User:Lethe/differential Galois theory posts

question
What's the antiderivative of exp(-x2)? of sin(x)/x? of x^x? -Matthew P Wiener (weemba@sagi.wistar.upenn.edu)

initial answer
These, and some similar problems, can't be done.

More precisely, consider the notion of "elementary function". These are the functions that can be expressed in terms of exponentals and logarithms, via the usual algebraic processes, including the solving (with or without radicals) of polynomials. Since the trigonometric functions and their inverses can be expressed in terms of exponentials and logarithms using the complex numbers C, these too are elementary.

The elementary functions are, so to speak, the "precalculus functions".

Then there is a theorem that says certain elementary functions do not have an elementary antiderivative. They still have antiderivatives, but "they can't be done". The more common ones get their own names. Up to some scaling factors, "erf" is the antiderivative of exp(-x^2) and "Si" is the antiderivative of sin(x)/x, and so on.

initial definitions
For those with a little bit of undergraduate algebra, we sketch a proof of these, and a few others, using the notion of a differential field. These are fields (F,+,.,1,0) equipped with a derivation, that is, a unary operator ' satisifying (a+b)'=a'+b' and (a.b)'=a.b'+a'.b. Given a differential field F, there is a subfield Con(F)={a:a'=0}, called the _constants_ of F. We let I(f) denote an antiderivative. We ignore +cs.

Most examples in practice are subfields of M, the meromorphic functions on C (or some domain). Because of uniqueness of analytic extensions, one rarely has to specify the precise domain.

Given differential fields F and G, with F a subfield of G, one calls G an algebraic extension of F if G is a finite field extension of F.

One calls G a logarithmic extension of F if G=F(t) for some transcendental t that satisfies t'=s'/s, some s in F. We may think of t as log s, but note that we are not actually talking about a logarithm function on F. We simply have a new element with the right derivative. Other "logarithms" would have to be adjoined as needed.

Similarly, one calls G an exponential extension of F if G=F(t) for some transcendental t that satisfies t'=t.s', some s in F. Again, we may think of t as exp s, but there is no actual exponential function on F.

Finally, we call G an elementary differential extension of F if there is a finite chain of subfields from F to G, each an algebraic, logarithmic, or exponential extension of the next smaller field.

The following theorem, in the special case of M, is due to Liouville. The algebraic generality is due to Rosenlicht. More powerful theorems have been proven by Risch, Davenport, and others, and are at the heart of symbolic integration packages.

A short proof, accessible to those with a solid background in undergraduate algebra, can be found in Rosenlicht's AMM paper (see references). It is probably easier to master its applications first, which often use similar techniques, and then learn the proof.

Main Theorem
MAIN THEOREM. Let F,G be differential fields, let a be in F, let y be in G, and suppose y'=a and G is an elementary differential extension field of F, and Con(F)=Con(G). Then there exist c1,...,cn in Con(F), u1,...,un, v in F such that


 * $$a = c_1\frac{u_1'}{u_1}+...c_n\frac{u_n'}{u_n}+v'$$

That is, the only functions that have elementary antiderivatives are the ones that have this very specific form. In words, elementary integrals always consist of a function at the same algebraic "complexity" level as the starting function (the v), along with the logarithms of functions at the same algebraic "complexity" level (the ui 's).

Examples of theorem
This is a very useful theorem for proving non-integrability. Because this topic is of interest, but it is only written up in bits and pieces, I give numerous examples. (Since the original version of this FAQ from way back when, two how-to-work-it write-ups have appeared. See Fitt & Hoare and Marchisotto & Zakeri in the references.)

In the usual case, F,G are subfields of M, so Con(F)=Con(G) always holds, both being C. As a side comment, we remark that this equality is necessary. Over R(x), 1/(1+x^2) has an elementary antiderivative, but none of the above form.

We first apply this theorem to the case of integrating f.exp(g), with f and g rational functions. If g=0, this is just f, which can be integrated via partial fractions. So assume g is nonzero. Let t=exp(g), so t'=g't. Since g is not zero, it has a pole somewhere (perhaps out at infinity), so exp(g) has an essential singularity, and thus t is transcendental over C(z). Let F=C(z)(t), and let G be an elementary differential extension containing an antiderivative for f.t.

Then Liouville's theorem applies, so we can write


 * $$ft=c_1\frac{u_1'}{u_1}+...+c_n\frac{u_n'}{u_n}+v'$$

with the ci constants and the ui and v in F. Each ui is a ratio of two C(z)[t] polynomials, U/V say. But (U/V)'/(U/V)=U'/U-V'/V (quotient rule), so we may rewrite the above and assume each u_i is in C(z)[t]. And if any u_i=U.V factors, then (U.V)'/(U.V)=U'/U+V'/V and so we can further assume each u_i is irreducible over C(z).

subsection 1
What does a typical u'/u look like? For example, consider the case of u quadratic in t. If A,B,C are rational functions in C(z), then A',B',C' are also rational functions in C(z) and


 * $$\frac{(At^2+Bt+C)'}{At^2+Bt+C} = \frac{A't^2+2At(gt)+B't+B(gt)+C'}{At^2+Bt+C}=\frac{(A'+2Ag)t^2+(B'+Bg)t+C'}{At^2+Bt+C}$$

(Note that contrary to the usual situation, the degree of a polynomial in t stays the same after differentiation. That is because we are taking derivatives with respect to z, not t.  If we write this out explicitly, we get (t^n)' = exp(ng)' = ng'.exp(ng) = ng'.t^n.)

