Solution
= Solution
For fixed $x>0$, integrate first over $y$:
$$
I=\int_0^\infty\left[\int_{x/2}^{x}xe^{-xy}\,dy\right]dx
=\int_0^\infty\left(e^{-x^2/2}-e^{-x^2}\right)dx.
$$
The integrand is nonnegative, so this order of the <double integral> is legitimate even before its finiteness is known. Applying the scaled <Gaussian integral> $\int_0^\infty e^{-ax^2}\,dx=\sqrt\pi/(2\sqrt a)$ yields
$$
\boxed{I=\frac{\sqrt\pi}{2}(\sqrt2-1).}
$$