The Duffy map turns a vertex singularity into an integrable radial weight

For $0<\epsilon <1$, restrict to $T_{\epsilon }=\{(x,y)\in T:x\ge \epsilon \}$. The polynomial map $D(r,s)=(r,rs)$ extends to the open set $U=(0,\mathrm{\infty })\times \mathbb{R}$, where it is injective with inverse $(x,y)\mapsto (x,y\mathrm{/}x)$ and determinant $r>0$. Its image $T_{\epsilon }$ is the compact polygon $\{\epsilon \le x\le 1,0\le y\le x\}$ and is Jordan measurable, since its boundary is a finite union of line segments. On this image $\mathrm{\parallel }(x,y){\mathrm{\parallel }}_2\ge \epsilon$, so $f$ and $\mathrm{\mid }f\mathrm{\mid }$ are continuous and bounded there and [F8] makes both Riemann integrable. Thus [F3] applies to each on the compact Jordan rectangle $[\epsilon ,1]\times [0,1]$. By [F1]–[F2] and [F6], its transformed integrand is $r(r\sqrt{1+s^2}{)}^{-\alpha }q(r,rs)=r^{1-\alpha }G(r,s)$, continuous on the truncated rectangle, so [F8] identifies its multiple integral with ${\int }_{\epsilon }^1{\int }_0^1{r}^{1-\alpha }G(r,s)dsdr$.

The function $G$ is continuous on $[0,1{]}^2$ by [F4] and [F7], so let $M=\max \mathrm{\mid }G\mathrm{\mid }<\mathrm{\infty }$. Since $1-\alpha >-1$, [F5] bounds the absolute transformed integral over $0<r<\epsilon$ by $M{\epsilon }^{2-\alpha }\mathrm{/}(2-\alpha )$, which tends to zero. Hence the truncated integrals have an absolutely convergent limit and the displayed identity follows from step 1.1.

The same bound applies to any neighborhood of the vertex contained in $0\le x<\epsilon$, and every sufficiently small Euclidean ball about $(0,0)$ is contained in such a set. Therefore the absolute integral near the vertex tends to zero independently of the particular shrinking neighborhood, establishing the usual improper-integral interpretation.