From: Michael Jørgensen <Michael.Joergensen@intel.com>
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<a<b<c, with c^2 = a^2 + b^2. Is there an algebraic (or
> 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 <Michael.Joergensen@intel.com>
To: Dave Rusin <rusin@math.niu.edu>
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