I have a complicated integral function that I'm trying to evaluate, which will have to be done numerically. I wanted to try a simpler one first, one that has an analytic form, because I could check the answer.

The version FT I defined with Integrate works fine.

w0 = Sqrt[2*Pi*zR/(Pi*k)];
zR = 1;
k = 2*Pi/(532*10^-9);
w[z_] := Sqrt[w0^2*(1 + (z/zR)^2)]
f[x_, y_, z_] := Sqrt[2/Pi]*1/w[z]*Exp[-(x^2 + y^2)/(w[z]^2)]

FT[kx_, ky_, z_] := 
  1/(2*Pi)*Integrate[f[x, y, z]*Exp[-I*kx*x - I*ky*y], {x, -∞, ∞}, {y, -∞, ∞}]
FT[1, 1, 1]

However, when I use NIntegrate in place of Integrate, I get an error and no answer.

FT[kx_, ky_, z_] :=  
  1/(2*Pi)*NIntegrate[f[x, y, z]*Exp[-I*kx*x - I*ky*y], {x, -∞, ∞}, {y, -∞, ∞}]    
FT[1, 1, 1]

Anyone know what the problem is?

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    – bbgodfrey
    Aug 23, 2015 at 23:17
  • 1
    $\begingroup$ What error does it give? (Include it in the question, so others searching the site with the same problem might benefit.) $\endgroup$
    – Michael E2
    Aug 24, 2015 at 1:29
  • 2
    $\begingroup$ The "problem" might be... read the docs? In any case, among many ways to resolve this, using MinRecursion -> 5 in the NIntegrate and Chop on the result gets the same end result... $\endgroup$
    – ciao
    Aug 24, 2015 at 2:53
  • $\begingroup$ As noted by @ciao, FT[kx_, ky_, z_] := 1/(2*Pi)*NIntegrate[ f[x, y, z] *Exp[-I*kx*x - I*ky*y], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}, MinRecursion -> 4] // Chop gives the same numerical value as the symbolic solution, 0.00023217. $\endgroup$
    – bbgodfrey
    Aug 24, 2015 at 3:50

1 Answer 1


Modifying your second definition of FT to

FT[kx_, ky_, z_] := 
      f[x, y, z]*Exp[-I*kx*x - I*ky*y], {x, -∞, ∞}, {y, -∞, ∞}, 
      MinRecursion -> 4] // Chop

solves your problem.


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