I want to scale the Gaussian function $\exp(-x^2)$ to the unit disc. In particular, I wish to represent $\int_0^\infty \exp(-x^2) dx$ as $\int_0^1 g(x) dx$, where $g$ should be the rescaled Gaussian function.
I read that $g(x)=\exp(-(1-x^2)^{-1})$ should do the job, but I can not see how to derive that or why it should be true.
I mean, a homeomorphism between the positive real line and the unit line is given by $x\mapsto \frac{x}{1+x}$.
But how can we conclude the above expression?
I would be glad if you could help me out, best regards
First, we can reduce the set of integration to $[0,\infty)$. Then defining $\phi(x):=\frac x{1+x}$, $x\geqslant 0$, we get a strictly increasing map from $[0,\infty)$ to $[0,1)$. Let $t:=\frac x{1+x}=1-\frac 1{1+x}$; then $1-t=\frac 1{x+1}$, that is $x+1=\frac 1{1-t}$ hence $x=\varphi^{-1}(t)=\frac{1-(1-t)}{1-t}=\frac t{1-t}$. This gives $\mathrm dx=\frac 1{(1-t)^2}\mathrm dt$, hence $$\int_{\mathbb R}e^{-x^2}\mathrm dx=2\int_0^1e^{-\frac{t^2}{(1-t)^2}}\frac 1{(1-t)^2}\mathrm dt.$$