Re: Definite integration question
- From: Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Thu, 09 Aug 2007 18:27:59 -0500
Theo Markettos <theom+news@xxxxxxxxxxxxxxxxxxxxxx> writes:
This isn't a homework problem, but something I've been puzzled by for a
while.
I'm trying to find an analytic form of the surface integral:
\int_S{ (l + h cos (phi))^-1 dh dphi }
over the circle h=0 to h0, phi=0 to 2*pi. The one bit of information I
don't know how to represent is that l>h. There's a pole in the system at
l=+/-h, so perhaps I need to include this somehow.
Doing this numerically works, and it converges.
My textbook suggests the substitution g=tan(0.5*x), but evaluating this
returns the answer:
[insert subs result]
which evaluates to zero. It's not as if we're introducing a pole at
g=tan(pi) since the integral will still evaluate.
[look at wikipedia closed form]
According to Maple, the integral over phi (for 0 < h < 1) is
2 pi/sqrt(1 - h^2). If you use the substitution g = tan(phi/2) for this,
be careful because that's undefined at phi = pi: you'll want to do
the intervals 0 < phi < pi and pi < phi < 2 pi separately (or just do one
of them and use symmetry).
The integral over h is then easy.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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