From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Help with solving integral Date: 22 Dec 1999 08:09:17 GMT Newsgroups: sci.math Keywords: reducing non-elementary antiderivatives to elliptic integrals In article <385E5878.C5AC8C6@viii.ntu-kpi.kiev.ua>, Oleg Kulygin wrote: >The problem is to solve this integral > > ( ---------------- > | / a*x*x+b*x+c / > | - / ------------- dx > | \/ x >) but in another article he posted a slightly different integral. That other integrand, sqrt(a*x^2+b*x+c)/x, has an elementary antiderivative, as was shown in a followup by Zdislav Kovarik. Evidently this is what was intended here as well: In article , Virgil wrote: >If b^2 > 4*a*c, try substitution > x = (-b + sec(theta)*sqrt(b^2-4*a*c) )/(2*a) > >If b^2 < 4*a*c, try substitution > x = (-b + tan(theta)*sqrt(4*a*c-b^2) )/(2*a) > >If If b^2 = 4*a*c, then a*x^2 + b*x + c = a*(x + b/(2*a))^2 The integral shown at top, however, is different. This present integrand is sqrt((a*x^2+b*x+c)/x), and has no elementary antiderivative. In each case, we are given a certain curve y = y(x) in the x-y plane and asked to compute an antiderivative integral( y dx ). In the "easy" case, the curve is y^2=(a*x^2+b*x+c)/x^2, which is parameterizable by rational functions. Indeed, one need only parameterize the conic Y^2=(a*x^2+b*x+c) by rational functions x=x(t), Y=Y(t) (say, by letting (x,Y) be the point where the conic meets the line (Y - sqrt(c)) = t*(x - 0) in the case c>0 ) and then letting y(t) = Y(t)/x(t), another rational function. It follows that the integral may be expressed as the integral of a rational function of t, and so an elementary antiderivative exists. (Evidently I am European-minded, according to Kovarik's characterization!) In the "hard" case, the curve is y^2=(a*x^2+b*x+c)/x, which is superficially very similar to the previous case, but in fact quite distinct. As in the previous case we see the ability to parameterize this curve would be sufficient for integration, and equivalent to the ability to parameterize the curve Y^2=a*x^3+b*x^2+c*x . But this curve is not a conic at all; rather it is an elliptic curve (unless b^2=4ac or c=0 or a=0) and so it is known that it cannot be parameterized by rational functions. (Other functions can be used but then the integral is at best described in terms of these; would an answer involving elliptic functions be suitable?) A formal proof that the "hard" integral cannot be given in terms of elementary functions requires some differential algebra. Perhaps it is sufficient to note that both the antiderivative and also elementary functions may be extended to the complex plane but none of the latter have symmetries quite like the former. Study of integrals like these (and the double periods of the integrals) led to the study of Riemann surfaces a century and a half ago. The (more or less) systematic recognition of integrals which can be rendered in closed form is a topic of symbolic algebra. dave Riemann surfaces: index/30-XX.html For symbolic integration see (somewhat incorrectly classified): index/28-XX.html ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Help with solving integral Date: 27 Dec 1999 11:07:07 GMT Newsgroups: sci.math In article <385E5878.C5AC8C6@viii.ntu-kpi.kiev.ua>, Oleg Kulygin wrote: >The problem is to solve this integral > > ( ---------------- > | / a*x*x+b*x+c / > | - / ------------- dx > | \/ x >) In response, I wrote > A formal proof that the integral cannot be given in terms of > elementary functions requires some differential algebra. Kulygin then asked, >But how to solve it using eliptical functions? I'm not sure I'm the best person to describe this, but I'll give it a shot. I'm working from memory here and really only savor the algebraic and geometric aspects of this; someone who really uses this material should correct and fill in the analytic parts. Suppose we wish to integrate 'expressions' which 'involve' sqrt(C(x)), where C is a rational function. (I'll treat the case in which C is a cubic polynomial here, since that covers the case at hand, but much of the general treatment carries over to other cases too.) We need to be more specific: let's assume that what we want to integrate are rational functions of x and y = sqrt(C(x)), that is, elements of the extension field R(x)[y]/(y^2 - C) of the field R(x) of rational functions. We may write every element