To: sci-math-research@moderators.isc.org
From: Alain Verberkmoes <verberkm@wins.uva.nl>
Newsgroups: sci.math.research
Subject: are these integrals known?
Date: 19 Nov 1998 13:57:56 +0100

While studying a problem in statistical mechanics I stumbled
across some integrals unknown to me.  I would like to know if
they can be expressed in terms of `known' functions.

Let b be a complex number and c its complex conjugate.  Define

          (z-1/z)-(b-1/b)  1/6
t(z) := ( --------------- )
          (z-1/z)-(c-1/c)

The integrals I need are:

        b
        /  t(z)+1/t(z)
(A)     \  ----------- dz
        /       z
        c

      -1/b
        /  t(z)+1/t(z)
(B)     \  ----------- dz
        /       z
        b

       inf
        /  t(z)+1/t(z)-1
(C)     \  ------------- dz
        /        z
        0

These integrals depend on the branch of t(z) and (the homology
of) the integration contour chosen, so I'll specify one of the
cases I am interested in.

Let Im b>0, |b|>1.  The branch cuts of t(z) are the arc of the
circle of centre 0 and radius |b| running counterclockwise from
b to c, and the arc of the circle of centre 0 and radius 1/|b|
running clockwise from -1/b to -1/c.  The branch of t(z) is
defined by t(0) = t(inf) = exp(pi i/3).  Note that for this
branch of t(z) the enumerator t(z)+1/t(z)-1 of the integrand in
(C) vanishes when z tends to zero or to infinity.

For (A) the integration contour shall be the arc of the circle
of centre 0 and radius |b| running counterclockwise from c to b.
For (B) and (C) I am only interested in the real part of the
integral, and as far as I can see this does not depend on the
integration contour (as long as it doesn't cross the branch
cuts).

Any help will be greatly appreciated.

Alain.
==============================================================================

From: Dave Rusin <rusin@math.niu.edu>
Date: Thu, 19 Nov 1998 11:20:01 -0600 (CST)
To: verberkm@wins.uva.nl
Subject: Re:  are these integrals known?

It won't be possible to express the integrals in terms of the standard
elementary functions. You're integrating  int(y dz)  where (y,z) lie on an
algebraic curve but this curve has genus  5 -- that is, if you look for a
polynomial which describes the relation between  y  and  z  (you can do it:
it's of degree  6  in y) then look at all complex pairs  (y,z)  satisfying
this polynomial, you get a surface in  C^2  which is topologically a 
5-holed torus. An antiderivative of  y  will thus have to exhibit a
natural transformation under the fundamental group of that surface, which 
none of the standard functions do. (For comparison, we can construct
log(z), arcsin(z) etc  also as  int(y dz) where  yz=1  or  x^2+z^2=1
respectively, but these describe curves of genus 0, and so have a periodicity
only under the action of the integers:  antilog(z)=antilog(z + 2 pi i),
sin(z)=sin(z+2 pi). We need new functions to describe the elliptic
integrals  int(y dz)  where  y^2+cubic(x) = 0  or  y^2+quartic(x) = 0;
these equations describe a curve of genus  1  -- a torus -- and so the
elliptic functions exhibit a regularity under the action of  Z x Z: they
are doubly periodic. You're looking for integrals which come from a
topologically distinct surface, hence they can't be given in terms of
these other functions.)
Of course you can try to evaluate the integrals numerically; if you have
questions about that, try sci.math.num-analysis.

Most of the questions about integrals work out the same way: if it isn't
obvious how to express the integral in closed form, then it isn't possible.
So what? Just define a new function to be whatever your integral is.
Consequently we usually don't carry integration posts in sci.math.research.
If you think this is an exception let me know and I'll reconsider.

Some other remarks on integration are available in
	index/28-XX.html

[deletia -- djr]
==============================================================================

From: Dave Rusin <rusin@math.niu.edu>
Date: Thu, 19 Nov 1998 11:27:22 -0600 (CST)
To: rusin@math.niu.edu
Subject: About that genus-5 thing

Actually I just used CASA to test for genus when b=5,c=7. Took a while!

In general, we use the given equations to eliminate  t(z)  and write the
integral as  int(F/z  dz)  where  F  satisfies a sextic; indeed,  F=G^2
where  G  satisfies  (G-1)^2(G-4) = k z^2/Prod(z-r_i)   (r = b, c, -1/b, -1/c)
(k = (1+1/P)^2(S^2-4P), S=b+c, P=bc.)  

==============================================================================

[ For the corresponding _definite_ integrals, see  math-atlas.org/99/period_matrix  -- djr]