Define $f: [0,infty[times [0,infty[ to bar{mathbb R}$, $f(x,y):= xe^{-x^{2}(1+y^{2})}$ it is clear that $f(x,y)geq 0$ $lambda-$a.e. Therefore, Tonelli’s theorem holds. This means:

$int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (x)) dlambda (y)=int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)) dlambda (x)$

Looking exclusively at $int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)$, the only thing that comes to mind is “Riemann Integration”, namely:

$int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)=lim_{nto infty}int_{[0,n]}xe^{-x^{2}(1+y^{2})}dy=lim_{nto infty}frac{1}{-x^2times2y}e^{-x^2y^2}vert_{0}^{n}$

Which gets me stuck as I have a $y$ in the denominator, however, I see no other way of doing it. Any other ideas?