In general, each u'/u is a ratio of polynomials of the same degree. We can, by doing one step of a long division, also write it as D+R/u, for some D in C(z) and R in C(z)[t], with deg(R)<deg(u).

By taking partial fractions, we can write v as a sum of a C(z)[t] polynomial and some fractions P/Q^n with deg(P)<deg(Q), Q irreducible, with each P,Q in C(z)[t]. v' will thus be a polynomial plus partial fraction like terms.

subsection 2
Somehow, this is supposed to come out to just f.t. By the uniqueness of partial fraction decompositions, all terms other than multiples of t add up to 0. Only the polynomial part of v can contribute to f.t, and this must be a monomial over C(z). So f.t=(h.t)', for some rational h. (The temptation to assert v=h.t here is incorrect, as there could be some C(z) term, cancelled by u'/u terms.  We only need to identify the terms in v that contribute to f.t, so this does not matter.)

Summarizing, if f.exp(g) has an elementary antiderivative, with f and g rational functions, g nonzero, then it is of the form h.exp(g), with h rational.

We work out particular examples, of this and related applications. A bracketed function can be reduced to the specified example by a change of variables.


 * $$e^{z^2}[\sqrt{z}e^z,e^z/\sqrt{z}]$$

subsection 3
Let h.exp(z^2) be its antiderivative. Then h'+2zh=1. Solving this ODE gives h=exp(-z^2)*I(exp(z^2)), which has no pole (except perhaps at infinity), so h, if rational, must be a polynomial. But the derivative of h cannot cancel the leading power of 2zh, contradiction.


 * exp(z)/z [exp(exp(z)),1/log(z)]

Let h.exp(z) be an antiderivative. Then h'+h=1/z. I know of two quick ways to prove that h is not rational.

One can explicitly solve the first order ODE (getting exp(-z)*I(exp(z)/z)), and then notice that the solution has a logarithmic singularity at zero. For example, h(z)->oo but sqrt(z)*h(z)->0 as z->0. No rational function does this.

subsection 4
Or one can assume h has a partial fraction decomposition. Obviously no h' term will give 1/z, so 1/z must be present in h already. But (1/z)'=-1/z^2, and this is part of h'. So there is a 1/z^2 in h to cancel this. But (1/z^2) is -2/z^3, and this is again part of h'. And again, something in h cancels this, etc etc etc. This infinite regression is impossible.


 * sin(z)/z [sin(exp(z))]
 * sin(z^2) [sqrt(z).sin(z),sin(z)/sqrt(z)]

Since sin(z)=%[exp(iz)-exp(-iz)] (where %=1/2i), we merely rework the above f.exp(g) result. Let f be rational, let t=exp(iz) (so t'/t=i) and let T=exp(iz^2) (so T'/T=2iz) and we want an antiderivative of either %f.(t-1/t) or T-1/T. For the former, the same partial fraction results still apply in identifying %f.t=(h.t)'=(h'+ih).t, which can't happen, as above. In the case of sin(z^2), we want %T=(h.T)'=(h'+2izh).T, and again, this can't happen, as above.

Although done, we push this analysis further in the f.sin(z)/z case, as there are extra terms hanging around. This time around, the conclusion gives an additional k/t term inside v, so we have -%f/t=(k/t)'=(k'-ik)/t. So the antiderivative of %f*(t-1/t) is h.t+k/t.

subsection 5
If f is even and real, then h and k (like t=exp(iz) and 1/t=exp(-iz)) are parity flips of each other, so (as expected) the antiderivative is even. Letting C=cos(z), S=sin(z), h=H+iF and k=K+iG, the real (and only) part of the antiderivative of f is (HC-FS)+(KC+GS)=(H+K)C+(G-F)S. So over the reals, we find that the antiderivative of (rational even).sin(x) is of the form (rational even).cos(x)+ (rational odd).sin(x).

A similar result holds for (odd).sin(x), (even).cos(x), (odd).cos(x). And since a rational function is the sum of its (rational) even and odd parts, (rational).sin integrates to (rational).sin + (rational).cos, or not at all.

Let's backtrack, and apply this to sin(x)/x directly, using reals only. If it has an elementary antiderivative, it must be of the form E.S+O.C. Taking derivatives gives (E'-O).S+(E+O').C. As with partial fractions, we have a unique R(x)[S,C] representation here (this is a bit tricky, as S^2=1-C^2: this step can be proven directly or via solving for t,1/t coefficients over C). So E'-O=1/x and E+O'=0, or O''+O=-1/x. Expressing O in partial fraction form, it is clear only (-1/x) in O can contribute a -1/x. So there is a -2/x^3 term in O'', so there is a 2/x^3 term in O to cancel it, and so on, an infinite regress. Hence, there is no such rational O.


