PROVING THE NON-EXISTENCE OF ELEMENTARY ANTI-DERIVATIVES FOR CERTAIN ELEMENTARY FUNCTIONS VIRAAJ JAYARAM Abstract. This paper addresses the question of when functions have anti- derivatives that can be expressed in a satisfactory form. We first pose this question in a mathematically precise and algebraic sense by defining the notion of an elementary function and then prove and utilise Liouville’s Theorem on elementary anti-derivatives to show that certain elementary functions do not possess elementary anti-derivatives. Contents 1. Introduction 1 2. Di↵erential Fields and Elementary Extensions 1 3. Elementary Extension Fields of C(x) and Elementary Functions 5 4. The Restrictions Imposed by Having an Elementary Anti-Derivative 8 5. Deriving a Useful Corollary to Liouville’s Theorem 12 6. Tackling Specific Examples 14 Acknowledgements 15 References 15 1. Introduction When one first learns how to compute integrals and anti-derivatives, one may hear that it is ‘impossible’ to find an anti-derivative of functions like sin(x2) or x2 1 e− or ln(x) . This seems to be a rather sad fact of life as the integrals of these functions play important roles in optics, statistics and number theory, respectively. However, of course it is not impossible to find an anti-derivative of these functions in an abstract sense.The question is whether we can find a ‘nice’ expression for the anti-derivative. The purpose of this paper is firstly to illustrate how one can pose this question in a precise mathematical sense and then to prove that functions like the three mentioned above do not, in fact, have ‘nice’ anti-derivatives. 2. Differential Fields and Elementary Extensions We shall see that in order to determine what we mean by ‘nice’ anti-derivatives, it is helpful to create algebraically abstract notions of the analytic concepts of di↵erentiation, exponentiation and the logarithm. We do this via the following definitions. 1 2VIRAAJJAYARAM Definition 2.1. We say a function f on a ring R satisfies the sum rule if for all a, b R, f(a + b)=f(a)+f(b) and the product rule if, for all a, b R, f(ab)=2f(a) b + f(b) a. 2 · · Definition 2.2. A di↵erential field is a field F paired with a function 0 from F to F satisfying the sum and product rules. We shall refer to this function 0 as the derivation on F . Definition 2.3. Let a and b belong to some di↵erential field F and b0 = a.We then say that a is the derivative of b and that b is an anti-derivative of a. Definition 2.4. An element c in a di↵erential field F is said to be a constant if c0 = 0. At this point we can notice several results. Proposition 2.5. The constants of a di↵erential field form a subfield Proof. Constants are clearly closed under addition and multiplication. We then note that 10 =(11)0 =10 1+1 10 =210 and thus 10 = 0.The same proof works for 0. · · · · We then have that 0 = 10 =( 1 1)0 = 2 ( 1)0 and thus ( 1)0 is also 0. Thus if − ·− 1 − · − 1 − 1 1 c is a constant, so is c. Regarding c− , note that (c c− )0 = c0 (c− )+c (c− )0 =0 1 −c0 · · · and thus (c− )0 = −c2 = 0. ⇤ Proposition 2.6. The usual quotient and power rules hold in a di↵erential field. a b0 a0 a0 b b0 a Proof. From the proof above, we see that ( b )0= a b2 + b = · b−2 · . For the n n 1− · power rule, we want to show that (a )0 = n a − a0. We can do this by induction. The n = 1 case is trivial. So assuming the· result· holds for n = k, we have that k+1 k k 1 k (a )0 = a0 a + a k a − a0 =(k + 1) a a0. Thus the rule holds for all positive integers.· It shall· · also be· in our interest· to· consider the case for negative integers and so for the sake of completeness, we shall address that case here as well. n (n+1) We want to prove that (a− )0 = n a0 a− . We have seen that the result − · · (k+1) holds for n = 1. Assuming the result holds for n = k, we have that (a− )0 = (k+1) 1 2 k (k+2) k a0 a− a− + a0 a− a− = (k + 1) a0 a− and hence the result holds− · for· negative· integers− as· well.