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Re: Incorrect integral in Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg91386] Re: Incorrect integral in Mathematica
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Tue, 19 Aug 2008 07:16:08 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <g8b8t8$8pq$1@smc.vnet.net>

Eran Mukamel wrote:

> I'm getting a nonsensical answer of 0 for the following integral in Mathematica 5.0 (Mac OSX):
> 
> Integrate[Exp[-x^2/(2w^2)]Exp[-(y - y0)^2/(2z^2)],
>   {x, -Infinity, Infinity}, {y, -Infinity, Infinity}]
> 
> The result should be (2 Pi w z).  Any ideas why this fails?

Seems to be a bug. Note that the result you expect is valid only for 
both w and z real and of same sign, so you may want to add some 
assumptions (either with Assuming[] or the option Assumptions -> ...)

Mathematica 6.0.3 gives a complete and correct solution.

In[1]:= Integrate[
  Exp[-x^2/(2 w^2)] Exp[-(y - y0)^2/(2 z^2)], {x, -Infinity,
   Infinity}, {y, -Infinity, Infinity}]

Out[1]= (1/Sqrt[(1/(z^2))])Sqrt[2 \[Pi]]
   If[Re[w^2] > 0, Sqrt[2 \[Pi]]/Sqrt[1/w^2],
   Integrate[E^(-(x^2/(2 w^2))), {x, -\[Infinity], \[Infinity]},
    Assumptions -> Re[w^2] <= 0]]

In[2]:= Assuming[Element[w | z, Reals],
  Integrate[
   Exp[-x^2/(2 w^2)] Exp[-(y - y0)^2/(2 z^2)], {x, -Infinity,
    Infinity}, {y, -Infinity, Infinity}]]

Out[2]= 2 \[Pi] Abs[w] Abs[z]

In[3]:= $Version

Out[3]= "6.0 for Mac OS X x86 (64-bit) (May 21, 2008)"

Regards,
-- Jean-Marc


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