From: Dave Rusin Subject: Re: Another integral to calculate revised Date: 14 Apr 2000 20:36:21 GMT Newsgroups: sci.math.num-analysis Summary: [missing] In article , germinal Magro wrote: >I'm sorry but I forget one constant in the expression of the integral. >The true integral is wrote: > >int(0 to x)(dx/sqrt(a*(1+x)^3+(1-a-b)*(1+x)^2+b)) >with a and b constant Since you're asking again, let me try to combine all the integration themes into one. I'll mostly discuss the symbolics problem, but I'm also asking about the numerical one. How would you compute int(0 to x)( dt/sqrt(1-t^2) ) ? The answer is arcsin(x), but why, exactly? The right way to do this is to view the problem as the evaluation of the function F( (x,y) ) = int((0,1) to (x,y) )( dt/u ) , a path integral over the curve u^2 = 1-t^2 (which of course is the unit circle). There is the problem that the path integral is not well defined, because different (homotopy classes of) paths will give a different value to the path integral; this is a consequence of the fact that the integral around the circle is 2 pi. So F isn't really a function taking the circle to the real numbers; it's a function taking the circle to the quotient group (quotient space) R / (2 pi Z). We can overcome the ambiguity by inverting F: G = F^{-1} is a function defined on R / (2 pi Z) taking values on the circle; equivalently, we can think about G as a function defined on the real line which is periodic: G(a) = G(a + 2 pi n) for integer n. While it's easy to see the geometry this way, there's really no difference if we treat all the variables as complex: we perform path integrals over the complex space defined by u^2 = 1 - t^2, getting complex numbers which are only defined modulo multiples of 2 pi (which is still the integral over the generating homotopy class of nontrivial loops on this surface.) Invert this F to get a function G defined on the complex plane, which admits a single periodicity, and in fact provides a parameterization of this complex surface. Perhaps this isn't the way you think of this integral, although the result itself shouldn't be surprising: the integral is the arcsine, its inverse is the sine, meaning G takes a real or complex number a to the point on the curve with first coordinate equal to sin(a) (and second coordinate equal to cos(a) or -cos(a), depending on which direction you use to parameterize the original curve). The reason this discussion is relevant is that it shows the importance of the geometry of the space defined by the equation u^2=1-t^2. The space u^2=1-t, for example, admits a simple change of variables to flat affine space, meaning that there are no interesting loops over which to integrate, giving G no periodicity at all. And, correspondingly, int(0 to x)( dt/sqrt(1-t) ) is expressed with no reference to sine/cosine. More to the point, your example shows the opposite problem. When f(t) is a typical cubic polynomial, the equation u^2=f(t) describes (in complex 2-space) a complex curve (=real surface) which looks like a torus; in particular there are two distinct nontrivial closed curves over which one might integrate (go once around the donut through the central hole, or else go all the way around the outer perimeter). This means that when we play the same game as above, G will be a function from the complex plane to this torus, having two (independent) periods: G(a + p1 n + p2 m) = 0 for any integers n and m, where p1 and p2 are the (complex) path integrals going around the two paths I just described. The reason this is relevant is that _you (probably) don't know any functions like this_. That is, since the usual elementary functions have at most single periodicity, this G we have just described cannot be expressed using simple algebraic combinations of those. Therefore, you cannot express in any simple formula what this G is, and thus likewise you can't expect to write down a formula for its inverse, F. Your integral (and its inverse) describe a _whole new class_ of functions. There is no way to rewrite the integral except to say it's "the function described by this integral". Your misprint prompted part of this discussion: you accidentally asked for an integral of the type above but with cubic of the form f(t) = t^3+t^2. This cubic has a double root at the origin, which gives the complex curve u^2=f(t) a singular point at the origin, which makes the set not look like a torus any more. Indeed, it is easy to parameterize _this_ set using rational functions of a single variable. (It's similar to the other kind of degeneracy, when f has a _triple_ root; the space given by u^2=t^3 is of course parameterized by u=s^3, t=s^2.) I'm glossing over quite a few technicalities here, but this is the fundamental problem in trying to give formulas for these antiderivatives: the antiderivatives are functions which a special symmetry determined by the geometry of the underlying surface (known as a Riemann surface); one coarse feature of these surfaces is the genus (the number of 'holes'), and unless the surface has genus zero, the functions cannot be expressed in terms of the usual elementary functions. (Actually the converse is true, too: genus=0 implies rationally parameterizable, implies there is a reduction of your integral to one of rational functions, implies the antiderivative is 'elementary'.) If you watch Maple try to resolve some of these integrals, you'll see it mention that it's computing the genus (of a certain extension field of a function field...) I was sort of hoping we would again hear the comment that this is a _numerical analysis_ newsgroup, because I would be interested to learn from the pros whether (hyper)elliptic integrals can be computed more efficiently than your run-of-the-mill function. Sure, we can use Simpson's Rule or whatever, but does the knowledge that the integrand has, say, the form 1/sqrt(f(x)) for a polynomial f give any extra information about how best to approximate the integral with such a sum, with improved error estimates perhaps? It seems like we could do an asymptotic analysis near the roots of f, and then away from the roots the integral is quite well behaved (indeed we can readily identify a finite number of regions on which the integrand is increasing/decreasing, concave up/down etc.) dave