At our course, the instructor discussed the Dirichlet problem $$\left\{ \begin{array}{ll} -\Delta u(x)=f(x) & x\in\mathbb{R}^N \\ \lim_{|x|\rightarrow\infty}u(x)=0 \end{array} \right.,$$ where $f(x)=e^{-|x|^2}$.
He tried to prove that $u=E*f$ (convolution) is the solution to this problem, where $E$ is the fundamental solution corresponding to Laplace operator: $$E(x)=\left\{ \begin{array}{ll} \frac{1}{2\pi}\ln{|x|} & N=2 \\ -\frac{1}{\omega_N|x|^{N-2}} & N\geq3 \end{array} \right.,$$ where $\omega_N$ is the surface area of the unit ball in $\mathbb{R}^N$.
The problem arose when he had to prove that $\lim_{|x|\rightarrow\infty}u(x)=0$. For $N\geq3$, he obtained $$u(x)=c_N\int_{\mathbb{R}^N}|y|^{2-N}e^{-|x-y|^2}dy=c'_N\int_0^\infty r\int_{S^{N-1}}e^{-|x|^2-r^2+2r\,x\cdot\sigma}d\sigma dr.$$
Using the majoration $|x\cdot\sigma|\leq|x|\cdot|\sigma|$, he said that the integral which is obtained (depending on $x$) converges to $0$ as $|x|\rightarrow\infty$. Later, he observed that this is not correct and we wonder how we can prove that $$\int_0^\infty r\int_{S^{N-1}}e^{-|x|^2-r^2+2r\,x\cdot\sigma}d\sigma dr\xrightarrow{|x|\rightarrow\infty}0.$$
Does anyone know what we should try?
Try this:
$$u(x) = c_N\int_{|y|\leq |x|/2} |y|^{2-N} e^{-|x-y|^2} \, dy + c_N\int_{|y| > |x|/2} |y|^{2-N} e^{-|x-y|^2} \, dy.$$
In the first integral you have $|x-y|\geq |x|/2$ and in the second $|y|^{2-N} \leq C|x|^{2-N}.$ This gives
$$|u(x)| \leq C\left(e^{-|x|^2/4} \int_{|y|\leq |x|/2} |y|^{2-N} \, dy + |x|^{2-N}\right).$$
Then compute $\int_{|y|\leq |x|/2} |y|^{2-N}\, dy= C|x|^2$ to get
$$|u(x)| \leq C( |x|^2e^{-|x|^2/4} + |x|^{2-N}).$$
Now you can send $|x|\to \infty$. Hope this helps.