· − · · ⇤ We can now begin to talk about exponentials and logarithms of elements of di↵erential fields. Definition 2.7. Let a and b belong to a di↵erential field F ,whereb is non-zero. b0 a We say the b is an exponential of a if b = a0. We shall use the notation b = e to denote the fact that b is an exponential of a. Definition 2.8. Let a and b belong to a di↵erential field F ,wherea is non-zero. a0 We say that b is a logarithm of a if b0 = a . We shall use the notation b =ln(a)to denote the fact that b is a logarithm of a. Definition 2.9. Let F be a di↵erential field. We say that adjoining an element t1 to create the field F (t1) is an elementary extension of F if t1 is either the exponential of an element of F , the logarithm of an element of F or is algebraic over F . ON ELEMENTARY FUNCTIONS AND THEIR ANTI-DERIVATIVES 3 Definition 2.10. We say the field F (t1,...,tn) is an elementary extension field of F if adjoining ti is an elementary extension of F (t1,...,ti 1) for each i. − The following proposition is an essential feature of elementary extension fields that we shall make use of time and time again. Proposition 2.11. Let F be a di↵erential field of characteristic 0 and F (f1,...,fn) be an elementary extension field of F . Then the derivation on F can be extended to a derivation on F (f1,...,fn) and this extension is unique. Proof. We shall prove this result by induction. To this end, note that it suffices to show that if F is a di↵erential field of characteristic 0 and the extension F (t)/F is an elementary extension, then the derivation on F can be extended uniquely to a derivation on F (t). This is because we can then apply this result to the extension F (f1,...fi)/F (f1,...,fi 1) by replacing F with F (f1,...,fi 1) and t with fi. For the case where −t is algebraic, we first show that if− such a structure were to exist, it is unique. Let h(x)= j a xi be the minimal polynomial of t in F [x]. i=0 i · We know that for any derivation 0,(h(t))0 = 0 and thus P j j i i 1 a0 (t) + t0 i a (t) − =0. j · · · i · i=0 i=0 X X j i 1 As the minimal polynomial is unique and i=0 i ai (t) − is the evaluation of a polynomial of degree less than that of the minimal· polynomial,· it does not equal 0 j i P i=0 aj0 (t) and hence t0 = −j · and so is uniquely determined. Once we have that t0 i a (t)i 1 iP=0 · i· − is uniquely determined,P this then uniquely determines the derivative of every other m i element in F (t) as any element can be written as i=0 ai t ,whereai F and m<j, and thus the sum and product rules determine its derivative.· We now2 need to show that we can indeed construct a derivation onP F (t) To construct this derivation, we shall first seek to construct a function that satisfies the sum and product rules on the polynomial ring F [x] and then use this to induce a derivation on F (t). To this end, we first define D0 from F [x]toF [x] such that i i D ( a x )= a0 x 0 i · i · i i X X and D1 from F [x]toF [x] such that i i 1 D ( a x )= i a x − . 1 i · · i · i i X X We can then define a function D from F [x]toF [x] such that D(p(x)) = D (p(x)) + g D (p(x)) 0 · 1 for some fixed g F [x]. Showing that D satisfies the sum and product rules 2 is equivalent to showing that D0 and D1 satisfy the sum and product rules. In n verifying this, the sum cases are immediate. For the product cases, if p = n an x m k · and q = m bm x , we have that the coefficient of the x term in their product is k · k k P equal to j=0 aj bk j. Applying D0, this becomes j=0 aj0 bk j + j=0 aj bk0 j, P · − · − · − P P P 4VIRAAJJAYARAM k which is exactly the coefficient of the x term of in D0(p) q + p D0(q). Applying D instead, the coefficient of the xk term in D (p) q + p ·D (q)is· 1 1 · · 1 k k (i + 1) ai+1 bk 1 + ai (k i + 1) bk i+1 · · − · − · − i=0 i=0 X X This can be rewritten as k k k (i ai+1 bk i i ai bk i+1)+(k + 1) ai bk i+1 + ai+1 bk i.