Solve a d-dimensional integral using coordinate transformation

Question

Let $c_n > 0, n \in \mathbb{N}$, $g:\mathbb{R} \to \mathbb{R}$ and $\varphi_n:\mathbb{R}^d \to \mathbb{R}$ defined by $$ g(t) = e^{-1/t}, t>0, $$ otherwise for $t\le0 \ \ g$ is equal to $0$, and $$ \varphi_n(x) = c_n g(1-n^2\|x\|^2).$$ By the definition of $g$ it follwos that $\|x\|<\frac1n$. Show that $c_n = c n^d$ with an independent constant $c$ when $$ \int_{\mathbb{R}^d} \varphi_n(x) dx =1 \ \ \forall n \in \mathbb{N}.$$ The lecturer gave us the hint to use a coordinate transformation to solve the integral. Since $\varphi_n$ is a function of $\|x\|$ I thought of using polar coordinates, however I didn´t know the d-dimensional variant. A quick Google search brought up the transformation rule, which then gives the following: $$\int_{\mathbb{R}^d} \varphi_n(x) dx = c_n\int_{0}^{\pi}\cdot\cdot\cdot\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{1/n} e^{-1/(1-n^2r^2)} r^{d-1}drd\varphi\prod_{j=1}^{d-2}(sin\theta_j)^jd\theta_j=c_n\cdot \frac1c \int_{0}^{1/n} e^{-1/(1-n^2r^2)} r^{d-1}dr$$ where $\frac1c$ is a consant given by the integration over all $d\varphi$ and $d\theta_j$ and $\|x\|=r$. Does anyone know how to solve the left over integral? The problem implies it should evaluate to $1/{n^d}$. Is there a coordinate transformation better suited to solve the integral? Help is very much appreciated, thanks in advance!

Answer 1

This is one of those times where doing the integral at all is unnecessary. From the first integral, use $y = nx$:

$$\int_{B\left(0,\frac{1}{n}\right)} g(1-n^2x^2)\:dx = \int_{B(0,1)} g(1-y^2)\: n^{-d }dy = cn^{-d}$$

since the integral converges and is independent of $n$ now.