Why (and how) to choose $a$ in $\varrho(t):=a 1_{[0,1]}(t)\exp\left(\frac 1{t^2-1}\right)$ such that $\int_{\mathbb{R}^n}\varrho(|x|)\;dx=1$?

Question

Let $$\varrho(t):=\begin{cases}\alpha\exp\left(\frac 1{t^2-1}\right)&\text{, if }t\in [0,1]\\ 0&\text{, otherwise}\end{cases}$$ Why (and how) can we choose $\alpha\in\mathbb{R}$ such that $$\int_{\mathbb{R}^n}\varrho\left(|x|\right)\;d\lambda_n(x)=1$$ (where $\lambda_n$ is the Lebesgue measure and $|x|$ is the Euclidean norm)?

Answer 1

We can choose $\alpha\in \mathbb{R}$ such that

$$\int_{\mathbb{R}^n} \varrho(|x|) dx =1$$

Because

$$\int_{\mathbb{R}^n} \varrho(|x|) dx = \alpha \int_{B(0,1)} \exp \left( \frac{1}{|x|^2-1} \right) dx $$

And consider

$$I =\int_{B(0,1)} \exp \left( \frac{1}{|x|^2-1} \right)dx $$

$I$ is well defined and different from 0, so you just need to choose $\alpha = I^{-1}$

Now if you want the exact value of $I$, it's another story, but do you really need it?