From: Michael Jørgensen Newsgroups: sci.math Subject: Re: Problem on Elliptic Functions Date: Sat, 02 Oct 1999 23:52:41 +0200 Just in case anyone cares... I figured out how to solve the problem myself, and I found the result I(a, i b) = I(b, c) - i I(a, c), c^2 = a^2 + b^2, i^2 = -1. -Michael. Michael Jørgensen wrote: > DEFINITION: > Define the function I(a,b) as the definite integral > I(a,b) = INT(0, infinity) [(a^2+t^2) * (b^2 + t^2)] ^(-1/2) dt > > PROPERTY: > The function I(a,b) may be written in terms of the complete elliptic > integral of the first kind K(k) as follows: > I(a,b) = 1/b * K(k), with k=sqrt(1-a^2 / b^2). > > PROBLEM: > Assume 0 otherwise simple) relation connecting the three values I(a,b), I(a,c), > and I(b,c)??? > > PARTIAL SOLUTION: > The Weierstrass elliptic function P(u) with periods w1 and w2 satisfies > y^2 = 4x^3 - g2 x - g3 = 4(x-x1)(x-x2)(x-x3), for (x,y) = (P(u), > P'(u)). > The periods w1 and w2 may be determined as elliptic integrals involving > the zeros x1, x2, and x3. If we assume the zeros are real and ordered > such that x1 < x2 < x3, then I find > w1 = 2 I(b,c), > w2 = 2 i I(a,c), > for a = sqrt(x2 - x1), b = sqrt(x3 - x2), and c = sqrt(x3-x1). They > obviously satisfy c^2 = a^2 + b^2. > > Now I would expect that I(a,b) is somehow related to w1 + w2. Is anyone > able to shed more light on this little home-made problem? > > Thanks for any suggestions, > > -Michael. ============================================================================== Date: Thu, 30 Sep 1999 14:33:07 +0200 From: Michael Jørgensen To: Dave Rusin Subject: Re: Problem on Elliptic Functions Keywords: Algebraic relations satisfied by elliptic functions Hi Dave, I think I've solved the problem myself, but it wasn't quite what I expected. If we assume a complex analytic continuation into the upper half plane, I arrive at the following result: I(a, ib) = I(b, c) - i I(a, c), when c^2 = a^2 + b^2, using the same notation as in the original post. To get this result I simply write out the integral and divide the integration interval into two parts: 0 to b and b to infinity. The function I(a,b) is related to the complete elliptic integral of the first kind K(k) as follows: I(a, b) = 1/a * K(k), with k^2 = 1- (b/a)^2 If b is replaced by ib as above, then k becomes larger than 1. The above result can thus be rewritten as an expression involving the continuation of K(k) for k>1. Explicitly, I get K(1/x) = x K(x) - i x K(sqrt(1-x^2)), where x = a/c < 1. I am not able to determine whether this last result is correct. Are you able to verify this? -Michael. Dave Rusin wrote: > I don't have a solution but I admire the problem :-) > If you get any responses could you send me a copy? Thanks. > If you don't get any responses in the near future you might consider > reposting to sci.math.research. > dave