Re: Can I solve these integrals?
- From: "G. A. Edgar" <edgar@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Mon, 20 Jun 2005 10:50:04 -0400
In article <d96brn$8a8$1@xxxxxxxxxxxx>, JH <nospamgunmm@xxxxxxxxxxx>
wrote:
> Hello,
>
>
> I have to compute these integrals (in order to evaluate a
> electromagnetic field), but I think it isn't possible :
>
> >> int( sin(x) / cos((B+x)/2) * exp(-(A*x)^2), x=0..PI)
>
>
> PI
> / 2 2
> | sin(x) exp(- A x )
> | ------------------- dx
> / / B x \
> 0 cos| - + - |
> \ 2 2 /
>
> >> int( sin(x) * cos((B+x)/2) * exp(-(A*x)^2), x=0..PI)
>
> PI
> /
> | / B x \ 2 2
> | sin(x) cos| - + - | exp(- A x ) dx
> / \ 2 2 /
> 0
>
>
>
> with A > 1. Mupad (and me to!;) cannot give me better than these.
>
> I can to it by numericals evaluations (under Matlab), but I'm searching
> for a simplification of these integrals (because numerical evaluation is
> time consuming!).
>
> Mathematica seems be able to evaluate these integral (with hypergeom
> functions) but I hope it can be resolved exactly... Please let me know
> yours advices.
>
>
Maple does not do the first one, but for the second one it yields
-1/8*Pi^(1/2)*exp(-1/16/A^2)*(2*I*erf(3/4*I/A)*exp(-1/2/A^2)*cos(1/2*B)+
2*I*erf(1/4*I/A)*cos(1/2*B)+I*erf(1/4*(4*A^2*Pi-3*I)/A)*exp(-1/2/A^2)*co
s(1/2*B)-erf(1/4*(4*A^2*Pi-3*I)/A)*exp(-1/2/A^2)*sin(1/2*B)+I*erf(1/4*(4
*A^2*Pi-I)/A)*cos(1/2*B)+erf(1/4*(4*A^2*Pi-I)/A)*sin(1/2*B)-I*erf(1/4*(4
*A^2*Pi+I)/A)*cos(1/2*B)+erf(1/4*(4*A^2*Pi+I)/A)*sin(1/2*B)-I*erf(1/4*(4
*A^2*Pi+3*I)/A)*exp(-1/2/A^2)*cos(1/2*B)-erf(1/4*(4*A^2*Pi+3*I)/A)*exp(-
1/2/A^2)*sin(1/2*B))/A
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
.
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