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Re: Please help!

  • To: mathgroup at smc.vnet.net
  • Subject: [mg28735] Re: Please help!
  • From: Otto Linsuain <linsuain+ at andrew.cmu.edu>
  • Date: Fri, 11 May 2001 03:38:51 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Dear "Rex",

Being a physicsist I can see what kind of integrals you are trying to
deal with, but to Mathematica you are just doing the integral:

Y[a_,b_,c_]:=Integrate[Exp[a x^2+b x+c], {x,-Infinity,+infinity}]

With the important assumption that a!=0, otherwise we are doing a
simpler  (but divergent) integral.

This integral can be inmediately recast into:

Y[a_,b_,c_]:=Exp[(4 a c-b^2)/(4 a)]/Sqrt[-a]
Integrate[Exp[-u^2],{u,(b/(2a)-Infinity) Sqrt[-a], (b/(2a)+Infinity)
Sqrt[-a]}]

 or if you prefer:

Y[a_,b_,c_]:=Exp[(4 a c-b^2)/(4 a)]/Sqrt[a]
Integrate[Exp[u^2],{u,(b/(2a)-Infinity) Sqrt[a], (b/(2a)+Infinity)
Sqrt[a]}]

I point out that this result SEEMS to depend of your choice of Sqrt[-a]
(there are two numbers that can be called Sqrt[-a], they have opposite
signs). The result in fact DOES NOT depend on that choice, since
although the coefficient in front of the integral changes sign, the
limits of integration change as well.

The integral that you have left to do is just a Gaussian

Integrate[Exp[-u^2],{u,-Infinity,+Infinity}], which is known to be some
number involving Sqrt[Pi]

Except that the limits of integration have changed. Instead of just the
real line (-Infinity,+Infinity), now you have added b/2a. This has the
effect of dragging the contour of integration up or down by an amount
Im[b/2a]. I believe this does not change the value of the integral. But
we are also multiplying the limits of integration by Sqrt[-a]. This has
the effect of tilting the contour of integration an angle. That will be
important, especially if the angle is 90 degrees, you end up with

Integrate[Exp[-u^2],{u,-i Infinity,+i
Infinity}]=Integrate[Exp[v^2],{v,-Infinity,+Infinity}] divergent!! (In
fact too divergent )

This can be understood more simply by noticing that in the original
integral "a" cannot be positive, since that would make the parabola grow
to +Infinity when x goes to +/-Infinity.

Have you told Mathematica whether your quantities (T0, w) are real or
not? Don't assume that T0^2  ( I don't know that Mathematica accepts the
underscore "_" in T_0 ) makes things positive, or that i w is imaginary.
It does work that way. T0^2 and I w (by the way I believe you must type
"I w" and not "i w") could have any phase, and if Mathematica cannot
find an answer that is true regardless of those phases, it will not give
you the answer, say, for real T0 and w. I tried this integral in version
4 and Mathematica is kind enough to answer with and If statement,
something like If phase of this is such and such, then this result,
otherwise, returns the integral unevaluated. What else can be asked!!

I hope this is helpful. Good luck! Otto Linsuain.


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