Factoid 1: If a characteristic function is infinitely differentiable at zero, all the moments of the corresponding random variable are finite.
Factoid 2: If all the moments of a random variable are finite, the corresponding characteristic function is infinitely differentiable everywhere on the real line.
Factoid 3: The function $t\mapsto|\cos(t)|$ is infinitely differentiable at $t=0$ but not everywhere on the real line, for example not at $t=\pi/2$.
Ergo.
For a random variable $X$ having even pdf, $\varphi_{|X|}(t)=\varphi_X(t)+\mathbf{i}\,H\varphi_{X}(t)$, where $H\varphi$ is the Hiblert transform of $\varphi$ (see, e.g., this note). Thus,
\begin{align}
\varphi_{|X|}(t)&=e^{-\frac{(\sigma t)^2}{2}}-\mathbf{i}\left(\mathbf{i}\,e^{-\frac{(\sigma t)^2}{2}}\operatorname{erf}\!\left(\mathbf{i}\,\frac{\sigma t}{\sqrt{2}}\right)\!\right) \\
&=e^{-\frac{(\sigma t)^2}{2}}\left(1+\operatorname{erf}\!\left(\mathbf{i}\,\frac{\sigma t}{\sqrt{2}}\right)\!\right) \\[0.4em]
&=2e^{-\frac{(\sigma t)^2}{2}}\Phi(\mathbf{i}\,\sigma t).
\end{align}
(the derivation of $H\varphi$ can be found in this note).
You can directly check that the difference between i.i.d. half-normal random variables $X_1$ and $X_2$ is not normally distributed (though, it's pdf is also even):
$$
\varphi_{X_1-X_2}(t)=\varphi_{X_1}(t)\varphi_{X_2}(-t)=e^{-(\sigma t)^2}(1-[\operatorname{erf}(\mathbf{i}\, t/\sqrt{2})]^2),
$$
which is not the c.f. of a normal distribution (with zero mean).
If you want to compute the c.f. of $|X|$ directly (assume, for simplicity, that $\sigma=1$), note that
$$
\varphi_{|X|}(t)=\frac{1}{\sqrt{2\pi}}\int_0^{\infty}e^{\mathbf{i}\,tx}e^{-\frac{x^2}{2}}\,dx+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^0 e^{-\mathbf{i}\,tx}e^{-\frac{x^2}{2}}\,dx.
$$
Consider the first integral (the second can be evaluated similarly):
$$
\frac{1}{\sqrt{2\pi}}\int_0^{\infty}e^{\mathbf{i}\,tx}e^{-\frac{x^2}{2}}\,dx=\frac{e^{-t^2/2}}{\sqrt{2\pi}}\int_0^{\infty}e^{-\frac{(x-\mathbf{i}\, t)^2}{2}}\,dx=\frac{e^{-t^2/2}}{2}(1+\operatorname{erf}(\mathbf{i}\, t/\sqrt{2})).
$$
Best Answer
$$ \phi(t)=\phi^2\left(\frac{t}{\sqrt{2}}\right)=... =\phi^{2^n}\left(\frac{t}{(\sqrt{2})^{n}}\right)=\left(1+\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1\right)^{\frac{1}{\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1} 2^n\left(\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1\right)} $$ As we know: $$ \lim_{n\to \infty} \left( 1+\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1 \right)^{\frac{1}{\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1}}=e $$ So $$ \phi(t)=e^{\lim_{n\to \infty}2^n\left(\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1\right)} $$ Since $\phi(t)=1+\frac{1}{2}\phi''(0)t^2+o(t^2)$, set $\frac{t}{(\sqrt{2})^{n}}=x$, then $n=\frac{2\ln(\frac{t}{x})}{\ln2}$. We have: $$ \lim_{n\to \infty}2^n\left(\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1\right)=\lim_{x\to 0} (\frac{t}{x})^2(\phi(x)-1) =\frac{1}{2}\phi'’(0)t^2 $$