Integral of the usual mollifier function: finding its necessary constant Context: I'm going through some notes on $L^p$ and Sobolev spaces, and - in a portion on approximating $L^p$ via $C^\infty$ - I came across the notion of a mollifier function. The standard example seems to be the function $\eta \in C^\infty(\Bbb R^n)$ given by
$$\eta(x) = \begin{cases}
\displaystyle c_n \exp \left( \frac{1}{|x|^2 - 1} \right) & |x| < 1 \\
0 & \text{otherwise}
\end{cases}$$
where $c_n \in \Bbb R_{>0}$ is a constant (presumably dependent on the dimension of $\Bbb R^n$) which ensures normalization, i.e.
$$\int_{\Bbb R^n} \eta = 1$$
Now, in some past MSE posts (e.g. here, here), people have asked my question of concern already: "what is this normalization constant precisely?" However, their primary concern seems to have been simply with "how do I verify that, indeed, $\int_{\Bbb R^n} \eta = 1$?" -- which is quite simply the definition of the constant, and is not what I'm concerned with.

The Question: My concern is simply one of curiosity: is there some nice closed form for the mollification constants for $\eta$ as given above? Not in the sense of "I need this for some application of the theory" -- indeed, I have no real reason to suspect one even exists, and I recognize no one really uses the actual value anyways -- but rather I'm just examining this as an interesting integral problem.
Clearly, we can write these in terms of the unit $n$-dimensional ball,
$$c_n = \frac{1}{\displaystyle \int_{B(0;1)} \exp \left( \frac{1}{|x|^2 - 1} \right) \, \mathrm{d}x}$$
but from there I'm stuck. Of course, $c_1$ is slightly easier to manage in that, through evenness,
$$c_1 = \frac{1}{2\displaystyle \int_{0}^1 \exp \left( \frac{1}{x^2 - 1} \right) \, \mathrm{d}x}$$
but that doesn't get us any further.
A couple of approximations, for what it's worth:

*

*$c_1 \approx 2.25228$

*$c_2 \approx 2.14357$
Really struggle with getting Wolfram to do much else for me, to be honest. Multidimensional integrals and series are not my forte.
Any thoughts would be appreciated!
 A: With $|\cdot|_p$ the $\ell^p$-norm on $\mathbb{R}^n$, and "good enough" $f:\mathbb{R}_+\to\mathbb{R}$, we have $$\int_{\mathbb{R}^n}f(|x|_p)\,dx=K_{n,p}\int_0^\infty nr^{n-1}f(r)\,dr,\quad K_{n,p}:=\frac{\big[2\Gamma(1+1/p)\big]^n}{\Gamma(1+n/p)}$$ (see e.g. this answer of mine). Now we may consider, more generally, $$\frac1{c_{n,p,q}}:=\int_{\substack{x\in\mathbb{R}^n\\|x|_p<1}}\exp\frac1{|x|_p^q-1}\,dx=K_{n,p}I_{n,q}\qquad(p,q>0)$$ (so that $c_n=c_{n,2,2}$) where, after $r=(1-t^{-1})^{1/q}$ and integration by parts, $$I_{n,q}:=\int_0^1 nr^{n-1}\exp\frac1{r^q-1}\,dr=\int_1^\infty(1-t^{-1})^{n/q}e^{-t}\,dt.$$ Thus, using Tricomi's confluent hypergeometric function $U$, $$I_{n,q}=e^{-1}\Gamma(1+n/q)U(1+n/q,2,1).$$

The case $[p={}]q=2$ (as in the question) results in expressions using the exponential integral $E_1(z)=\int_z^\infty\frac{e^{-t}}{t}\,dt$ (if $n$ is even) or the modified Bessel functions $K_0(z),K_1(z)$ (if $n$ is odd).
Specifically, the result is as follows. Let \begin{align*}X_0&=(1,0),&X_1&=(1,1),&X_{n+1}&=\frac{2n+1}{n}X_n-X_{n-1}\\Y_0&=(1,-1),&Y_1&=(1,1),&Y_{n+1}&=\frac{4n}{2n-1}Y_n-Y_{n-1}\end{align*} so that $X_n=(A_n,B_n)$ and $Y_n=(C_n,D_n)$; then $$I_{2n,2}=A_n e^{-1}-B_n E_1(1),\quad I_{2n-1,2}=\frac{C_n K_1(1/2)-D_n K_0(1/2)}{2e^{1/2}}.$$
Note that oeis.org has $(n-1)!A_n=\texttt{A321942}$ and $(n-1)!B_n=\texttt{A000262}$ listed.
