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6

The solution is a straightforward application of Integrate. Integrate[Exp[-((x - x0)^2 + (y - y0)^2)/(2 c) - I (kx x + ky y)], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}, Assumptions -> c > 0] (* 2 c E^(-(1/2) c (kx^2 + ky^2) - I (kx x0 + ky y0)) π *)


5

Examine your integrand (which is suggested by the error, after all). PiecewiseExpand will collect all terms under one piecewise function. c*h[c, k1, t1]*(1 - H[c, k2, t2])*(1 - H[c, k3, t3]) // PiecewiseExpand (* Power::infy, Infinity::indet errors... *) You can see that the function does not have numeric values for c > 1. How to fix it is ...


4

I would suggest adding option GenerateConditions->False to Integrate to speed up the integration. Then, instead of D, use Derivative. Then, to generate a SeriesData apply Series: f[x_] := 1/x; max = 4; em[n_Symbol] := Series[Integrate[f[x], {x, 1, n}, GenerateConditions -> False] + (f[1] + f[n])/2 + Sum[BernoulliB[2 k]/(2 k)! (Derivative[2 ...


4

Perhaps useful I think this substitution provides a preferable form for the numerical integration: exp = ((w E^(-w/a) Sin[((w - s)^2)/2])/((w - s)^2)/2 /. w -> -a Log[g]) D[-a Log[g], g] f[b_?NumericQ] := Block[{a = 10^-6, s = -10^(-1) Sqrt[10^(-3)^2 + b^2]}, - a^2 NIntegrate[exp/a^2, {g, 0, 1}, WorkingPrecision -> 30, ...


4

Evaluating the integral analytically seems to take far too long on my machine, so I aborted it. Since you mention that you studied the integrated expression, perhaps you could consider showing result you got from the integration in your question. Having said that, numerical integration suggests that the values of the integral are very small over a wide ...


2

Some of the definitions in the original question were problematic. I edited the question to have more consistent code. In order to plot the function numerical values are needed, so using Integrate is not necessary. We can use NIntegrate instead. The plot is produced within 30 seconds on my laptop with Mathematica 10.3.1. Here is the function redefined: ...


1

If numerical results are acceptable; Needs["NumericalCalculus`"] func[θ1_,n_]:= 1/2 ((n Cos[θ1] - 0.2 Sqrt[1 - 25. n^2 Sin[θ1]^2])^2/(n Cos[θ1] + 0.2 Sqrt[1 - 25. n^2 Sin[θ1]^2])^2 + (-0.2 Cos[θ1] + n Sqrt[1 - 25. n^2 Sin[θ1]^2])^2/(0.2 Cos[θ1] + n Sqrt[1 - 25. n^2 Sin[θ1]^2])^2) int[n_?NumberQ] := int[n] = NIntegrate[func[\[Theta]1, n], ...



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