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From: Walter Roberson on 20 Apr 2010 01:42
Walter Roberson wrote:
> xiscopolo Polo wrote:

> So, rewriting from the mess you posted into matlab, and guessing that
> you mean implied multiplication in some cases:
>
> Zf1 = 1/sqrt(cos(x)^2+i*sin(x)*cos(y)*4*beta) / ke + (2*beta/ke)^2 );
>
> which is a syntax error, as it has too many ')'.
>
> beta = 0.884
>
> Zf2 = 1/2*pi * int(Zf1,y,0,2*pi)
> K = int(2*sin(x) / real(Zf1), x, 0, pi/2)
>
> One might almost suspect that
> the expressions are wrong and that K should reference Zf2 instead of Zf1.

I'm not sure that I trust the result, but I found via Maple that Zf2 can
be transformed to be dependently only on ke, and independent of x. That
then becomes a constant as far as the integral in K is concerned,
leaving you with a trivial integral (provided that you use Zf2 instead
of Zf1 for K.)

The steps that you would do in Maple to achieve this independence upon x
and y, is:

Zf2 := 1/2*Pi * int(convert(Zf1,`1F1`),y=0..2*Pi);

That is, Zf1 gets rewritten in terms of 1F1 forms such as GAMMA and
exp(Pi*I*something) . A fair number of these cancel in the expression.
The integral that results for Zf2 is 1/2*Pi * (97682/15625)*Pi/ke^2

Then in K, provided that ke is a real non-zero number, Re() of this Zf2
will be the same as Zf2, eliminating a nasty computational complexity.
int(2* sin(x) / constant, x = 0..Pi/2) is 2/constant which is
(31250/48841)*ke^2/Pi^2 .

I do not yet trust that the integral of the 1F1 converted form really
does eliminate both x and y from consideration... it doesn't if I simply
the expression before doing the integral, for example. I will need to
cross-check for bugs in the symbolic package. _After_ I get some sleep!
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