 * arcsin(z)/z [z.tan(z)]

We consider the case where F=C(z,Z)(t) as a subfield of the meromorphic functions on some domain, where z is the identify function, Z=sqrt(1-z^2), and t=arcsin z. Then Z'=-z/Z, and t'=1/Z. We ask in the main theorem result if this can happen with a=t/z and some field G. t is transcendental over C(z,Z), since it has infinite branch points.

So we consider the more general situation of f(z).arcsin(z) where f(z) is rational in z and sqrt(1-z^2). By letting z=2w/(1+w^2), note that members of C(z,Z) are always elementarily integrable.

subsection 6
Because x^2+y^2-1 is irreducible, C[x,y]/(x^2+y^2-1) is an integral domain, C(z,Z) is isomorphic to its field of quotients in the obvious manner, and C(z,Z)[t] is a UFD whose field of quotients is amenable to partial fraction analysis in the variable t. What follows takes place at times in various z-algebraic extensions of C(z,Z) (which may not have unique factorization), but the terms must combine to give something in C(z,Z)(t), where partial fraction decompositions are unique, and hence the t term will be as claimed.

Thus, if we can integrate f(z).arcsin(z), we have f.t = sum of (u'/u)s and v', by the main theorem.

The u terms can, by logarithmic differentiation in the appropriate algebraic extension field (recall that roots are analytic functions of the coefficients, and t is transcendental over C(z,Z)), be assumed to all be linear t+r, with r algebraic over z. Then u'/u=(1/Z+r')/(t+r). When we combine such terms back in C(z,Z), they don't form a t term (nor any higher power of t, nor a constant).

Partial fraction decomposition of v gives us a polynomial in t, with coefficients in C(z,Z), plus multiples of powers of linear t terms. The latter don't contribute to a t term, as above.

If the polynomial is linear or quadratic, say v=g.t^2 + h.t + k, then v'=g'.t^2 + (2g/Z+h').t + (h/Z+k'). Nothing can cancel the g', so g is just a constant c. Then 2c/Z+h'=f or I(f.t)=2c.t+I(h'.t). The I(h'.t) can be integrated by parts. So the antiderivative works out to c.(arcsin(z))^2 + h(z).arcsin(z) - I(h(z)/sqrt(1-z^2)), and as observed above, the latter is elementary.

If the polynomial is cubic or higher, let v=A.t^n+B.t^(n-1)+...., then v'=A'.t^n + (n.A/Z+B').t^(n-1) +.... A must be a constant c.  But then nc/Z+B'=0, so B=-nct, contradicting B being in C(z,Z).

subsection 7
In particular, since 1/z + c/sqrt(1-z^2) does not have a rational in "z and/or sqrt(1-z^2)" antiderivative, arcsin(z)/z does not have an elementary integral.


 * z^z

In this case, let F=C(z,l)(t), the field of rational functions in z,l,t, where l=log z and t=exp(z.l)=z^z. Note that z,l,t are algebraically independent. Then t'=(l+1).t, so for a=t in the main theorem, the partial fraction analysis shows that the only possibility is for v=w.t+... to be the source of the t term on the left, with w in C(z,l).

So this means, equating t coefficients, 1=w'+(l+1)w. This is a first order ODE, whose solution is w=I(z^z)/z^z. So we must prove that no such w exists in C(z,l). So suppose w=P/Q, with P,Q in C[z,l] and no common factors. Then z^z=(z^z*P/Q)'=z^z*[(1+l)PQ+P'Q-PQ']/Q^2, or Q^2=(1+l)PQ+P'Q-PQ'. So Q|Q', meaning Q is a constant, which we may assume to be one. So we have it down to P'+P+lP=1.

Let P=Sum[P_i l^i], with P_i, i=0...n in C[z]. But then in our equation, there's a dangling P_n l^(n+1) term, a contradiction.

tangent
On a slight tangent, this theorem of Liouville will not tell you that Bessel functions are not elementary, since they are defined by second order ODEs. This can be proven using differential Galois theory. A variant of the above theorem of Liouville, with a different normal form, does show however that J_0 cannot be integrated in terms of elementary methods augmented with Bessel functions.

sketch of proof: lemmas
What follows is a fairly complete sketch of the proof of the Main Theorem. First, I just state some easy (if you've had Galois Theory 101) lemmas. Throughout the lemmas F is a differential field, and t is transcendental over F.

Lemma 1: If K is an algebraic extension field of F, then there exists a unique way to extend the derivation map from F to K so as to make K into a differential field.

Lemma 2: If K=F(t) is a differential field with derivation extending F's, and t' is in F, then for any polynomial f(t) in F[t], f(t)' is a polynomial in F[t] of the same degree (if the leading coefficient is not in Con(F)) or of degree one less (if the leading coefficient is in Con(F)).

Lemma 3: If K=F(t) is a differential field with derivation extending F's, and t'/t is in F, then for any a in F, n a positive integer, there exists h in F such that (a*t^n)'=h*t^n. More generally, if f(t) is any polynomial in F[t], then f(t)' is of the same degree as f(t), and is a multiple of f(t) iff f(t) is a monomial.

