Can the number m be chosen so that the improper integral
Can the number $m$ be chosen so that the improper integral: $$ \iint \frac{dx\,dy}{\sqrt{(x^2+y^2)^m}}, $$ taken over the entire $xOy$ plane, converges?
Explanation of Method
Unlike an integral over a bounded region (where only the origin matters) or an integral over an unbounded region excluding the origin (where only infinity matters), an integral over the entire plane must contend with both potential trouble spots at once: the origin and infinity. In polar coordinates the radial integral $\int_0^\infty r^{1-m}\,dr$ splits into a piece near $0$ and a piece near $\infty$, and these two pieces demand exactly opposite conditions on $m$ — so no single value of $m$ can satisfy both.
- Convert to polar coordinates.
- Split the radial integral at $r=1$ and analyze each piece separately.
- Show the two convergence conditions are mutually exclusive.
Step 1 · Convert to polar coordinates
Since $\sqrt{(x^2+y^2)^m}=r^m$, and $dx\,dy=r\,dr\,d\theta$:
$$ I = \int_0^{2\pi}\int_0^\infty \frac{r}{r^m}\,dr\,d\theta = 2\pi\int_0^\infty r^{1-m}\,dr. $$
Step 2 · Split and analyze each piece
$$ \int_0^\infty r^{1-m}\,dr = \int_0^1 r^{1-m}\,dr + \int_1^\infty r^{1-m}\,dr. $$
Convergence near $r=0$: $\displaystyle\int_0^1 r^{1-m}\,dr$ converges iff $1-m>-1$, i.e. $m<2$.
Convergence near $r=\infty$: $\displaystyle\int_1^\infty r^{1-m}\,dr$ converges iff $1-m<-1$, i.e. $m>2$.
Step 3 · Show the two conditions are incompatible
Convergence of the full integral requires both pieces to converge simultaneously, i.e. both $m<2$ and $m>2$. No real number $m$ satisfies both strict inequalities at once.
Checking the boundary case $m=2$ directly: $\displaystyle\int_0^\infty r^{-1}\,dr$ diverges at both ends (logarithmically at $r=0$ and at $r=\infty$), so $m=2$ does not rescue convergence either.
Additional Notes
- This is the two-dimensional instance of a general obstruction: for $\displaystyle\iint_{\mathbb R^2}\dfrac{dx\,dy}{(x^2+y^2)^\alpha}$ (with $\alpha=m/2$), convergence at the origin needs $\alpha<1$ while convergence at infinity needs $\alpha>1$ — the same irreconcilable pair.
- The resolution used throughout the rest of this collection is to restrict to one trouble spot at a time: over a bounded disk (Problems 103–107), only the origin's behavior matters, and $\alpha<1$ suffices; over an unbounded region excluding a neighborhood of the origin (Problems 91, 94, 96, 97), only decay at infinity matters, and $\alpha>1$ suffices. A radially symmetric power-law singularity can never simultaneously be tame at the origin and integrable at infinity when the whole plane is included.
Result
No such $m$ exists: