Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 12 / a

The region is an unbounded wedge in the first quadrant, lying between the rays $y=x/2$ and $y=x$. Both boundaries and the origin are included. For integration in the opposite order, the same region is $y\ge 0$ with $y\le x\le 2y$.

For fixed $x>0$, integrate first over $y$:The integrand is nonnegative, so this order of the double integral is legitimate even before its finiteness is known. Applying the scaled Gaussian integral ${\int }_0^{\infty }{e}^{-a{x}^2}dx=\sqrt{\pi }/(2\sqrt{a})$ yields

Reverse the order in the nonnegative double integral, using $0\le y<\infty$ and $y\le x\le 2y$. For $y>0$, an antiderivative in $x$ is $-(x/y+1/y^2)e^{-xy}$. ThusLet $J$ denote the requested improper integral. Its numerator is $y^2+O(y^4)$ near zero, so its apparent singularity is removable; it also decays at infinity. We may therefore separate the preceding expression into the convergent integralsIt follows thatThe two terms divided separately by $x^2$ would each have a divergent integral at zero; the cancellation in their difference is essential.