These are all fairly elementary. For example, (a*t^n)'=(a'+at'/t)*t^n in lemma 3. The final 'iff' in lemma 3 is where transcendence of t comes in. Lemma 1 in the usual case of subfields of M is an easy consequence of the implicit function theorem.

main theorem: restatement
MAIN THEOREM. Let F,G be differential fields, let a be in F, let y be in G, and suppose y'=a and G is an elementary differential extension field of F, and Con(F)=Con(G). Then there exist c_1,...,c_n in Con(F), u_1,...,u_n, v in F such that

u_1'           u_n' a = c_1 --- + ... + c_n --- + v'. u_1            u_n

In other words, the only functions that have elementary antiderivatives are the ones that have this very specific form.

main theorem: proof
Proof:

By assumption there exists a finite chain of fields connecting F to G such that the extension from one field to the next is given by performing an algebraic, logarithmic, or exponential extension. We show that if the form (*) can be satisfied with values in F2, and F2 is one of the three kinds of allowable extensions of F1, then the form (*) can be satisfied in F1. The form (*) is obviously satisfied in G: let all the c's be 0, the u's be 1, and let v be the original y for which y'=a. Thus, if the form (*) can be pulled down one field, we will be able to pull it down to F, and the theorem holds.

So we may assume without loss of generality that G=F(t).

Case 1: t is algebraic over F. Say t is of degree k.  Then there are polynomials U_i and V such that U_i(t)=u_i and V(t)=v. So we have

U_1(t)'           U_n(t)' a = c_1 -- + ... + c_n -- + V(t)'. U_1(t)            U_n(t)

Now, by the uniqueness of extensions of derivatives in the algebraic case, we may replace t by any of its conjugates t_1,..., t_k, and the same equation holds. In other words, because a is in F, it is fixed under the Galois automorphisms. Summing up over the conjugates, and converting the U'/U terms into products using logarithmic differentiation, we have

[U_1(t_1)*...*U_1(t_k)]' k a = c_1 --- + ...  + [V(t_1)+...+V(t_k)]'. U_1(t_1)*...*U_n(t_k)

But the expressions in [...] are symmetric polynomials in t_i, and as they are polynomials with coefficients in F, the resulting expressions are in F. So dividing by k gives us (*) holding in F.

Case 2: t is logarithmic over F. Because of logarithmic differentiation we may assume that the u's are monic and irreducible in t and distinct. Furthermore, we may assume v has been decomposed into partial fractions. The fractions can only be of the form f/g^j, where deg(f)<def(g) and g is monic irreducible. The fact that no terms outside of F appear on the left hand side of (*), namely just a appears, means a lot of cancellation must be occuring.

Let t'=s'/s, for some s in F. If f(t) is monic in F[t], then f(t)' is also in F[t], of one less degree. Thus f(t) does not divide f(t)'. In particular, all the u'/u terms are in lowest terms already. In the f/g^j terms in v, we have a g^(j+1) denominator contribution in v' of the form -jfg'/g^(j+1). But g doesn't divide fg', so no cancellation occurs. But no u'/u term can cancel, as the u's are irreducible, and no (**)/g^(j+1) term appears in a, because a is a member of F. Thus no f/g^j term occurs at all in v.  But then none of the u's can be outside of F, since nothing can cancel them. (Remember the u's are distinct, monic, and irreducible.) Thus each of the u's is in F already, and v is a polynomial. But v' = a - expression in u's, so v' is in F also. Thus v = b t + c for some b in con(F), c in F, by lemma 2. Then

u_1'           u_n'     s'          a =  c_1 --- + ... + c_n --- + b --- + c'                  u_1             u_n      s

is the desired form. So case 2 holds.

Case 3: t is exponential over F. So let t'/t=s' for some s in F.  As in case 2 above, we may assume all the u's are monic, irreducible, and distinct and put v in partial fraction decomposition form. Indeed the argument is identical as in case 2 until we try to conclude what form v is. Here lemma 3 tells us that v is a finite sum of terms b*t^j where each coefficient is in F. Each of the u's is also in F, with the possible exception that one of them may be t. Thus every u'/u term is in F, so again we conclude v' is in F. By lemma 3, v is in F. So if every u is in F, a is in the desired form. Otherwise, one of the u's, say u_n, is actually t, then

u_1' a = c_1 --- + ... + (c_n s + v)' u_1

is the desired form. So case 3 holds.

post 2
[                                                                        ] [META-THEOREM: If you can't see how to integrate it easily, it probably   ] [	can't be integrated in closed form. ] [                                                                        ] [Here is a sample question of this type, a summary of why such things are ] [	not integrable in closed terms, and a bibliography. ] [A somewhat revised version of Wiener's long post is also available; see ] [      http://www.math-atlas.org/97/nonelem_integr2         -- djr       ] [                                                                        ] &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

From: mlerma@carl.ma.utexas.edu (Miguel Lerma) Newsgroups: sci.math Subject: Re: CAN YOU SOLVE THIS INTEGRAL, PLEASE? Date: 24 Aug 1995 23:37:19 GMT

