Absolute value of standard normal random variable is not infinitely divisible Suppose that $X$ is standard normal random variable. The problem is to show that the distribution $\mu$ of $|X|,$ is not infinitely divisible.
I've learned that in the case it is infinitely divisible, its characteristic function must be of a form
$\phi_\mu(\xi)=\exp(im\xi+\int^\infty_0(e^{i\xi x}-1)M(dx)),$
where $m\geq 0,$ and $\int^\infty_0 \min(x,1) M(dx)<\infty.$ I can show that in this case $m=0,$ hence
$\sqrt{2/\pi} \int^\infty_0 e^{i\xi x-\frac{x^2}{2}}dx=\exp(\int^\infty_0(e^{i\xi x}-1)M(dx)).$
Moreover, I can show that M has all moments, but it doesn't help me much to get the contradiction.
 A: This is probably a stupid approach, but it yields the result. Since we consider distributions on $[0,\infty)$, take the Laplace transform of $|X|$:
$$
F(\lambda)= \sqrt{2/\pi}\int_0^\infty e^{-\lambda x-x^2/2}\,dx = 
e^{\lambda^2/2}(1-\operatorname{erf}(\lambda/\sqrt{2}))
\tag{1}$$
If $|X|$ were infinitely divisible, then $\frac{d}{d\lambda}(-\log F)$ would be completely monotone. In particular, in the expansion 
$$
-\log F = -\frac{\lambda^2}{2} +\sum_{n=1}^\infty \frac{1}{n}(\operatorname{erf}(\lambda/\sqrt{2}))^n
\tag{2}$$
the coefficient of $\lambda^n$ should have the same sign as $(-1)^{n-1}$, at least starting with $n=1$. I claim that this fails for $n=5$, i.e., the coefficient of $\lambda^5$ is negative. 
For convenience let $z=\lambda/\sqrt{2}$ and use the Maclaurin series
$$
\operatorname{erf} (z) = \frac{2}{\sqrt{\pi}} \sum_{k=0}^\infty \frac{(-1)^k z^{2k+1}}{k!(2k+1)}=  \frac{2}{\sqrt{\pi}}\left( z -\frac{1}{3}z^3+\frac{1}{10}z^5-\dots\right)
\tag{3}$$
The coefficient of $z^5$ in the sum $\sum_{n=1}^\infty \frac{1}{n}(\operatorname{erf}(z))^n$ comes only from the terms
$$
\operatorname{erf}(z)+\frac{1}{3}(\operatorname{erf}(z))^3+\frac{1}{5}(\operatorname{erf}(z))^5
\tag{4}$$
and is equal to 
$$
c_5=\frac{2}{\sqrt{\pi}}\frac{1}{10} +\frac{1}{3} \frac{8}{{\pi}^{3/2}} \binom{3}{1}\left(-\frac{1}{3}\right) +\frac{1}{5} \frac{32}{{\pi}^{5/2}} 
\tag{5}$$
The latter simplifies to 
$$
c_5 = \frac{3\pi^2-40\pi+96}{15\pi^{5/2}} <0
\tag{6}$$
proving the claim.
Inequality (6) is a close call. The polynomial $3x^2-40x+96$ has roots $(20\pm 4\sqrt{7})/3$, the smaller of which is $3.138998\dots$
A: Thank you for this solution. I've done it already in another way by exploiting the equality $\sqrt{2/\pi} \int^\infty_0 e^{i\xi x-\frac{x^2}{2}}dx=\exp(\int^\infty_0(e^{i\xi x}-1)M(dx)).$ From this equality it is seen that function $\int^\infty_0(e^{i\xi x}-1)M(dx)$ is infinitely differentiable at zero. Hence, for all $n\geq 1$ $\int^\infty_0 x^n M(dx)<\infty.$ Moreover, by differentiating first equality at zero it can be seen that for $n\geq 1$
$\int^\infty_0 x^n M(dx) \leq \sqrt{2/\pi} \int^\infty_0 x^n e^{-\frac{x^2}{2}}dx.$ Hence, for all $\lambda >0$ $\int^\infty_0(e^{\lambda x}-1)M(dx)<\infty$ and using analytic continuation we get that $\sqrt{2/\pi} \int^\infty_0 e^{\lambda x-\frac{x^2}{2}}dx=\exp(\int^\infty_0(e^{\lambda x}-1)M(dx)).$ 
Hence, $\int^\infty_0(e^{\lambda x}-1)M(dx)=\frac{\lambda^2}{2}+\ln(\sqrt{2/\pi} \int^\infty_{-\lambda} e^{-\frac{x^2}{2}}dx).$
If we divide by $\lambda^3$ and tend $\lambda\to \infty$ we'll get that $\int^\infty_0 x^3 M(dx)=0$ and $M=0.$
