Is $\int_{\mathbb{R}^2} e^{-u} \Delta u < \infty$? Question. Let $u : \mathbb{R}^2 \to \mathbb{R}$ be a smooth function such that
$$
u(x, y) > 0 \quad\text{and}\quad \Delta u(x, y) > 0
$$
for all $(x, y) \in \mathbb{R}^2$, where $\Delta := \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$ is the Laplacian. Is the integral
$$
\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-u(x,y)} \Delta u(x, y) \; dx\, dy
$$
finite?
I tried to no avail to find a counterexample where $u$ is rotationally symmetric, i.e. $u(x, y) = f(x^2 + y^2)$ for some $f$. The problem reduces to the following:
Subquestion.
Let $f : [0, \infty) \to \mathbb{R}$ be a smooth function such that $f(t) > 0$ and $t f''(t) + f'(t) > 0$ for all $t \in [0, \infty)$.
Is $$
\int_0^\infty e^{-f(t)} (t f''(t) + f'(t)) dt
$$
finite?
But it is still not clear to me whether or not this initegral will be finite under the given conditions. I tried many different $f$, but the integral always converge.
 A: 1. For the original question, consider the function
$$u(x, y) = x^2 + 1.$$
Then we have both $u > 0$ and $\Delta u = 2 > 0$. However,
$$ \int_{\mathbb{R}^2} e^{-u} \Delta u \, \mathrm{d}x\mathrm{d}y
= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} 2e^{-x^2-1} \, \mathrm{d}x\mathrm{d}y = +\infty. $$

2. Given that the answer to the original question is negative, the subquestion becomes more interesting. Although I have no decisive answer, I suspect that the answer is yes. Here is my approach so far:

*

*Write $\psi(t) = tf'(t)$ and note that the statement is equivalent to the existence of a smooth function $\psi : [0, \infty) \to \mathbb{R}$ such that $\psi(0) = 0$, $\psi'(t) > 0$, and
$$ I := \int_{0}^{\infty} \exp\left(-\int_{0}^{t} \frac{\psi(s)}{s} \, \mathrm{d}s \right) \, \mathrm{d}\psi(t) < +\infty. $$
Then $ I \leq \int_{0}^{\infty} \mathrm{d}\psi(t) = \lim_{t\to\infty} \psi(t) $, and so, any counter-example must satisfy $\psi(t) \to +\infty$ as $t \to +\infty$. For this reason, we may add this condition to the assumption on $\psi$.


*Define $\Phi(t) = \int_{0}^{t} s \, \mathrm{d}\log\psi^{-1}(s) $. Then $\Phi$ is also smooth, and it is strictly increasing if and only if $\psi$ is strictly increasing. Also, by substituting $t \mapsto \psi^{-1}(t)$, we get
$$ \int_{\varepsilon}^{\infty} \frac{\Phi(t)}{t^2} \, \mathrm{d}t = \frac{\Phi(\varepsilon)}{\varepsilon} + \log\left(\frac{\psi^{-1}(\infty)}{\psi^{-1}(\varepsilon)}\right) \qquad \text{and} \qquad \int_{0}^{\infty} e^{-\Phi(t)} \, \mathrm{d}t = I $$
for each $\varepsilon > 0$.
Summarizing, the answer to the subquestion will be yes if for any smooth, strictly increasing function $\Phi : [0, \infty) \to \mathbb{R}$ such that $\Phi(0) = 0$ and $\int_{\varepsilon}^{\infty} \frac{\Phi(t)}{t^2} \, \mathrm{d}t = +\infty$ for any $\varepsilon > 0$, we have
$$ \int_{0}^{\infty} e^{-\Phi(t)} \, \mathrm{d}t < +\infty. $$
