# Convergence of the Double Integral of a Polynomial Reciprocal

Let $f \in \mathbb{R}[x,y]$ be a polynomial satisfying the following conditions:

(i) $f(\mathbb{R}^2) \subset [a,\infty)$ where $a>0$;

(ii) $f$ is non-degenerate, in the sense that there isn't a non-singular change of variables that turns $f$ into a function of one variable.

(iii) The leading homogeneous part has degree at least $4$.

Question: Does the double integral $$\iint_{\mathbb{R}^2} \frac{1}{f}$$ converge?

Comment: It is easy to see from (i) that the leading homogeneous part of $f$ must be of even degree, and cannot have in its factorization over $\mathbb{R}$ any linear factors of odd exponent.

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The condition (ii) is not sufficient. Consider $f=a+(x+y^2)^2$, for example. –  Alex Gavrilov Apr 5 '14 at 12:58
Thanks Alex, I've edited the question. –  Siksek Apr 6 '14 at 11:12

No. Take $f(x) = 1 + y^2 + (xy-1)^2$. Let $$D = \{ (x,y) : 0 \leq x,\ 0 \leq y \leq 2,\ xy \leq 2 \}.$$ Then $f(x,y) \leq 3$ on $D$, so the integral of $1/f$ is bounded below by $(1/3) \mathrm{Area}(D)$. But the area of $D$ is infinite.

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That's neat David. Thanks! –  Siksek Apr 6 '14 at 15:07