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Re: A beginer's simple question about Mathematica...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg63618] Re: A beginer's simple question about Mathematica...
  • From: Maxim <m.r at inbox.ru>
  • Date: Sat, 7 Jan 2006 03:13:51 -0500 (EST)
  • References: <dpg11e$pm4$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On Wed, 4 Jan 2006 08:29:34 +0000 (UTC), <leelsuc at yahoo.com> wrote:

> I try to run
>
> Integrate[x*Exp[-((x-2*a)^2/(8*a)+(y-b)^2/(2*b)+(z*x-y-c)^2/(2*c))],{x,-\[infinity],0},{y,-\[infinity],+\[infinity]}]
>

>
> What does this result mean?
>
> Thanks a lot for replying.
>

I'll try to illustrate a couple of subtle points on a simpler example.  
With the default setting GenerateConditions -> Automatic Mathematica  
returns a conditional answer only for the outer integral:

In[1]:= Integrate[Exp[-a*x^2 - 2*b*x*y - c*y^2],
     {x, -Infinity, Infinity}, {y, -Infinity, Infinity}] //
   Cases[#, If[a_, __] -> a, {0, -1}]&

Out[1]= {Re[b^2/c] < Re[a]}

In[2]:= Integrate[Exp[-a*x^2 - 2*b*x*y - c*y^2],
     {x, -Infinity, Infinity}, {y, -Infinity, Infinity},
     GenerateConditions -> True] //
   Cases[#, If[a_, __] -> a, {0, -1}]&

Out[2]= {Re[b^2/c] < Re[a], Re[c] > 0}

Obviously, if the second condition is not satisfied, the integral will  
diverge. Further, the conditions in Out[2] only tell us when the repeated  
integral Integrate[Integrate[..., y], x] converges. This is not the same  
as the convergence of the double integral in the usual sense. E.g., if we  
take a = -1, b = 2I, c = 1, the double integral clearly diverges but the  
repeated integral exists:

In[3]:= % /. {a -> -1, b -> 2*I, c -> 1}

Out[3]= {True, True}

In[4]:= Integrate[Exp[-a*x^2 - 2*b*x*y - c*y^2] /.
     {a -> -1, b -> 2*I, c -> 1},
   {x, -Infinity, Infinity}, {y, -Infinity, Infinity}]

Out[4]= Pi/Sqrt[3]

The double integral converges when the matrix {{a, b}, {b, c}} is positive  
definite. Its hermitian part is Re[{{a, b}, {b, c}}], therefore that  
condition is equivalent to

In[5]:= {Re[a] > 0, Det[Re[{{a, b}, {b, c}}]] > 0}

Out[5]= {Re[a] > 0, -Re[b]^2 + Re[a]*Re[c] > 0}

Not surprisingly, what we obtained is a subset of the conditions from  
Out[2]:

In[6]:= Reduce[ForAll[{a, b, c}, And @@ %5, And @@ %2]]

Out[6]= True

Maxim Rytin
m.r at inbox.ru


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