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 All Discussions : sci.math : Topic : Message < previous || Message: Repost: Integral of x^x Subject: Repost: Integral of x^xAuthor: G. A. Edgar Date Posted: Aug 29 2001 1:23:40:000PM ```From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener) Newsgroups: sci.math Subject: Re: Repost: Integral of x^x Date: 30 Nov 1997 20:42:28 GMT What's the antiderivative of exp(-x^2)? of sin(x)/x? of x^x? ------------------------------------------------------------------------ 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. ------------------------------------------------------------------------ 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. 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 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 u_i 's). ------------------------------------------------------------------------ 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 u_1' u_n' f.t = c_1 --- + ... + c_n --- + v' u_1 u_n with the c_i constants and the u_i and v in F. Each u_i 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). 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 (A.t^2+B.t+C)' A'.t^2 + 2At(gt) + B'.t + B.(gt) + C' ------------- = ------------------------------------- A.t^2+B.t+C A.t^2 + B.t + C (A'+2Ag).t^2 + (B'+Bg).t + C' = ----------------------------- . A.t^2 + B.t + 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)oo but sqrt(z)*h(z)->0 as z->0. No rational function does this. 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. 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. 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). 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. ------------------------------------------------------------------------ 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. ======================================================================== 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. 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. ------------------------------------------------------------------------ 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) Suggestion Box || Home || The Math Library || Help Desk || Quick Reference || Search  