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

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.

Scale the ellipsoid to a unit ball usingwith $0\le r\le 1$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. This is spherical coordinates followed by a diagonal linear transformation, so the Jacobian determinant is $abc{r}^2\sin \theta$. The required volume integral is

The ellipsoid and the exponential factor are invariant under separate reflections of any coordinate. Split the integrand into its two polynomial terms. The term $x^4{y}^{2}z$ times that exponential is odd in $z$, so its volume integral vanishes by reflecting $z\mapsto -z$. The other term $x{y}^2{z}^4$ times the exponential is odd in $x$, so its volume integral vanishes by reflecting $x\mapsto -x$. Both integrands are continuous on a bounded region. Therefore odd-integrand cancellation by reflection gives