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MathGroup Archive 1998

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Re: Integral?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg14180] Re: Integral?
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Wed, 30 Sep 1998 19:42:15 -0400
  • Organization: University of Western Australia
  • References: <6ud2f0$17f@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Torben Mikael Hansen wrote:

> How do I make Mathematica perform the double integral
>
> Integrate[Integrate[Exp[x]*Exp[y]*p[x,y],{x,-Infinity,Infinity}],
> {y,-Infinity,Infinity}]
> 
> where
>
> p[x_,y_]:=1/(2 Pi s^2 Sqrt[1-r^2])*Exp[-(x^2-2 r s x y +y^2)/
> (2 s^2 (1-r^2))]

For your p[x,y] (Daniel entered this incorrectly),

In[1]:=
                    2    2
                   x  + y  - 2 r s x y
             Exp[-(-------------------)]
                              2
                         1 - r
p[x_, y_] := ---------------------------
                      2           2
                2 Pi s  Sqrt[1 - r ]

we suppress the integrate conditions,

In[2]:= SetOptions[Integrate, GenerateConditions -> False];

and compute the integral over x: 

In[3]:= Integrate[Exp[x] Exp[y] p[x, y], {x, -Infinity, Infinity}]

Out[3]=
       2           2               2
      y          (r  - 2 s y r - 1)
Exp[------ + y - -------------------]
     2                   2
    r  - 1           4 (r  - 1)
-------------------------------------
                        2
            2 Sqrt[Pi] s

To simplify this result we "complete the square" as follows:

In[4]:= % /. Exp[a_] :> Exp[Collect[a, y, Simplify]] Out[4]=
          2  2   2
    (1 - r  s ) y                  1       2 Exp[-------------- + (r s +
1) y + - (1 - r )]
         2                         4
        r  - 1
----------------------------------------------
                            2
                2 Sqrt[Pi] s

In[5]:=
           2                            b  2                    b  2 %
/. (a_) y  + (b_) y + (c_) :> a (y + ---)  + Simplify[c - a (---) ]
                                       2 a                     2 a
Out[5]=
                   2
          2  2   (r  - 1) (r s + 1)     2
    (1 - r  s ) (------------------ + y)
                           2  2               2
                   2 (1 - r  s )             r  - 1
Exp[------------------------------------- + ---------]
                    2                       2 r s - 2
                   r  - 1
------------------------------------------------------
                                2
                    2 Sqrt[Pi] s

Since we are integrating with respect to y over -Infinity to Infinity we
do a simple change of variables:

In[6]:=
                2
              (r  - 1) (r s + 1)
% /. y -> y - ------------------
                        2  2
                2 (1 - r  s )
Out[6]=
          2  2   2     2
    (1 - r  s ) y     r  - 1
Exp[-------------- + ---------]
         2           2 r s - 2
        r  - 1
-------------------------------
                     2
         2 Sqrt[Pi] s

and then do the integration:

In[7]:= Integrate[%, {y, -Infinity, Infinity}] Out[7]=
         2
        r  - 1
   Exp[---------]
       2 r s - 2
--------------------
           2  2
   2      r  s  - 1
2 s  Sqrt[---------]
            2
           r  - 1

As a numerical check,

In[8]:= % /. {r -> 0.2, s -> 1.2}
Out[8]= 0.659048

which agrees with the direct numerical integration:

In[9]:= NIntegrate[Evaluate[Exp[x] Exp[y] p[x, y] /. 
	{r -> 0.2, s -> 1.2}], {x, -Infinity, Infinity}, 
		{y, -Infinity, Infinity}, AccuracyGoal -> 4] Out[9]= 0.65907

Cheers,
	Paul 
 
____________________________________________________________________ 
Paul Abbott                                   Phone: +61-8-9380-2734
Department of Physics                           Fax: +61-8-9380-1014
The University of Western Australia            Nedlands WA  6907       
mailto:paul at physics.uwa.edu.au  AUSTRALIA                       
http://www.physics.uwa.edu.au/~paul

            God IS a weakly left-handed dice player
____________________________________________________________________


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