Re: Can I get ComplexExpand to really work?

*To*: mathgroup at smc.vnet.net*Subject*: [mg14634] Re: Can I get ComplexExpand to really work?*From*: "Kevin J. McCann" <kevinmccann at Home.com>*Date*: Wed, 4 Nov 1998 13:47:03 -0500*Organization*: @Home Network*References*: <719f5p$lc6@smc.vnet.net> <71ee5p$7to$1@dragonfly.wolfram.com>*Sender*: owner-wri-mathgroup at wolfram.com

You can also try using Upset to tell Mathematica that a is real: Im[a]^=0;Re[a]^=a; Do a Help on Upset or on ^= Then Integrate[E^(I*a*x^2), {x, -Infinity, Infinity}] produces what you expect (I think) (Sqrt[Pi/2]*(1 + I*Sign[a]))/(a^2)^(1/4) If you know that a is positive, then an additional Upset: Sign[a]^=1; Gives the following answer ((1/2 + I/2)*Sqrt[2*Pi])/(a^2)^(1/4) Cheers, Kevin Hans Staugaard Nielsen wrote in message <71ee5p$7to$1 at dragonfly.wolfram.com>... >Try this > > Integrate[E^(I a >x^2),{x,-Infinity,Infinity},Assumptions->{Im[a]==0,a>0}] > > >Hans > > >Topher Cawlfield wrote: > >> Hi, >> >> I'm having lots of problems getting Mathematica to make simplifying >> assumptions. It always seems to want to produce horribly complex >> results because it assumes every variable is complex. I wish I had >> better control of that. In fact, it would also be nice if I could >> assure Mathematica that certain variables were positive as well. >> >> It sounds like the function ComplexExpand should do the trick, at least >> by assuming that variables are real unless otherwise specified. But it >> doesn't really seem to work for me. Here's an example: >> >> ComplexExpand[Integrate[E^(I a x^2), {x, -Infinity, Infinity}]] >> >> produces: >> >> If[Im[a] == 0, Sqrt[Pi/2] (1 + I Sign[a]) / (a^2)^(1/4), Integrate[E^(I >> a x^2), {x, -Infinity, Infinity}]] >> >> But if it really was assuming that 'a' was real, then it should know >> that Im[a] == 0! If I could also tell it that 'a' was positive, the >> answer would be: >> >> (1 + I) Sqrt[Pi/a] or better still, Sqrt[2 Pi I / a] >> >> This is much simpler, and is the answer I want. >> >> Of course, my real application of this problem is much more complicated, >> but ultimately comes down to doing that integral (several times over). >> The right answer should be just about that simple, but instead >> Mathematica gives me about 5 pages of output. >> >> Is there any hope of getting reasonable symbolic results here? >> >> - Topher Cawlfield > > > >