of this field as a sum P(x) + Q(x) y but it seems to be a little easier technically to use {1, 1/y} as a basis, that is, to write every element as a sum P(x) + Q(x)/y for some rational functions P and Q. Note that (writing D for the differentiation operator d/dx ) D( P(x) + Q(x)/y ) = P'(x) + [Q'(x) -(1/2)Q(x)C'(x)/C(x)]/y so that the two copies of R(x) in this field are actually D-invariant. This helps with antidifferentiation: we are looking for some functions in a larger set whose derivatives include all the P + Q/y, but it is clear that we need only include a few new functions to account for the cokernel of D. For example, in order to antidifferentiate the rational functions P(x) themselves, we first look to see how large is the family of derivatives of rational functions. Using a partial fractions decomposition, we know every rational function is a unique linear combination of powers of x and powers of terms 1/(x-r) for various real r. But each power of x is already the derivative of another power of x (up to a constant), and likewise each power of 1/(x-r) -- after the first -- is the derivative of another such power. So the image of D on R(x) is the complement of the span of the functions 1/(x-r). Our survey of antidifferentiation is then complete when we add to our supply of functions some new functions L_r(x) which have (L_r)'(x) = 1/(x-r). Of course these functions are just L_r(x) = log(x-r), although what, exactly, that means depends on the context of the problem. (Is x real or complex? Do we really want log(|x-r|) ? Is L_r supposed to be single-valued? ...) There are problems with L_r which are perhaps best dispensed with by treating L_r as a (local?) inverse to the very nice function X_r(x) = r + exp(x) which, significantly, has period 2 pi i, that is, it is appropriate to view X_r as a function on the cylinder C / (2 pi i) Z formed by identifying points in the complex plane which differ by an integer multiple of 2 pi i. This little review of antidifferentiation of rational functions sets the stage for the antidifferentiation of the other functions Q(x)/y. We first look to see how much of this space we can obtain as the derivatives of other functions in this same space. Now I will want to be explicit and write C(x) = c(x-r1)(x-r2)(x-r3) for some complex c, r1, r2, r3. (We may assume in our context that the r_i are distinct.) Then (*) D( Q(x)/y ) = [Q'(x) -(1/2)Q(x) (1/(x-r1) + 1/(x-r2) + 1/(x-r3)) ]/y As before it will suffice to consider those Q(x) in a basis of R(x) over R. Taking Q(x) = 1/(x-r)^n for some r not equal to any of the three r_i, then formula (*) and more applications of partial-fractions decompositions shows that the image of D includes the difference between [1/(x-r)^(n+1)]/y and a sum of multiples of [1/(x-r)^m]/y (with m <= n) and 1/(x-r_i)/y. Repeating this procedure, we "reduce" an integral of 1/(x-r)^(n+1) / y to such integrals with (n+1)=1 only. (This method is of course just integration by parts, if it helps to think of it that way). Applying (*) to Q(x) = 1/(x-r1)^n, say, we see the image of D contains [1/(x-r1)^n][ (2n+1)/(x-r1) + 1/(x-r2) + 1/(x-r3) ]/y which, again using partial fractions, is (1/y) times a linear combination of 1/(x-r2), 1/(x-r3), and powers of 1/(x-r1) up to and including the (n+1)st power. So as in the previous paragraph we may "reduce" an integral of 1/(x-r_i)^(n+1) / y to one with lower powers of 1/(x-r_i), except for the integrals 1/(x-r_i)/y themselves. Finally, rather than applying (*) to Q(x)=x^n we simply compute D( Q(x)/y ) = [x^(n-1)][ n C(x) - (1/2) x C'(x) ]/C(x) /y and compare degrees of polynomials in x to see the image of D contains x^(n-1) plus a polynomial of lower degree plus linear combinations of the 1/(x-r_i)/y. Hence we may again "reduce" integrals x^(n-1)/y to simpler ones. So we have shown that antidifferentiation of terms Q(x)/y will be complete once we have included some new functions E_r for each constant r having (E_r)'(x) = 1/(x-r)/y (although in the previous paragraphs I took advantage of the derivative of every rational-function multiple of 1/y except 1/y itself; computing its derivative we see E_r1 + E_r2 + E_r3 must, if defined, differ by a constant from 2/y). Well, these E_r are the elliptic functions, more or less (I can't quite recall the normalizations used, but the point is that this is how many additional functions