Ernesto Alfonso Piwonka Carrasco (eapiwonk@malloco.ing.puc.cl) wrote:
 * Dear reader:
 * I can see that you have been interested in
 * solving an integral. So, I'll tell you that I've been trying to solve it
 * by transforms of coordinates, variable changing and with a large list of
 * tricks I know, and I'm still trying. Maybe, because I'm not an expert in
 * Calculus yet: I learnt Calculus in school (very good) and I'm now starting
 * to learn integration in UNiversity. But you are here to solve this integral,
 * so here I go:
 * so here I go:


 * $$\int\sqrt{1+9x^4}\,dx$$


 * So, try it! Please, If you find the integral, send
 * me the answer by e-mail to the following electronic adress:


 * eapiwonk@ing.puc.cl


 * Remember: the integral is INT(SQRT(1+9x^4))dx. I
 * say it because I think that somebody can read bad and not see the square
 * root.


 * Ernesto Alfonso Piwonka Carrasco
 * Ingenieria Civil
 * Pontificia Universidad Catolica de Chile

Chebyshev' theorem
That integral is non elementary by Chebyshev's theorem. In general, the integral of


 * $$\int x^p (a + b x^r)^q\, dx$$

is elementary if and only if at least one of  (p+1)/r,  q,  or  (p+1)/r + q  is an integer (see Marchisoto & Zakeri: "An Invitation to Integration in Finite Terms", The College Mathematics Journal, Vol. 25, No. 4, Sept.1994, pp.295-308).

Miguel A. Lerma 8/24/95

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener) Newsgroups: sci.math Subject: Re: integrate x^x ? Date: 21 Feb 1995 20:13:46 GMT

In article <3icq8l$df@news.kth.se>, tordm@vana (Tord G Malmgren) writes: >>I could really use a solution to the problem of integrating x^x dx >>(x to the xth). It has been keeping me up late at the math library >>for days, & I can not find a solution or a good reference/article on >>this.

> hmm.. isn't this extremly basic? you might be killing a mosquito >with a sledgehammer. Couldn't you write x^x=exp(xlnx).

I really really really ought to polish this up for FAQ inclusion. I am combining two old articles of mine, the first giving and sketching the general Liouville theory, the second applying this theory to x^x:

section
We give a fairly complete sketch of the proof that certain functions, including the asked for one, are not integrable in elementary terms. The central theorem is due to Liouville in 1835. His proof was analytic. The sketch below is mostly algebraic and is due to Maxwell Rosenlicht. See his papers in the _Pacific Journal of Mathematics_, 54, (1968) 153-161 and 65, (1976), 485-492.

WARNING: Prerequisites for understanding the proof is a first year graduate course in algebra, and a little complex analysis. No deep results are used, but I cannot take the time to explain standard notions or all the deductions.

Notation: a^n is "a power n", a_n is "a sub n". C is the complex numbers, for fields F, F[x] is the ring of polynomials in x OR an algebraic extension of F, F(x) is the field of rational functions in a transcendental x, M is the field of meromorphic functions in one variable. If f is a complex function, I(f) will denote an antiderivative of f.

A differential field is a field F of characteristic 0 with a derivation. Thus, in addition to the field operations + and *, there is a derivative mapping ':F->F such that (a+b)'=a'+b' and (ab)'=a'b+ab'. Two standard examples are C(z) and M with the usual derivative map. Notice a basic identity (logarithmic differentiation) holds:

[(a_1 ^ k_1) * ... * (a_n ^ k_n)]'     a_1'             a_n' - = k_1 --- + ... + k_n --- (a_1 ^ k_1) * ... * (a_n ^ k_n)       a_1              a_n

The usual rules like the quotient rule also hold. If a in F satisfies a'=0, we call a a constant of F. The set of constants of F is called Con(F), and forms a subfield of F.

The basic idea in showing something has no elementary integral is to reduce the problem to a sequence of differential fields F_0, F_1, etc., where F_0 = C(z), and F_(i+1) is obtained from F_i by adjoining one new element t. t is obtained either algebraically, because t satisfies some polynomial equation p(t)=0, or exponentially, because t'/t=s' for some s in F_i, or logarithmically, because t'=s'/s is in F_i. Notice that we don't actually take exponentials or logarithms, but only attach abstract elements that have the appropriate derivatives. Thus a function f is integrable in elementary terms iff such a sequence exists starting with C(z).

Just so there is no confusion, there is no notion of "composition" involved here. If you want to take log s, you adjoin a transcendental t with the relation t'=s'/s. There is no log function running around, for example, except as motivation, until we reach actual examples.

We need some easy lemmas. Throughout the lemmas F is a differential field, and t is transcendental over F.

Lemma 1: If K is an algebraic extension field of F, then there exists a unique way to extend the derivation map from F to K so as to make K into a differential field.

Lemma 2: If K=F(t) is a differential field with derivation extending F's, and t' is in F, then for any polynomial f(t) in F[t], f(t)' is a polynomial in F[t] of the same degree (if the leading coefficient is not in Con(F)) or of degree one less (if the leading coefficient is in Con(F)).

