT m−1 +(m−1)am−1 (x)T m−2 +· · ·+2a2 (x)T +a1 (x)
with degree m − 1 satisfies P 0 (fn ) 6= 0. But
X
X
X
0 = P (fn )0 =
jaj (x)fnj−1 fn0 +
a0j (x)fnj = P 0 (fn )fn0 +
a0j (x)fnj ,
j>0
0
so P (fn )fn0
we get fn0 ∈
0
j
j<m aj (x)fn
j<m
j<m
=−
∈ K0 (fn ). Since P (fn ) 6= 0 and P 0 (fn ) ∈ K0 (fn ),
K0 (fn ) by division.
P
0
Remark 3.7. A field K of meromorphic functions that is closed under differentiation is called a differential field. The preceding theorem says that elementary fields are examples of differential fields, but the method of proof shows
more: if K = K0 (f ) with K0 any differential field and f either algebraic over
K0 or an exponential or logarithm of an element of K0 then K is a differential
field. The field C(x, sin(x), cos(x)) is a differential field, since sin0 (x) = cos(x)
and cos0 (x) = − sin(x). (Alternatively, this field coincides with C(x, eix ), and
(eix )0 = ieix .) In contrast, C(x, sin(x)) is not a differential field. More specifically,
sin0 (x) = cos(x) but cos(x) is not an element of C(x, sin(x)); that is, cos(x) does not
admit a rational expression in terms of x and sin(x) (with coefficients in C). The
verification of this “obvious” fact requires Lemma 5.1 below and the irreducibility
of U 2 + V 2 − 1 in C(U )[V ]; it is left as an exercise. Liouville’s work on integration
in elementary terms is best viewed as a contribution to the theory of differential
fields, and this theory is the point of departure in [1].
Definition 3.8. A meromorphic function f can be integrated in elementary
terms if f = g 0 for an elementary function g (and so f is necessarily elementary, by
Theorem 3.6).
Definition 3.8 captures any reasonable intuitive notion of an “elementary formula” for an anti-derivative of a function of the sort considered in calculus, though
the use of C-valued functions permits a much wider class of anti-derivatives to be
considered as elementary than one may have wanted to allow in the classical setting
of R-valued functions. What is more important is that if we can prove an elemen2
tary function such as e−x cannot be integrated in elementary terms in the sense
IMPOSSIBILITY THEOREMS FOR ELEMENTARY INTEGRATION
7
of the preceding definition (that allows C-valued functions) then it is certainly not
susceptible to “elementary” integration in any reasonable R-valued sense! That is,
by proving an impossibility theorem with the preceding definitions we are proving
something even stronger than we might have hoped to be true in the R-valued
setting.
In case the reader is still disturbed by our use of C-valued functions, we should
point out that using only R-valued functions in Definition 3.8 would give the wrong
concept of elementary integrability. For example, under any reasonable definition we would want to say that 1/(1 + x2 ) admits an elementary integral (such as
tan−1 (x)). This is the case in the C-valued setting, since we know that tan−1 (x)
is obtained from a logarithm of (x + i)/(x − i) (see Example 2.1). However, if we
work in the R-valued setting and permit only the operations of exponentiation, logarithm, and solving of algebraic equations (as in Definition 3.8, via Definition 3.3)
then it can be proved that 1/(1 + x2 ) is not integrable (over R) in such elementary
terms. (See [1, p. 968] for a rigorous proof.) A way around this technical glitch
in the R-valued case is to incorporate all of the usual trigonometric functions and
their inverses (and not merely exponentials and logarithms) in an R-valued definition of “integration in elementary terms”. Unfortunately, this change in definitions
is disastrous for the attempt to push through an R-valued analogue of Liouville’s
results because such trigonometric functions and their inverses are not solutions to
simple first-order differential equations. Since our main interest is in impossibility
results, Liouville’s work in the C-valued setting will give what we require.
