I have a little doubt: if I have the following distribution $$\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi\sigma}}e^{-\frac{1}{2\sigma}(|x|-\mu)^2}$$ $1)$ can I say that it correspond to the $N(\mu,\sigma)$, also if I have |x| (but integral starting from $0$).
$2)$ for solving the integral how could I procede: I could start in this way $$2\int_{0}^\infty \frac{1}{\sqrt{2\pi\sigma}}e^{-\frac{1}{2\sigma}(x-\mu)^2} $$ but from here I can only deduce that the previous is $\le1$, right?
If you are asking whether $$\frac{1}{\sqrt{2\pi\sigma}}e^{-\frac{1}{2\sigma}(|x|-\mu)^2} I_{x>0}$$ is the density of a normal random variable, it is not. This density vanishes on $(-\infty, 0)$ so it is not normal.
$\int_0^{\infty} e^{-\frac 1 {2\sigma} (x-\mu)^{2}}dx=\int_{-\mu}^{\infty} e^{-\frac 1 {2\sigma} x^{2}}dx$ which can be expressed in terms of the standard normal distribution function $\Phi (x)$: it is $\sqrt {\frac 2 {\pi}} [{1-\Phi(-\frac {\mu} {\sqrt \sigma})}]$.