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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