4. Integrability criterion and applications
Liouville’s main theorem asserts that if an elementary function f is integrable
in elementary terms then there are severe constraints on the possible form of an
elementary anti-derivative of f :
Theorem 4.1 (Liouville). Let f be an elementary function and let K be an
elementary field containing f . The function f can be integrated in elementary terms
if and only if there exist nonzero c1 , . . . , cn ∈ C, nonzero g1 , . . . , gn ∈ K, and an
element h ∈ K such that
X gj0
f=
cj + h0 .
gj
The key point
P is that the gj ’s and h can be found in any elementary field K
containing f ;
cj log(gj ) + h is then an elementary integral of f .
2
Example 4.2. Consider f = e−x . This lies in the elementary field K =
2
C(x, e−x ). Hence, Liouville’s theorem says that an elementary anti-derivative of
P
2
f must have the special form
cj log gj + h for some h ∈ C(x, e−x ) and nonzero
2
cj ∈ C and gj ∈ C(x, e−x ). It is not obvious how to prove the non-existence of such
h and gj ’s, but this still represents a significant advance over the problem of contemplating all elementary functions as candidates for elementary anti-derivatives of
2
e−x . We will soon see that the possible form of such an elementary anti-derivative
2
of e−x can be made even more special, and so it becomes a problem that we can
solve without too much difficulty.
Example 4.3. There is a very interesting class of integrals for which Liouville’s
result in the above form is immediately applicable without an extra simplification:
8
BRIAN CONRAD
elliptic integrals. Just as trignometric Rfunctions
√ may be introduced through inversion of integral functions of the form dx/ x2 − 1 that arise from calculation of
arc length along a unit circle, the theory ofR elliptic
p functions grew out of a study of
inversion of integral functions of the form dx/ P (x) for certain cubic and quartic polynomials P (X) ∈ R[X] without repeated roots; such integrals arise in the
calculation of arc length along an ellipse. In general, if P (X) ∈ R[X] Ris anypmonic
polynomial with degree ≥ 3 and no repeated roots
√ then we claim that dx/ P (x)
is not an elementary function. Since K = C(x, P ) is an elementary field, by the
criterion in Theorem 4.1 it suffices to prove that there does not exist an identity of
the form
X gj0
1
p
=
cj + h0
gj
P (x)
with nonzero c1 , . . . , cn ∈ C, nonzero g1 , . . . , gn ∈ K, and h ∈ K. Such impossibility
is a consequence of general facts from the theory of compact Riemann surfaces.
More specifically, for the advanced reader who knows a bit about this theory, the
above identity is equivalent to the equality of meromorphic 1-forms
dx X dgj
=
cj
+ dh
y
gj
on the compact Riemann surface C associated to the equation y 2 = P (x), and for
deg(P ) > 2 the left side is a nonzero holomorphic 1-form on C. But a nonzero
holomorphic 1-form on a compact Riemann surface never admits an expression
as a linear combination of logarithmic meromorphic differentials dg/g and exact
meromorphic differentials dh.
The proof of Liouville’s theorem rests on an inductive algebraic study of differential fields. We refer the reader to [1, §4–§5] for the proof of Theorem 4.1, and we
now focus our attention on using and proving the following criterion that emerges
as a consequence:
Theorem 4.4. Choose f, g ∈ C(X) with f 6= 0 and g nonconstant. The
function f (x)eg(x) can be integrated in elementary terms if and only if there exists
a rational function R ∈ C(X) such that R0 (X) + g 0 (X)R(X) = f (X) in C(X).
