Function which is not in $L^2(R^n)$ I want to prove that if $V$ is a polynomial in $R[X_1,...,X_n]$ such that the limit of $V(x)$ doesn't exist as $|x|\rightarrow \infty$ then 
$\exp^{-V(x)} \notin L^2(R^n)$ .
Can someone help ? Thanks in advance.
 A: This is false: In $\mathbb R^2,$ define $v(x,y) =x^2y^2(x-y)^2.$ Note that $v(x,0) \equiv 0,$ while $v(x,-x) = 4 x^6,$ which $\to \infty$ as $x \to \infty.$ Thus $\lim_{(x^2+y^2)^{1/2}\to\infty} v(x,y)$ fails to exist.
Nevertheless, $e^{-v} \in L^2(\mathbb R^2).$ To see this let's go to polar coordinates, where we have
$$v(r\cos t,r\sin t) = r^6[\cos^2 t \sin^2 t(\cos t - \sin t]^2.$$
Let's call the term in brackets $p(t).$ For $t \in [0, 2\pi),$ $p$ is positive except for $t= \pi/2, 3\pi/2, 0, \pi, \pi/4, 5\pi/4.$ At each of these six points, $p$ has a zero of order $2.$ Let's look at what is happening near $t=0.$ In some $(-a,a)$ for a small $a$ we have $p(t)\ge ct^2$ for some $c>0.$ Integrating in the sector $\{re^{it}: t\in (-a,a), r > 0,$ we see the integral of $e^{-2v}$ is bounded above by
$$\int_{-a}^a\int_0^\infty e^{-2cr^6t^2}r\,dr \, dt.$$
In the inner integral above, let $r=s/|t|^{1/3}.$ We get
$$\int_{-a}^a (1/|t|^{2/3})\int_0^\infty e^{-2cs^6}s\,ds \, dt.$$
That's finite. The argument for the other zeros of $p$ is the same. Thus the integral of $e^{-2v}$ over six sectors like the one above is finite, and away from these sectors, the integrability of $e^{-2v}$ is easy.
