Re: Definite integration question

On Thu, 9 Aug 2007 16:20:20 -0400, "Zdislav V. Kovarik"
<kovarik@xxxxxxxxxxx> wrote:



On Thu, 9 Aug 2007, Theo Markettos wrote:

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]


The tan(x/2) substitution, when used carelessly, is a poison -
and you saw why.

After some manipulation, you can get the indefinite integral of


Slavek, when you have time, I would appreciate it if you would show in
some detail how you got from here ...

1/(1 + h * cos(phi)) d(phi)

as a function continuous for all real phi. The tricks are called "adding
zero" and "multiplyin by one", and the result is, after we define

k = h / (1 + (1 - h^2)^(1/2)),


.... to here.

1/(1-h^2)^(1/2) * (phi - 2 * arctan (k * sin(phi) / (1 + k * cos(phi))))

Now for 0<h<1, you have 0<k<1, so the continuity of the result is evident.

The outer integral, by h, is then familiar.

(Work out the details, and at least check the indefinite integral!)

Cheers, ZVK(Slavek)

.

Relevant Pages

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  • Re: Definite integration question
    ... I'm trying to find an analytic form of the surface integral: ... It's not as if we're introducing a pole at ... be careful because that's undefined at phi = pi: ... of them and use symmetry). ...
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  • Definite integration question
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