Consider the function : $$G(x,t)=(4\pi t)^{-\frac{n}{2}}e^{-\frac{\mid x\mid^2}{4t}}$$ where $x \in \Bbb R^n$, $t\geq 0$ and $\mid .\mid$ denote the euclidean norm on $\Bbb R^n$. In my course, it is written :
$G_1(x)=G(x,1)$ is integrable because the integral converges absolutely.
I don't understand this at all. If we would like to show that $G_1$ is integrable on $\Bbb R^n$, what do we have to do ?
$\int_{\mathbb R^{n}} e^{-c|x|^{2}}dx= \prod _i\int_{\mathbb R} e^{-cx_i^{2}}dx_i=(\int_{\mathbb R} e^{-t^{2}}dt )^{n} <\infty$ for any $c >0$.