The content of the criterion in this theorem is not that the differential equation
R0 (x) + g 0 (x)R(x) = f (x) has a solution as a C-valued differentiable function of
x (since we can always write down a simple integral formula for a solution via
integrating factors), but rather than there is a solution with the special property
that it is a rational function in x. (For any such R ∈ C(X), R(x)eg(x) is an
elementary anti-derivative of f (x).) In specific examples, as we shall see, it can be
proved that there does not exist a rational function in x that solves the differential
equation R0 (x) + g 0 (x)R(x) = f (x); this is how we will verify that some functions
of the form f (x)eg(x) cannot be integrated in elementary terms. The deduction of
Theorem 4.4 from Theorem 4.1 is given in §5, and the remainder of this section
is devoted to applying Theorem 4.4 to the problem of computing the Gaussian
integral and the logarithmic integral in elementary terms.
Remark 4.5. In what follows, we will systematically work with the field C(X)
as an “abstract” field (also identified in this obvious manner with the more concrete
field C(x) of rational C-valued functions on an open interval in R). On this field
the operation of differentiation is defined by the standard formula on C[X] and
is extended to C(X) via the quotient rule; one checks without difficulty that this
IMPOSSIBILITY THEOREMS FOR ELEMENTARY INTEGRATION
9
is a well-defined operation on C(X) and that it satisfies all of the usual formulas
with respect to sums, products, and quotients. The purpose of this algebraic point
of view is to permit the use of differentiation in algebraic identities in C(X) that
we may then specialize at points z ∈ C that are possibly not in R. (The theory
of differentiation for C-valued functions of a complex variable has some surprising
features; we do not require it for our purposes.)
2
Example 4.6. We now prove that e−x cannot be integrated in elementary
terms. Taking f = 1 and g = −x2 in Theorem 4.4, we have to prove that the
differential equation R0 (X) − 2XR(X) = 1 in C(X) has no solution (in C(X)).
The method of integrating
factors gives a formula for the general function solution,
2 R
2
namely Rc (x) = −ex ( e−x dx + c) with c ∈ C, but we cannot expect to show by
inspection
that this is never a rational function since we have no way to describe
R −x2
e
dx in the first place! However, this formula for the general solution still
provides a genuine simplification on the initial problem because the necessary and
sufficient condition that the function Rc (x) is a rational function in x for some
2
choice of constant c says exactly that if e−x is to have an elementary anti-derivative
2
then there must be such an anti-derivative having the form e−x r(x) for a rational
function r ∈ C(X). This is a severe constraint on the possible form of an elementary
2
anti-derivative. (Explicitly, the condition that e−x r(x) be an anti-derivative to
2
e−x is precisely the condition that r0 (x) − 2xr(x) = 1.)
To prove the non-existence of solutions to R0 (X) − 2XR(X) = 1 in C(X),
we shall argue by contradiction. If R ∈ C(X) is such a solution, then certainly
R is non-constant and we claim that R cannot be a polynomial (in X). Indeed,
if R(X) is a polynomial with some degree n > 0 then R0 (X) − 2XR(X) is a
polynomial with degree n + 1 and hence it cannot equal 1. Thus, in reduced
form we must have R(X) = p(X)/q(X) for nonzero relatively prime polymomials
p(X), q(X) ∈ C[X] with q(X) nonconstant. The identity R0 (X) − 2XR(X) = 1 in
C(X) says (p(X)/q(X))0 − 2X(p(X)/q(X)) = 1.
Since q(X) is a nonconstant polynomial, by the Fundamental Theorem of Algebra it has a root z0 ∈ C; of course, usually z0 is not in R. Relative primality
of p and q in C[X] implies p(z0 ) 6= 0. Hence, if z0 is a root of q with multiplicity
µ ≥ 1 then p(X)/q(X) = h(X)/(X − z0 )µ with h(X) ∈ C(X) having numerator
and denominator that are non-vanishing at z0 . Differentiation gives
0
p(X)
−h(X)
h0 (X)
=
+
,
q(X)
µ(X − z0 )µ+1
(x − z0 )µ
so as z → z0 in C we see that (p(X)/q(X))0 |X=z has absolute value that blows up
like A/|z − z0 |µ+1 with A = |h(z0 )/µ| =
6 0. But | − 2z · (p(z)/q(z))| has growth
bounded by a constant multiple of 1/|z − z0 |µ as z → z0 in C, so
0
!
p(X)
p(X)
A
− 2X ·
∼
X=z
|z − z0 |µ+1
q(X)
q(X)
as z → z0 in C. This contradicts the identity (p(X)/q(X))0 − 2X(p(X)/q(X)) = 1.
