MathGroup Archive 2006

[Date Index] [Thread Index] [Author Index]

Search the Archive

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.


  • Prev by Date: Re: How to hide a cell?
  • Next by Date: Evaluating x^0
  • Previous by thread: Re: A beginer's simple question about Mathematica...
  • Next by thread: Re: A beginer's simple question about Mathematica...