Lemma 3: If K=F(t) is a differential field with derivation extending F's, and t'/t is in F, then for any a in F, n a positive integer, there exists h in F such that (a*t^n)'=h*t^n. More generally, if f(t) is any polynomial in F[t], then f(t)' is of the same degree as f(t), and is a multiple of f(t) iff f(t) is a monomial.

These are all fairly elementary. For example, (a*t^n)'=(a'+at'/t)*t^n in lemma 3. The final 'iff' in lemma 3 is where transcendence of t comes in. Lemma 1 in the usual case of subfields of M can be proven analytically using the implicit function theorem.

MAIN THEOREM. Let F,G be differential fields, let a be in F, let y be in G, and suppose y'=a and G is an elementary differential extension field of F, and Con(F)=Con(G). Then there exist c_1,...,c_n in Con(F), u_1,...,u_n, v in F such that u_1'           u_n' a = c_1 --- + ... + c_n --- + v'. u_1            u_n

In other words, the only functions that have elementary anti-derivatives are the ones that have this very specific form.

This is a very useful theorem for proving non-integrability. In the usual case, F,G are subfields of M, so Con(F)=Con(G)=C always holds.

Proof: By assumption there exists a finite chain of fields connecting F to G such that the extension from one field to the next is given by performing an algebraic, logarithmic, or exponential extension. We show that if the form (*) can be satisfied with values in F2, and F2 is one of the three kinds of allowable extensions of F1, then the form (*) can be satisfied in F1. The form (*) is obviously satisfied in G: let all the c's be 0, the u's be 1, and let v be the original y for which y'=a. Thus, if the form (*) can be pulled down one field, we will be able to pull it down to F, and the theorem holds.

So we may assume without loss of generality that G=F(t).

Case 1: t is algebraic over F. Say t is of degree k.  Then there are polynomials U_i and V such that U_i(t)=u_i and V(t)=v. So we have

U_1(t)'           U_n(t)' a = c_1 -- + ... + c_n -- + V(t)'. U_1(t)            U_n(t)

Now, by the uniqueness of extensions of derivatives in the algebraic case, we may replace t by any of its conjugates t_1,..., t_k, and the same equation holds. In other words, because a is in F, it is fixed under the Galois automorphisms. Summing up over the conjugates, and converting the U'/U terms into products using logarithmic differentiation, we have

[U_1(t_1)*...*U_1(t_k)]' k a = c_1 --- + ...  + [V(t_1)+...+V(t_k)]'. U_1(t_1)*...*U_n(t_k)

But the expressions in [...] are symmetric polynomials in t_i, and as they are polynomials with coefficients in F, the resulting expressions are in F. So dividing by k gives us (*) holding in F.

Case 2: t is logarithmic over F. Because of logarithmic differentiation we may assume that the u's are monic and irreducible in t and distinct. Furthermore, we may assume v has been decomposed into partial fractions. The fractions can only be of the form f/g^j, where deg(f)<def(g) and g is monic irreducible. The fact that no terms outside of F appear on the left hand side of (*), namely just a appears, means a lot of cancellation must be occuring.

Let t'=s'/s, for some s in F. If f(t) is monic in F[t], then f(t)' is also in F[t], of one less degree. Thus f(t) does not divide f(t)'. In particular, all the u'/u terms are in lowest terms already. In the f/g^j terms in v, we have a g^(j+1) denominator contribution in v' of the form -jfg'/g^(j+1). But g doesn't divide fg', so no cancellation occurs. But no u'/u term can cancel, as the u's are irreducible, and no (xx)/g^(j+1) term appears in a, because a is a member of F. Thus no f/g^j term occurs at all in v.  But then none of the u's can be outside of F, since nothing can cancel them. (Remember the u's are distinct, monic, and irreducible.) Thus each of the u's is in F already, and v is a polynomial. But v' = a - expression in u's, so v' is in F also. Thus v = b t + c for some b in con(F), c in F, by lemma 2. Then

u_1'           u_n'     s'          a =  c_1 --- + ... + c_n --- + b --- + c'                  u_1             u_n      s

is the desired form. So case 2 holds.

Case 3: t is exponential over F. So let t'/t=s' for some s in F.  As in case 2 above, we may assume all the u's are monic, irreducible, and distinct and put v in partial fraction decomposition form. Indeed the argument is identical as in case 2 until we try to conclude what form v is. Here lemma 3 tells us that v is a finite sum of terms b*t^j where each coefficient is in F. Each of the u's is also in F, with the possible exception that one of them may be t. Thus every u'/u term is in F, so again we conclude v' is in F.  By lemma 3, v is in F.  So if every u is in F, a is in the desired form. Otherwise, one of the u's, say u_n, is actually t, then

u_1' a = c_1 --- + ... + (c_n s + v)' u_1

is the desired form. So case 3 holds. --QED-- This proof, by the way, is a LOT easier than it looks. Just work out some examples, and you'll see what's going on. (If this were a real expository paper, such examples would be provided. Maybe it's better this way.  Indeed, if anybody out there takes the time to work some out and post them, I would be much obliged.)

