Author 
Comment/Response 
Fabrice P. Laussy

11/13/99 07:15am
>I really would appreciate your help in solving this integral and if it's possible show the solution step by step. >
>I wonder if the integral could be plotted, in order to find a simpler function that closely approximates the given function. The integral is a function of z` and z (butI wanted to hold z constant at first).
>
>E=[1+ (1/k^2)(2nd parital diff eq respect to z)]{2 Integral{sin[k(Lz`)] * Exp {jkSqrt[(zz`)^2 + a^2]} / Sqrt[(zz`)^2 + a^2]}dz`}
> The limits are from 0 to L.
>
>where,
> z = L/2 (a try)
> lamda = 3 meters
> L = lamda/4 = 0.75 meters
> a = 0.03 meters
> k = 2 Pi/lamda
> E = Electric field in the gap
>
>I split the integral into the real and imaginary parts. I attempted to integral the imaginary part first.
>I used Taylor series to approximate the function for L=0.4. It turns out that a fourth order expansion is a good approximation. Mathematica was able to integral it. It seems unable to differentiate the integral and add to it the original integral. I would really appreciate your help with this difficult problem.
>I have a copy of Mathematica for students version 3.0.0, license: L27091961, running on a Pentium 333MHz with a Windows 98 OS.
First, you shouldn't bother with all those constants L, a, k, ect... At least, at first, you should reduce your integral to its simplest looking, putting all constants to 1. Later, if it works, you can come back to these.
Then. What is it about? I guess your problem is actually those of solving a complex integral (I suppose your ''j'' is I). But the title says ''electric field of a dipole''. Well, what I want to say is what I had done, namely, add two scalar real fields, one being the potential of the first charge, and the other those of the second charge of the dipole. The gradient of which would have brought me back to the electic field. It's all real. All numerical. All neat.
Try to state better your problem. You have no need for complex integration to arise. You can compute the potential in one Mathematica line.
Laussy.
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