Change of variables (i don,t understand)

  • Thread starter eljose
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  • #1
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let be the integral:

[tex]\int_1^{\infty}\int_{-\infty}^{\infty}f(x,y)dydx [/tex]

i make the change of variable xy=u y=v whose Jacobian is 1/v but then what would be the new limits?...
 

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  • #2
OlderDan
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eljose said:
let be the integral:

[tex]\int_1^{\infty}\int_{-\infty}^{\infty}f(x,y)dydx [/tex]

i make the change of variable xy=u y=v whose Jacobian is 1/v but then what would be the new limits?...
v is y, so u = xv, and x can never be less than 1. What does that tell you about the possible values of u for any given v?
 
  • #3
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could you write the new limits...i can,t work it out the new values of v for v gives me (-8,8) (here 8 means infinite but for u i got... (0,8) is that true?..what would happen if y choose the change of variable x/y=u y=v? thanx
 
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OlderDan
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eljose said:
could you write the new limits...i can,t work it out the new values of v for v gives me (-8,8) (here 8 means infinite but for u i got... (0,8) is that true?..
From your original integral

[tex]\int_1^{\infty}\int_{-\infty}^{\infty}f(x,y)dydx [/tex]

you have y ranging over all reals and x ranging from +1 to infinity. With v = y, v will range over all reals and with u = xy = xv, u will range from -infinity to v when v is negative and from v to infinity when v is positive. Looks like that gives you

[tex]\int_{-\infty}^{0}\int_{-\infty}^{v}g(u,v)dudv + \int_{0}^{\infty}\int_v^{\infty}g(u,v)dudv [/tex]

g(u,v) includes the contribution from f(x,y) and the Jacobian, and you need to be careful with the signs.
 

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