Compute $ \iint_{\mathbb{R}^2} \ln(\sqrt{x^2+y^2})\Delta(e^{-3x^2-4y^2})\,dx\,dy$ Compute $$ \iint_{\mathbb{R}^2} \ln(\sqrt{x^2+y^2})\Delta(e^{-3x^2-4y^2})\,dxdy$$ knowing that $\Delta \ln(\sqrt{x^2+y^2})=0$ on $\mathbb{R}^2\setminus \{(0,0)\}$.
I defined $g(x,y)=\ln(\sqrt{x^2+y^2})$ and $f(x,y)=e^{-3x^2-4y^2}$ and after some calculations I have found:
$$g \Delta f= \nabla \cdot (g \nabla f) - \nabla \cdot (f \nabla g)+f \Delta g$$
I don't know how to proceed. I tried using divergence theorem but I failed to apply it.
 A: The equation 
$$g \Delta f= \nabla \cdot (g \nabla f) - \nabla \cdot (f \nabla g)+f \Delta g$$
applies here with $g(x,y)=\ln r$ and $f(x,y)=\exp(-3x^2-4y^2)$. Integrating over a region $R$,
$$ \iint_R (\ln r)\Delta e^{-3x^2-4y^2}\,\mathrm{d}x\mathrm{d}y$$
becomes
$$ =\iint_R \left[\nabla\cdot(\ln r\,\nabla e^{-3x^2-4y^2})-\nabla\cdot(e^{-3x^2-4y^2}\,\nabla \ln r)+e^{-3x^2-4y^2}\Delta \ln r \right] \, \mathrm{d}x\mathrm{d}y $$
Now, $\nabla \ln r=0$ and the other two are susceptible to the divergence theorem,
$$ = \oint_{\partial R} \left[\ln r\nabla e^{-3x^2-4y^2}-e^{-3x^2-4y^2}\nabla \ln r\right]\cdot \mathrm{d}\mathbf{n}$$ 
The integrand is $\exp(-3x^2-4y^2)$ times some other stuff, and the exponential term will dominate so we can expect this to tend to $0$ exponentially as $R$ expands. More precisely, the integral is
$$ \oint_{\partial R} \left[(\ln r)(-6x,-8y)-\frac{1}{r^2}(x,y) \right]e^{-3x^2-4y^2}\cdot \mathrm{d}\mathbf{n} $$
Let $\partial R$ be a circle of radius $a$, so $r=a$, the maximum of $\|(-6x,-8y)\|$ is $8a$, the maximum of $\|(x,y)\|$ is $a$, and the minimum of $3x^2+4y^2$ is $3a^2$, so this integral is bounded above in magnitude by
$$ \left( 8a \ln a+\frac{1}{a}\right)e^{-3a^2} $$
which indeed tends to $0$ as $a\to\infty$.