So how to you actually go about using this theorem? Suppose you want to integrate f*exp(g) for f,g in C(z), g non zero. [This isn't yet the asked for problem.] Let t=exp(g), so t'/t=g'. Let F=C(z)(t), G=any differential extension field containing an antiderivative of f*t. [Note that t is in fact transcendental over C(z): g is rational and non-zero, so it has a pole (possibly at infinity) and so t has an essential singularity and can't be algebraic over C(z).] Is G an elementary extension? If so, then

u_1'           u_n' f*t = c_1 --- + ... + c_n --- + v'                       u_1             u_n

where the c_i, u_i, and v are in F. Now the left hand side can be viewed as a polynomial in C(z)[t] with exactly one term. We must identify the coefficient of t in the right hand side and get an equation for f. But the first n terms can be factored until the u_i's are linear (using the logarithmic differentiation identity to preserve the abstract from). As for the v' term, long divide and use partial fractions to conclude v is a sum of monomials: if v had a linear denominator other than t, raised to some power in its partial fraction decomposition, its derivative would be one higher power, and so cannot be cancelled with anything from the u_i terms. (As in the proof.) If w is the coefficient of t in v, we have f=w'+wg' with w in C(z). Solving this first order ODE, we find that w=exp(-g)*I(f*exp(g)). In other words, if an elementary antiderivative can be found for f*exp(g), where f,g are rational functions, then it is of the form w*exp(g) for some rational function w. [Notice that the conclusion would fail for g equal to 0!]

For example, consider f=1 and g=-z^2. Now exp(z^2)*I(exp(-z^2)) has no poles (except perhaps at infinity), so if it is a rational function, it must be a polynomial. So let (p(z)*exp(-z^2))'=exp(-z^2). One quickly verifies that p'-2zp=1. But the only solution to that ODE is the error function I(exp(-z^2)) itself (within an additive constant somewhere)! And the error function is NOT a polynomial! (Proof? OK, for one thing, its Taylor series obtained by termwise integration is infinite.  For another, its derivative is an exponential.)

As an exercise, prove that I(exp(z)/z) is not elementary. Conclude that neither is I(exp(exp(z))) and I(1/log(z)).

For a slightly harder exercise, prove that I(sin(z)/z) is not elementary. Conclude that neither is I(sin(exp(z))).

Finally, we consider the case I(z^z).

So this time, let F=C(z,l)(t), the field of rational functions in z,l,t, where l=log z and t=exp(zl)=z^z. Note that z,l,t are algebraically independent. (Choose some appropriate domain of definition.) Then t'=(1+l)t, so for a=t in the above situation, the partial fraction analysis (of the sort done in the previous posts) shows that the only possibility is for v=wt+... to be the source of the t term on the left, with w in C(z,l).

So this means, equating t coefficients, 1=w'+(l+1)w. This is a first order ODE, whose solution is w=I(z^z)/z^z. So we must prove that no such w exists in C(z,l). So suppose (as in one of Ray Steiner's posts) w=P/Q, with P,Q in C[z,l] and having no common factors. Then z^z= (z^z*P/Q)'=z^z*[(1+l)PQ+P'Q-PQ']/Q^2, or Q^2=(1+l)PQ+P'Q-PQ'. So Q|Q', meaning Q is a constant, which we may assume to be one. So we have it down to P'+P+lP=1.

Let P=Sum[P_i l^i], with P_i, i=0...n in C[z]. But then in our equation, there's a dangling P_n l^(n+1) term, a contradiction. -- -Matthew P Wiener (weemba@sagi.wistar.upenn.edu)

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Newsgroups: sci.math Date: Wed, 29 Apr 92 22:44:04 -0400 From: tga@math.appstate.edu (Terry Anderson)

The book "DEs with Applications and Historical Notes" by George F. Simmons (2nd edition, McGraw Hill) contains a note on Joseph Liouville (1809-1882), who studied the question of integration in finite terms.

References given are:

D. G. Mead, "Integration," American Mathematical Monthly, vol. 68, pp. 152-156 (1961). (QA 1 .A515)

G. H. Hardy, The Integration of Functions of a Single Variable, Cambridge Univ ersity Press,  London, 1916.

J. F. Ritt, Integration in Finite Terms,  Columbia University Press, New York, 1948. (QA 308   .R5)

Simmons' book also has extensive historical notes on Newton, Gauss, and Euler as well as short notes on Fermat, the Bernoulli family, Riemann, Laplace, Abel, Poincare', and others. More recent references on the solution of DEs in cl osed form are given in

"Computer Algebra: Past and Future," by B. F. Caviness, Journal of Symbolic Computation (1986), 217-236 (see pp. 225-6, in particular).

J. J. Kovacic, "An Algorithm for solving 2nd order linear homogeneous DEs," manuscript (1979). Also in J. Symbolic Computation 2, (1986), 3-43.

M. F. Singer, "Functions satisfying elementary relations," Transactions of the AMS 227                       (1977), 185-206.

M. F. Singer, "Liouvillian solutions of nth order homogeneous linear DEs,"  Amer. J. Math. 103                (1981), 661-682.

M. F. Singer, B. D. Saunders, and B. F. Caviness, "An extension of Liouville's theorem on integration in finite terms,"  SIAM J. Comput. 14 (1985), 966-990.