R
R xExample 4.7. Consider the logarithmic integral dt/ log(t), or equivalently
(e /x)dx (with x = log t). Taking f = 1/x and g = x in Theorem 4.4, to prove
that the logarithmic integral cannot be expressed in elementary terms it suffices to
show that the differential equation R0 (X) + R(X) = 1/X does not have a solution
10
BRIAN CONRAD
in C(X). If there exists R ∈ C(X) such that R0 (X) + R(X) = 1/X then obviously
R(X) cannot be a polynomial in X. Writing R = p/q in reduced form with q ∈ C[X]
a nonconstant polynomial, q has a root of order µ ≥ 1 at some z0 ∈ C. Thus, R0 (X)
has a zero of order µ + 1 at z0 and so as with the previous example we see that
1/X = R0 (X) + R(X) considered as a function of z ∈ C has absolute value that
explodes like a nonzero constant multiple of 1/|z − z0 |µ+1 as z → z0 . But the only
w ∈ C for which |1/z| has explosive growth in absolute value as z approaches w in
C is w = 0, so z0 = 0. Hence, as z → 0 it follows that |1/z| grows like a nonzero
constant multiple of 1/|z − z0 |µ+1 = 1/|z|µ+1 , a contradiction since µ + 1 ≥ 2.
5. Proof of integrability criterion
We now show how to deduce Theorem 4.4 from Theorem 4.1. Assuming f eg
admits an elementary antiderivative, we have to find R ∈ C(X) such that R0 (X) +
g 0 (X)R(X) = f (X). Theorem 4.1 with K = C(x, eg ) implies
f (x)eg(x) =
(5.1)
X
cj
gj0 (x)
+ h0 (x)
gj (x)
(on a suitable non-empty open interval in R) with nonzero cj ∈ C and nonzero
gj ∈ C(x, eg ), and an element h ∈ C(x, eg ). We need to massage this expression to
get it into a more useful form. Note that each gj may be taken to be non-constant,
as otherwise gj0 = 0 and so gj0 /gj does not contribute anything for such j.
Choose nonzero pj , qj ∈ C[X, Y ] such that gj (x) = pj (x, eg(x) )/qj (x, eg(x) ).
Since
gj0 (x)
pj (x, eg(x) )0
qj (x, eg(x) )0
=
−
,
gj (x)
pj (x, eg(x) )
qj (x, eg(x) )
we have
X
cj
gj0 (x) X pj (x, eg(x) )0 X
qj (x, eg(x) )0
=
cj
+
−c
.
j
gj (x)
pj (x, eg(x) )
qj (x, eg(x) )
j
j
We write this as a sum over twice as many indices to reduce to the case when
each gj has the form pj (x, eg(x) ) with a nonzero polynomial pj (X, Y ) ∈ C[X, Y ].
We can assume each pj is nonconstant, as otherwise pj (x, eg(x) )0 = 0 and so
this term contributes nothing. ConsideringQpj (X, Y ) as a nonconstant element
in C(X)[Y ], we may factor pj as a product k rkj (X, Y ) of irreducible monics in
C(X)[Y
]Q
and (possibly)
a nonconstant element of C(X). Using the general identity
Q
P
( hk )0 / hk = h0k /hk for meromorphic functions hk (on a common non-empty
open interval in R), we get
X rkj (x, eg(x) )0
pj (x, eg(x) )0
=
.
g(x)
pj (x, e
)
rkj (x, eg(x) )
k
Upon renaming terms and indices once again we may suppose that each pj ∈
C(X)[Y ] is either a monic irreducible over C(X) or lies in C(X). It can also
be assumed that the pj ’s are pairwise distinct in C(X)[Y ] by collecting repeated
appearances of the same term and multiplying cj by a positive integer that counts
the number of repetitions.
