How Do You Evaluate an Integral Using a Jacobian and Change of Variables?

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SUMMARY

The discussion focuses on evaluating the integral I = ∫₀^∞ ∫₀^∞ (x² + y²)/(1 + (x² - y²)²) e^(-2xy) dx dy using a change of variables defined by u = x² - y² and v = 2xy. The participants emphasize the necessity of calculating the Jacobian determinant, |J|, which is derived as |J| = 1/(4(x² + y²)). The correct evaluation of the integral leads to the conclusion that I = π/4. Additionally, the discussion addresses the challenges of determining the bounds for the new integral and suggests using geometric methods for clarity.

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quantumkiko
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I was self-studying the Jacobian and the change in variables when I came upon the following problem:

In the integral

[tex]I = \int_0^\infty \int_0^\infty \frac{x^2 + y^2}{1 + (x^2 - y^2)^2} e^{-2xy} \, dx dy ,[/tex]

make the change of variables

[tex]u = x^2 - y^2 ,[/tex]
[tex]v = 2xy ,[/tex]

and evaluate I.

I know that I have to use the formula

[tex]I = \int_0^\infty \int_0^\infty |J| f(u, v) \, du dv[/tex]

where |J| is the determinant of the Jacobian. But I don't know how to calculate the elements of the Jacobian. For example, how do I calculate [tex]{\partial x}/{\partial u}[/tex] if x is an implicit variable in the equation for u? Do I use implicit differentiation (if there is one) or do I take the positive square root of x^2 then take its partial derivative?

Finally, how do I set up the bounds for the new integral? My way of thinking is if x = 0, then u = -y^2 and if x = infinity then u = infinity. I really feel there is something wrong with this. Also, some people would do it geometrically and graph to see the bounds. Is there any way to get the new bounds algebraically? Thank you very much! ^^

(By the way, the answer at the back is [tex]\pi / 4[/tex]). =P
 
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Do the Jacobian the other way:
[tex]\frac{\partial u,v}{\partial x,y}= \left|\begin{array}{cc}2x & -2y \\ 2y & 2x\end{array}\right|[/tex]
[tex]= 4x^2+ 4y^2[/tex]
so that
[tex]|J|= \frac{\partial x,y}{\partial u,v}= \frac{1}{4}\frac{1}{x^2+ y^2}[/tex]

Fortunately, you have [itex]x^2+ y^2[/itex] in your integral to cancel that!
 

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