The density of the simultane r.v. $(X,Y)$ is given by
$$f(x, y)=\left\{\begin{array}{ll} \frac{16 y}{3 x} & \text { for } 0<y<1,\ 0<x<\frac{1}{y} \\ 0 & \text { else } \end{array}\right. $$
I have to show that the integral over $A = \{ (x,y) \in \mathbb R^2 | 0<y<1,\ 0<x<\frac{1}{y} \}$ is $1$. So my integral expression with the limits are
$$ \int_0^1\int_0^{1/y} \frac{16}{3}x^{-1}y \:\: dx dy $$
When I calculate this integral I get $\infty$. Actually, I do not even have to calculate the integral to realise that it is $\infty$. I can realise this just by drawing A and observering that $f(x,y) \ge 0, \forall (x,y)$.
I have also checked this in Maple. The integral is indeed $\infty$. I'm very confused. I do not know how to do this integral. I was thinking about parameterizing $A$ and then insert this in $f$ and then integrate. But to be honest I do not know how to do this. This idea accurred to me because I was surfing the internet for solution to my problem.
Help me understand why my method for doing this integral and which way is the right way to do it.
Kind regards,