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The density function of some random vector $(X,Y)$ is: $f(x,y)= \frac{c}{1+x^2+x^2y^2+y^2}$ Compute the constant $c$.

We know we can compute the constant from the identity

$$\int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y)dxdy=1$$

How do I compute this definite integral when it is defined everywhere except $1+x^2+x^2y^2+y^2 \neq 0$?

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    Hint: $$f(x,y)= c\frac1{1+x^2}\frac1{1+y^2}$$ – 2017-01-11
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    Thank you. Now I see it :) – 2017-01-11
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    It is thus the "tensor" product of two Cauchy distributions. – 2017-01-11

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$$\int^{\infty}_{-\infty} \int^{\infty}_{-\infty}dx dy= \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}c\frac1{1+x^2}\frac1{1+y^2}dx dy$$ $$1=c\tan^{-1}x \bigg|^{\infty}_{-\infty}\int^{\infty}_{-\infty}\frac{1}{1+y^2}dy$$ $$c=\frac{1}{\pi^2}$$