Coefficient correlation when $f(x,y)=e^{-(x+y)}$ and $x,y>0$

Question

Coefficient correlation when $f(x,y)=e^{-(x+y)}$

I did this:

$Cov(x,y)=E[XY]-E[X]E[Y]$

$\rho=Cov(x,y)/(\sigma_x\sigma_y)$

$f(x)=e^{-x}, exp(1)$ and $f(y)=e^{-y}, exp(1)$ so $E[X]=E[Y]=1$

$f(x,y)=\int_{i=0}^\infty\int_{i=0}^\infty{e^{-x}e^{-y}dxdy}=1$

then I need $E[XY]$

$E[XY]=\int_{0}^\infty\int_{0}^\infty{(xy)e^{-x}e^{-y}dxdy}=...$

I don't know how is the result because of the "$^\infty$" limits, is there a solution?

Answer 1

A separable PDF means $X,\,Y$ are independent, so have correlation $0$. The upper limit of $\infty$ shouldn't prevent you from evaluating the integrals, e.g. $E[XY]=\int_0^\infty xe^{-x}dx\cdot\int_0^\infty ye^{-y}dy=1!^2=1$.