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MathGroup Archive 2007

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  • To: mathgroup at
  • From: djokic at
  • Date: Fri, 28 Sep 2007 02:14:55 -0400 (EDT)

A big HI to everybody !!!

A problem I am faced with is about numerically solving principal value
of a 3D integral whose integrand contains an infinity number of
singularities on a surface. So, in order not to be explaining that in
detail, here my problem is:

Q[y_] := (52.*y)/ArcTanh[(2.*y)/5.];
g[y_] := FindRoot[Q[x] == y, {x, 5., 2.}];
R[y_] := Re[g[y][[1]][[2]]];
J[x_, y_, z_] := 4.92*(Cos[x] + Cos[y] + Cos[z]) +
    3.88*(Cos[(x + y)/2.] + Cos[(y + z)/2.] + Cos[(z + x)/2.]);
j[x_, y_, z_] := 3.88*(Cos[(x - y)/2.] + Cos[(y - z)/2.] + Cos[(z - x)/
W[x_, y_, z_] := Sqrt[(39.942 + J[x, y, z] - j[x, y, z])*
     (39.942 - J[x, y, z] - j[x, y, z])];
=CE=A9[K_, T_, =CF=89_] := ((2.*K^2*R[T]^3)/Pi^3)*
    NIntegrate[(((39.942 - j[x, y, z])*(39.942 - j[x, y, z]))/W[x, y,
      (1/(=CF=89^2 - 4.*R[T]^2*W[x, y, z]^2 + 0.000001))*
      Coth[(R[T]*W[x, y, z])/(1.44*T)], {x, -Pi, Pi}, {y, -Pi, Pi},
     {z, -Pi, Pi}];
s = Table[=CE=A9[K = 2., T, =CF=89 = 50.], {T, 1., 152., 2.}]

When obtaining the function =CE=A9[T] and drawing it depending on parameter
T, the data result achieved in such a way does not look so smooth,
particularly in the case when =CF=89 <200. Well, how to get this function
as smoothed as possible?

All the best,

Dejan Djokic

P=2E S.  0.000001 has ben put in the integration to avoid
singularities on the surface.

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