Writing h(x) = p(x, eg(x) )/q(x, eg(x) ) for relatively prime p, q ∈ C(X)[Y ] with
q having leading coefficient equal to 1 as a polynomial in Y with coefficients in
IMPOSSIBILITY THEOREMS FOR ELEMENTARY INTEGRATION
11
C(X), (5.1) now says
f (x)eg(x)
(5.2)
0
=
X
cj
pj (x, eg(x) )0
+
pj (x, eg(x) )
=
X
cj
q(x, eg(x) )p(x, eg(x) )0 − p(x, eg(x) )q(x, eg(x) )0
pj (x, eg(x) )0
+
.
pj (x, eg(x) )
q(x, eg(x) )2
p(x, eg(x) )
q(x, eg(x) )
Recall that our goal is to find R ∈ C(X) such that R0 (X) + g 0 (X)R(X) = f (X)
in C(X), and a key step in this direction is to reformulate (5.2) in more algebraic
terms. This rests on a lemma that passes between functional identities and algebraic
identities:
Lemma 5.1. For nonconstant g(X) ∈ C(X), if H1 , H2 ∈ C[X, Y ] satisfy
H1 (x, eg(x) ) = H2 (x, eg(x) ) as functions of x in a non-empty open interval I in
R then H1 = H2 in C[X, Y ].
Proof. Passing to H1 −H2 reduces us to showing that if H ∈ C[X, Y ] satisfies
H(x, eg(x) ) = 0 for x ∈ I then H = 0 in C[X, Y ]. Suppose H 6= 0. Certainly H has
to involve Y , as otherwise H(X, 0) = H(X, Y ) is a nonzero polynomial in X and
the condition H(x, 0) = H(x, eg(x) ) = 0 says that the nonzero polynomial H(X, 0)
vanishes at the infinitely many points of I, an absurdity. Viewing H as an element
in C[X, Y ] ⊆ C(X)[Y ] and (without loss of generality) dividing through by the
coefficient in C(X) for the top-degree monomial in Y appearing in H, we get a
relation of the form
(5.3)
eng(x) + an−1 (x)e(n−1)g(x) + · · · + a1 (x)eg(x) + a0 (x) = 0
for all x in a non-empty open interval, with a0 , . . . , an−1 ∈ C(X) and some n > 0.
We consider such a relation with minimal n ≥ 0. Obviously we must have n ≥ 1,
and some aj ∈ C(X) must be nonzero.
Differentiating (5.3) gives
ng 0 (x)eng(x) + (a0n−1 (x) + (n − 1)g 0 (x)an−1 (x))e(n−1)g(x) + . . .
+(a01 (x) + g 0 (x)a1 (x))eg(x) + a00 (x) = 0,
yet ng 0 (X) ∈ C(X) is nonzero (as g(X) 6∈ C), so dividing by ng 0 (x) gives
X a0j (x) + jg 0 (x)aj (x)
a00 (x)
jg(x)
(5.4)
eng(x) +
e
+
=0
ng 0 (x)
ng 0 (x)
n−1
j=1
for x in a non-empty open interval in R on which (5.3) also holds.
We have now produced two “degree n” relations (5.3) and (5.4) for eg with
coefficients in C(X) and the same “degree n” term eng , so taking the difference gives
a relation of degree ≤ n−1 in eg with coefficients in C(X). By the minimality of n, it
follows that all C(X)-coefficients in this difference relation must be identically zero
(as otherwise we could divide by the nonzero C(X)-coefficient corresponding to the
highest degree in eg to get a polynomial relation for eg over C(X) with degree < n
in eg and some nonzero coefficient in C(X), contrary to the minimality with respect
to which n was chosen). Hence, for 0 ≤ j ≤ n − 1 we must have a0j + jg 0 aj = ng 0 aj
in C(X). That is, a0j = (n − j)g 0 aj in C(X) for all 0 ≤ j ≤ n − 1.
