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