B. D. Saunders, "An implementation of Kovacic's algorithm for solving second order linear homogeneous DEs,"   in Proc. 1981 ACM Symp. on Symbolic and Algebraic Computation (editor P. S. Wang), pp. 105-108.

R. H. Risch - several papers on integration in finite terms.

M. Rosenlicht, "On Liouville's theory of elementary functions," Pacific J. Math. 65 (1976), 485-492.

B. W. Char, "Using Lie transformation groups to find closed form solutions to first order ordinary differential equations,"  in Proc. 1981 ACM Symp. on Symbolic and Algebraic Computation (editor P. S. Wang), pp. 44-50.

M. J. Prelle and M. F. Singer, "Elementary first integrals of DEs," Trans. AMS 279 (1983), 215-229. Kovacic's algorithm has been implemented in MACSYMA and MAPLE.

See Saunders above for MACSYMA and the following for MAPLE.

B. W. Char, G. J. Fee, K. O. Geddes, G. H. Gonnet, and M. B. Monagan, "A tutorial introduction to MAPLE,"  J. Symbolic Computation 2 (1986), 179-200.

C. Smith,  "A discussion and implementation of Kovacic's algorithm for ODEs,"   University of Waterloo Computer Science Dept. Research Report CS-84-35 (1984).

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Newsgroups: sci.math Date: Wed, 29 Apr 92 11:20:48 -0500 From: Bruce E Litow  [excerpt]

Indeed, not so elementary. Do the names Abel, Galois and Liouville mean anything? A good survey is in "Computer Algebra", by Davenport, Siret and Tournier, Academic Press, 1988

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From: edgar@math.ohio-state.edu (Gerald Edgar) Newsgroups: sci.math Subject: Re: On elementary functions... Date: Tue, 08 Nov 1994 09:28:14 +0000

Introductory papers, aimed at undergraduates: A.D. Fitt & G.T.Q. Hoare, "The closed-form integration of arbitrary functions". Mathematical Gazette (1993) 227--236. E. Marchisotto & G. Zakeri, "An invitation to integration in finite terms". College Math. J. 25 (1994) 295--308. . . . . Gerald A. Edgar         Internet:edgar@math.ohio-state.edu Department of Mathematics The Ohio State University  telephone: 614-292-0395(Office) Columbus, OH 43210                  292-4975 (Math. Dept.)

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From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Newsgroups: sci.math Subject: Re: Integration of algebraic functions Date: 31 Mar 1998 16:45:44 -0500

In article <3520AB8E.7EA7C649@student.luth.se>, Erik Stenelund  wrote:
 * Hi !
 * I am a student of mathematics at Lulea University in Sweden and I wonder
 * if anyone could mail me some papers about symbolic integration of
 * algebraic functions
 * algebraic functions

An older book (1981) is this (and it has a large bibliography, including three classical articles by Liouville):

James Harold Davenport: On the Integration of Algebraic Functions Lecture Notes in Computer Science No. 102 Springer-Verlag Berlin-Heidelberg-New York ISBN 0-387-10290-6 (New York) ISBN 3-540-10290-6 (Berlin)

Good luck, ZVK (Slavek).

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From: Nissim Broudo  Newsgroups: sci.math,sci.math.symbolic,sci.logic,sci.math.research Subject: non-elementary integrals Date: Wed, 29 Apr 1998 17:22:45 -0400

[deletia -- djr]

A good reference on non-elementary integrals is American Mathematical Monthly Februrary 1961 "Integration" by D.G. Mead p 152-156.

This article proves that the indefinite integrals of certain elementary functions are not themselves composed of elementary functions. That is, the space of elementary functions is not closed under 'antiderivation.'

[deletia -- djr]

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Newsgroups: sci.math.symbolic,sci.math From: kogeddes@daisy.uwaterloo.ca (Keith O. Geddes) Subject: Re: Symbolic integral of x^x Date: Wed, 23 Sep 1998 19:58:49 GMT

In article <3608984C.8AB73A3@ccis.adisys.com.au>, Dion Mendel  wrote: >Hi all > >Does anyone know if there is a symbolic solution to the integration >of x to the power x? This was a puzzle posed to me by a neighbour >which I have been unable to solve. I've tried querying MAPLE with >no luck. > >Any help with this puzzle would be most appreciated. > >TIA > >Dion Mendel.

Well, Maple actually invoked an algorithm (the Risch algorithm) which is able to prove that:

Within the class of elementary functions, the function x^x does not have an anti-derivative.

To see how such a statement can be proved see, for example:

K.O. Geddes, S.R. Czapor, and G. Labahn, Algorithms for Computer Algebra. Kluwer Academic Publishers, Boston, 1992, 585 pages. ISBN 0-7923-9259-0

In particular, Example 12.16 (page 560) handles this example.

--- Professor Keith Geddes Symbolic Computation Group Department of Computer Science University of Waterloo Waterloo ON  N2L 3G1 CANADA

E-mail: kogeddes@daisy.uwaterloo.ca URL:   http://daisy.uwaterloo.ca/~kogeddes ---