There is some j0 such that aj0 6= 0 in C(X), and for such a j0 we have
(n − j0 )g 0 (X) =
a0j0 (X)
aj0 (X)
12
BRIAN CONRAD
Q
in C(X). By the Fundamental Theorem of Algebra, aj0 (X) = c (X − ρk )ek for
some nonzero c ∈ C, nonzero integers ek , and pairwise distinct ρk ∈ C, so
a0j (X) X ek
=
.
(n − j0 )g 0 (X) = 0
aj0 (X)
X − ρk
Since g is nonconstant and n−j0 6= 0, so (n−j0 )g 0 (X) 6= 0, there must be some ρk ’s
and hence g 0 (x) behaves like a nonzero constant multiple of 1/|x − ρ1 | as x → ρ1 .
But this inverse-linear growth for the derivative of a nonconstant rational function
is impossible: we may write g(X) = (X − ρ1 )µ G(X) in C(X), where µ ∈ Z and
the nonzero G(X) ∈ C(X) has numerator and denominator that are non-vanishing
at ρ1 , and by separately treating the cases µ < 0, µ = 0, and µ > 0 we never get
inverse-linear growth for |g 0 (x)| as x → ρ1 . This is a contradiction.
By Lemma 5.1, if H ∈ C[X, Y ] is nonzero then 1/H(x, eg(x) ) makes sense as
a meromorphic function. Hence, for any rational function F = G(X, Y )/H(X, Y )
with G, H ∈ C[X, Y ] and H 6= 0, we get a well-defined meromorphic function
F (x, eg(x) ) = G(x, eg(x) )/H(x, eg(x) ) for x in a suitable non-empty open interval
in R. We now use this to recast (5.2) is more algebraic terms by eliminating the
appearance of exponentials as follows. Observe that (eg(x) )0 = g 0 (x)eg(x) , so more
generally for any h(X, Y ) ∈ C(X)[Y ] we have h(x, eg(x) )0 = (∂h)(x, eg(x) ) where
∂ : C(X)[Y ] → C(X)[Y ] is the operator
X
X
∂(
rj (X)Y j ) =
(rj0 (X) + jg 0 (X)rj (X))Y j .
This leads to the following reformulation of (5.2):
Lemma 5.2. The identity
X ∂(pj (X, Y )) q(X, Y )∂(p(X, Y )) − p(X, Y )∂(q(X, Y ))
(5.5) f (X)Y =
cj
+
pj (X, Y )
q(X, Y )2
j
holds in the two-variable rational function field C(X, Y ).
Proof. By converting both sides into functions of x upon replacing Y with
eg(x) we get the equality (5.2) since (∂h)(x, eg(x) ) = h(x, eg(x) )0 for any h(X, Y ) ∈
C(X)[Y ]. Thus, after cross-multiplying by a common denominator on both sides
we may use Lemma 5.1 to complete the proof.
Our aim is to use the identity (5.5) to find some R ∈ C(X) such that f (X) =
R0 (X) + g 0 (X)R(X) in C(X). Since (5.5) is an algebraic identity that does not
involve the intervention of exponentials, this is a purely algebraic problem and so
is good progress! The idea for solving the algebraic problem is to analyze how the
terms on the right side of (5.5) can possibly have enough cancellation in the denominators so that the right side can be equal to the left side whose denominator lies
in C[X] (i.e., it involves no Y ’s). Once we take into account how such cancellation
can arise, a comparison of Y -coefficients on both sides of (5.5) will give the desired
identity f (X) = R0 (X) + g 0 (X)R(X) for a suitable R ∈ C(X).
