Solving for C when we have $C\int_0^\infty \int_0^\infty \frac{e^\frac{-(x_1+x_2)}{2}}{x_1+x_2} \,dx_1 \,dx_2=1$ Solving for C when we have $C\int_0^\infty \int_0^\infty \frac{e^\frac{-(x_1+x_2)}{2}}{x_1+x_2} \,dx_1 \,dx_2=1$
$$\int_0^\infty \int_0^\infty \frac{e^\frac{-(x_1+x_2)}{2}}{x_1+x_2} \,dx_1 \,dx_2$$
Let $u=x_1+x_2$ and $\,du=\,dx_1$.
$$\int_0^\infty \int_{x_2}^\infty \frac{e^\frac{-u}{2}}{u} \,du \,dx_2 $$
How do I compute the inner integral?

Integration by parts:))
$$
\int_0^{\infty}\int_{x_2}^{\infty}\frac{e^{-\frac u2}}udu\, dx_2=\int_0^{\infty}x_2\frac{e^{-\frac {x_2}2}}{x_2}dx_2=2.
$$
