For which values of p,q>0 does the following improper
For which values of $p,q>0$ does the following improper double integral converge? $$ \iint_{|x|+|y|\geqslant1} \frac{dx\,dy}{|x|^p+|y|^q}. $$
Explanation of Method
By symmetry across both axes, it suffices to analyze the integral over the first quadrant, $x,y\geqslant0$, $x+y\geqslant1$. The strategy is to integrate first over one variable across its entire unbounded range for a fixed value of the other, using a scaling substitution to reduce the inner integral to a constant (convergent only for an exponent $>1$) times a power of the outer variable; the remaining single-variable integral then gives the true convergence condition. Carrying this out symmetrically in both orders confirms a single, self-consistent criterion.
- Reduce to the first quadrant by symmetry.
- Fix $x$ and integrate over $y\in(0,\infty)$ via the scaling substitution $y=x^{p/q}s$.
- Integrate the resulting power of $x$ and read off the joint condition on $p,q$.
Step 1 · Reduce to the first quadrant
The integrand $\dfrac1{|x|^p+|y|^q}$ and the region $|x|+|y|\geqslant1$ are both symmetric under $x\to-x$ and $y\to-y$, so:
$$ I = 4\iint_{x,y\geqslant0,\ x+y\geqslant1} \frac{dx\,dy}{x^p+y^q}. $$
Removing the bounded strip near the origin where $x+y<1$ does not affect convergence at infinity, so it suffices to study $\displaystyle\iint_{x,y\geqslant0}\frac{dx\,dy}{x^p+y^q}$ restricted to $x\geqslant x_0>0$ (the behavior near the axes, where the integrand is bounded, is never the obstruction).
Step 2 · Integrate over $y$ for fixed $x$
Fix $x>0$ and substitute $y=x^{p/q}s$, so $dy=x^{p/q}\,ds$ and $y^q=x^p s^q$:
$$ \int_0^\infty \frac{dy}{x^p+y^q} = \int_0^\infty \frac{x^{p/q}\,ds}{x^p+x^ps^q} = x^{p/q-p}\int_0^\infty\frac{ds}{1+s^q}. $$
The remaining constant, $\displaystyle\int_0^\infty\frac{ds}{1+s^q}=\dfrac{\pi/q}{\sin(\pi/q)}$, is finite precisely when $q>1$ (for $q\leqslant1$, $\frac1{1+s^q}$ does not decay fast enough as $s\to\infty$, so the $y$-integral over all of $(0,\infty)$ already diverges for every fixed $x$).
Step 3 · Integrate the resulting power of $x$
Assuming $q>1$, the $x$-integral becomes:
$$ \int_{x_0}^\infty x^{p/q-p}\,dx, $$
which converges if and only if the exponent satisfies $\dfrac pq-p<-1$, i.e.:
$$ p\left(1-\frac1q\right)>1 \quad\Longleftrightarrow\quad \frac1p < 1-\frac1q \quad\Longleftrightarrow\quad \frac1p+\frac1q<1. $$
Step 4 · Confirm symmetrically
Repeating Steps 2–3 with the roles of $x$ and $y$ exchanged (fixing $y$, substituting $x=y^{q/p}s$, requiring $p>1$ for the $x$-integral over $(0,\infty)$ to converge, then integrating in $y$) yields the exponent condition $\dfrac qp-q<-1$, which rearranges to the identical criterion $\dfrac1p+\dfrac1q<1$. The two orders of integration agree, confirming the criterion.
Additional Notes
- The condition $\dfrac1p+\dfrac1q<1$ automatically forces $p>1$ and $q>1$: if $p\leqslant1$ then $\dfrac1p\geqslant1$, so $\dfrac1p+\dfrac1q\geqslant1+\dfrac1q>1$, violating the criterion. So the single inequality $\dfrac1p+\dfrac1q<1$ already encodes both individual requirements — there is no need to state them separately.
- Verified numerically by truncating the first-quadrant region at $x,y\leqslant R$ and letting $R\to\infty$: for $(p,q)=(3,3)$ and $(2.5,2.5)$ (where $\frac1p+\frac1q<1$) the truncated integral stabilizes ($\approx2.42$ and $\approx3.9$ respectively); for $(p,q)=(2,2)$ (the boundary case, $\frac1p+\frac1q=1$) it grows slowly (logarithmically) without bound; and for $(p,q)=(1.5,1.5)$ (where $\frac1p+\frac1q>1$) it grows rapidly without bound. The asymmetric case $(p,q)=(3,1.2)$, with $\frac1p+\frac1q\approx1.167>1$, also diverges, as does its mirror $(1.2,3)$ — consistent with the symmetric criterion.
Result
The integral converges if and only if: