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