PRINCIPAL VALUE + 3D NINTEGRATE+IMPROVED

*To*: mathgroup at smc.vnet.net*Subject*: [mg81585] PRINCIPAL VALUE + 3D NINTEGRATE+IMPROVED*From*: djokic at phy.bg.ac.yu*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)/ 2=2E]); 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, z])* (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.