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[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?...

- Thread starter eljose
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[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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OlderDan

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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?eljose said:

[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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OlderDan

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From your original integraleljose 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?..

[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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