Re: A beginer's simple question about Mathematica...
- To: mathgroup at smc.vnet.net
- Subject: [mg63561] Re: A beginer's simple question about Mathematica...
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Thu, 5 Jan 2006 03:12:32 -0500 (EST)
- Organization: The Open University, Milton Keynes, UK
- References: <dpg11e$pm4$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
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]}]
>
> get
>
> \!\(\(\(1\/\@\(1\/b +
> 1\/c\)\)\((\@\(2\ p\)\ If[Re[z] > \(-\(1\/2\)\) && Re[\(b + c +
> 4\ a\ z\
> \^2\)\/\(8\ a\ b +
> 8\ a\ c\)] >
> 0, \(-\(\(2\ \[ExponentialE]\^\(1\/2\ \((\(-a\) - b -
> \
> c)\)\)\ \((\(-\[ExponentialE]\^\(\(a\ \((b + c)\)\ \((1 + 2\
> z)\)\^2\)\/\(2\ \
> \((b + c + 4\
> a\ z\^2)\)\)\)\)\ \@\(2\ p\)\ \((1 + 2\ z)\) + 2\
> \@\(\(b + c \
> + 4\ a\ z\^2\)\/\(a\ b + a\ c\)\) + \[ExponentialE]\^\(\(a\ \((b +
> c)\)\ \((1 \
> + 2\ z)\)\^2\)\/\(2\ \((b + c + 4\ a\ z\^2)\)\)\)\ \@\(2\ p\)\ \((1 +
> 2\ z)\)\
> \ Erf[\(1 + 2\ z\)\/\(\@2\ \@\(\(b + c + 4\ a\ z\^2\)\/\(a\ b + a\ \
> c\)\)\)])\)\)\/\((\(b + c + 4\ a\ z\^2\)\/\(a\ b + a\
> c\))\)\^\(3/2\)\)\),
> Integrate[\(-\[ExponentialE]\^\(-\(\(4\ a\^2\ \((b
> + c)\) \
> + \((b + c)\)\ x\^2 + 4\ a\ \((b\^2 + c\^2 +
> x\^2\ z\^2 - c\ \((\(-x\) - 2\ x\ z)\) + b\ \((2\ c
> + x\ \
> \((1 + 2\ z)\))\))\)\)\/\(8\ a\ \((b + c)\)\)\)\)\)\ x, {x, 0, 8}, \
> Assumptions -> Re[z] = \(-\(1\/2\)\) || \((Re[z] > \(-\(1\/2\)\) &&
> Re[\(b + \
> c + 4\ a\ z\^2\)\/\(8\ a\ b + 8\ a\ c\)] = 0)\)]])\)\)\)
>
>
> What does this result mean?
Have a look at
http://documents.wolfram.com/mathematica/Built-inFunctions/NumericalComputation/Integration/FurtherExamples/Integrate.html
Basically, what is returned in this case is a conditional result since
the value of the integral depends on the values of several parameters.
1 b + c + 4 a z
If Re[z] > -(-) && Re[--------------] > 0 then the integral is equal to
2 8 a b + 8 a c
1
----------- Sqrt[2 Pi]
1 1
Sqrt[- + -]
b c
otherwise, it is equal to whatever expression is between the comma after
the if statement and the closing parentheses, times the common factor
1/Sqrt[1/b + 1/c]
EXCEPT if
1 1 b + c + 4 a z
Re[z] <= -(-) OR (Re[z] > -(-) AND Re[--------------] <= 0)
2 2 8 a b + 8 a c
since in this case Mathematica do not knoe how to compute this integral.
In[1]:=
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}]
Out[1]=
2
1 1 b + c + 4 a z
----------- (Sqrt[2 Pi] If[Re[z] > -(-) && Re[--------------] > 0,
1 1 2 8 a b + 8 a c
Sqrt[- + -]
b c
1 1/2 (-a - b - c)
-(------------------- (2 E
2
b + c + 4 a z 3/2
(--------------)
a b + a c
2 2
(a (b + c) (1 + 2 z) )/(2 (b + c + 4 a z ))
(-E Sqrt[2 Pi] (1
+ 2 z) +
2 2
2
b + c + 4 a z (a (b + c) (1 + 2 z) )/(2 (b + c
+ 4 a z ))
2 Sqrt[--------------] + E
Sqrt[2 Pi]
a b + a c
1 + 2 z
(1 + 2 z) Erf[----------------------------]))),
2
b + c + 4 a z
Sqrt[2] Sqrt[--------------]
a b + a c
2 2
Integrate[-Power[E, -((4 a (b + c) + (b + c) x +
2 2 2 2
4 a (b + c + x z - c (-x - 2 x z) + b (2 c + x (1 + 2
z))))/(8 a (b + c)))] x,
2
1 1
b + c + 4 a z
{x, 0, Infinity}, Assumptions -> Re[z] <= -(-) || (Re[z] > -(-)
&& Re[--------------] <= 0)]]
2 2
8 a b + 8 a c
)
Hope this helps,
/J.M.