Change of variable in integral of product of exponential and gaussian functions


by galuoises
Tags: exponential, functions, gaussian, integral, product, variable
galuoises
galuoises is offline
#1
Dec24-12, 07:46 PM
P: 8
I have the integral

[itex]\int_{-\infty}^{\infty}dx \int_{-\infty}^{\infty}dy e^{-\xi \vert x-y\vert}e^{-x^2}e^{-y^2}[/itex]

where [itex]\xi[/itex] is a constant. I would like to transform by some change of variables in the form

[itex]\int_{-\infty}^{\infty}dx F(x) \int_{-\infty}^{\infty}dy G(y)[/itex]

the problem is that due to absolute value in the integral one must take in account where x is greater or less than y,

can someone help me, please?
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lurflurf
lurflurf is offline
#2
Dec25-12, 04:18 AM
HW Helper
P: 2,148
First observe that

[itex]e^{-\xi \vert x-y\vert}e^{-x^2}e^{-y^2}=e^{-\xi \vert x-y\vert}e^{-(x-y)^2/2}e^{-(x+y)^2/2}[/itex]

Then you can either change variables such as
u=(x+y)/sqrt(2)
v=(x-y)/sqrt(2)
or break into two regions
x<y
x>y
JJacquelin
JJacquelin is offline
#3
Dec26-12, 03:06 AM
P: 744
Hi !

the clolsed form of the integral involves a special function (erf).
Attached Thumbnails
Exp Erfc.JPG  

galuoises
galuoises is offline
#4
Dec26-12, 08:32 AM
P: 8

Change of variable in integral of product of exponential and gaussian functions


Nice trick! Thank you so much!


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