are needed to be able to integrate all rational expressions involving sqrt(C).) The situation is more complex than with the introduction of the L_r, however. Let me point out some differences: (1) All the L_r were "the same": clearly the integral of 1/(x-r) may be transformed to that of 1/x by a linear substitution, allowing the definition L_r(x) = L_0(x-r) so that only a single additional function is needed, up to composition. On the other hand, the E_r are "all different"; no such composition rule is possible. This is clear since the various (E_r)' all share three poles at the r_i and yet have a distinct fourth pole. Moreover, the E_r depend not just on this parameter r but on the parameters used to define y in the first place. (It turns out we _can_ make some simple substitutions to assume r1=0, r2=1 say, but we still have a third parameter r3 with is essentially impossible to change to another value under simple substitution.) (2) We may define the E_r as is done for the L_r, namely as path integrals of the function we are attempting to antidifferentiate. As we have already seen for the L_r, it is necessary to specify the path used since the integrals are path-dependent. But the E_r are already more complicated, since in order to integrate 1/(x-r)/y, we must first be able to evaluate y itself! This is problematic since y involves a square root, and there are difficulties extracting the square root consistently around the complex plane. We can fix a path joining r1 to r2 and another path joining r3 to the point at infinity on the Riemann sphere; then y may be defined unambiguously on the complement of these two paths and remains analytic. (3) With the modifications of (2), we have at least a well-defined analytic integrand for E_r just as for L_r. But both integrals are still not quite well-defined since integrals around closed loops need not be zero. In the case of the integral of 1/(x-r), the only non-zero loop integrals are those which wrap n times around the point x=r (counterclockwise, say); these integrals evaluate to (2 pi i)n. Thus L_r is only well-defined up to multiples of this complex number. In the case of 1/(x-r), we consider only loops which do not cross the marked paths joining r1 and r2, or r3 and infinity. Such a loop will wrap, say, n times around one of those two arcs and m times around the other. We find the integral over such a loop to be of the form (alpha) m + (beta) n for two fixed complex numbers alpha and beta. Thus E_r will only be well-defined up to _linear combinations_ of these numbers (unless of course paths or simply-connected domains are specified). (4) We noted earlier that L_r is nicely viewed as the inverse of the entire function X_r, and the non-well-definedness of L_r is essentially the periodicity of X_r. Well, the same approach applies to the E_r: let Y_r be the inverse of E_r, so that Y_r is _doubly_ periodic with periods alpha and beta. The appropriate domain for Y_r is then the quotient space C/(Z alpha + Z beta) formed by identifying points in the complex plane which differ by an integer linear combination of alpha and beta. It's easy to see what this quotient space looks like: the complex numbers r alpha + s beta for real r, s in [0,1] form a little parallelogram, copies of which tile the complex plane, with points in distinct tiles being equivalent in the quotient space iff they are in the same relative position in their respective tiles. There must be a little allowance for identifications along the boundary, however, so that the quotient space is formed simply by taking one tile, glueing its left edge to its right, and then glueing its bottom edge to its top. What remains is a torus. This is the appropriate domain for the inverses of the elliptic functions. The theory of elliptic functions is quite classical and falls under the rubric of "special functions". Much of the general theory applies equally well to integration of more general algebraic functions, giving rise to Riemann Surfaces, although it must be admitted that these are less in demand for integration problems; moreover, when the genus is greater than one (e.g. when taking C(x) to be a quintic polynomial in the discussion above) we find there are no interesting counterparts to the best sections of this theory (e.g. there are no triply-periodic analytic functions, except the constants of course!) dave