We now look at denominators on the right side of (5.5). The jth term in the
sum on the right side contributes a possible denominator of pj (X, Y ) when pj (X, Y )
is a monic irreducible in C(X)[Y ]. (Recall that each pj (X, Y ) is an irreducible in
C(X)[Y ] that is monic in Y or is a nonconstant element of C(X)). The reason we
speak of a “possible” denominator pj (X, Y ) for such j is that we have to take into
account that such an irreducible pj might divide ∂(pj (X, Y )) in C(X)[Y ]. Suppose
IMPOSSIBILITY THEOREMS FOR ELEMENTARY INTEGRATION
13
such divisibility occurs for some j0 . Writing pj0 = Y m +rm−1 (X)Y m−1 +· · ·+r0 (X)
with rk (X) ∈ C(X) and m > 0, we compute
0
∂(pj0 ) = mg 0 Y m + (rm−1
+ (m − 1)g 0 rm−1 )Y m−1 + · · · + (r10 + g 0 r1 )Y + r00
and so consideration of Y -degrees (and the nonvanishing of both g 0 ∈ C(X) and m)
implies that the irreducible pj0 divides ∂(pj0 ) in C(X)[Y ] if and only if ∂(pj0 ) =
mg 0 pj0 . This says mg 0 rk = (rk0 − kg 0 rk ) for all 0 ≤ k < m, or in other words
rk0 = (m − k)g 0 rk for all such k. As we saw in the proof of Lemma 5.1, since
g ∈ C(X) is nonconstant and m > 0 such a condition with all rk ∈ C(X) can
only hold if rk = 0 for all k < m, which is to say that the monic irreducible
pj0 ∈ C(X)[Y ] is equal to Y .
To summarize, for each j0 such that pj0 ∈ C(X)[Y ] is a monic irreducible
distinct from Y , the sum over j on the right in (5.5) does contribute a denominator
of pj0 to the first power. There is no such term in the denominator on the left side
of (5.5), so if such pj0 ’s arise then they must cancel against a contribution from a
factor of pj0 in the denominator q in the final term on the right in (5.5). But if any
such factor appears in q with some multiplicity µ ≥ 1 then in the reduced form of
the ratio −p∂(q)/q 2 this term appears in the denominator with multiplicity exactly
µ + 1 (as we have just seen above that pj0 does not divide ∂(pj0 ) for such j0 , and
the operator ∂ satisfies the “Leibnitz rule” ∂(G1 G2 ) = G1 ∂(G2 ) + G2 ∂(G1 )). This
multiplicity of order µ + 1 is not cancelled on the right side of (5.5), and so this
contradicts the absence of such a term in the denominator on the left side of (5.5).
Hence, we have proved that (i) there is at most one pj in the sum over j on the
right side of (5.5) that is not in C(X), with the unique exceptional term (if any
occurs) equal to Y , and (ii) the monic q ∈ C(X)[Y ] cannot have any irreducible
factor other than Y . Hence, q = a(X)Y n with n ≥ 0 and a(X) ∈ C(X) a nonzero
element.
P
We conclude that h = p/q = k sk Y k with sk ∈ C(X) and k ranging through
a finite set of integral values (possibly negative). Among the terms pj that arise, the
ones that lie in C(X) do not contribute to any Y -terms, and since the only other
possible term pj = Y satisfies cj ∂(pj )/pj = cj g 0 (X) it also does not contribute to
any Y -terms. Hence,
P the left side f (X)Y in (5.5) must coincide with the linear
Y -term in ∂(h) = k (s0k + kg 0 sk )Y k , which is to say f (X)Y = (s01 + g 0 s1 )Y in
C(X)[Y ]. This gives s01 + g 0 s1 = f with s1 ∈ C(X), so R(X) = s1 (X) ∈ C(X)
satisfies R0 (X) + g 0 (X)R(X) = f (X).
References
[1] M. Rosenlicht, Integration in finite terms, American Math. Monthly 79 (1972), 963–972.
Department of Mathematics, University of Michigan, Ann Arbor, MI 48109-1043
E-mail address: [email protected]