From Euclidean correlator in CFT to time-ordered correlator in real time: how do I solve this ambiguity? The vacuum two-point function for a primary operator $O$ of weight $\Delta$ in a CFT$_2$ reads
$$
s(\tau,x)\equiv\langle O(\tau,x)O(0) \rangle = \frac{1}{(\tau^2+x^2)^{\Delta}}\cdot
$$
This is a Schwinger function (in the sense that it is an euclidean correlator). 
How do I obtain the corresponding Wightman function $w(t,x)$ and the time-ordered function $\tau(t,x)$?
Naively setting $\tau=it$ into $s$ is ambiguous (at least for non-integer $\Delta$), since for $x^2<t^2$ we will have different answers if we take $\tau\to i t+0^+$ or $\tau\to i t+0^-$. Note that both limits yields a result invariant under $(t,x)\mapsto(-t,-x)$ for any values of $t$ and $x$, hence none can be the Wightman function (which has this invariance only when $x^2>t^2$ due to local causality).
A natural guess would be that one of the two limits gives $\tau$ and the other gives the anti-time ordered correlator. Is this guess correct? If yes, what is the proof?
Another question is: what happens for integer $\Delta$? It seems like there is no ambiguity: what is the physical meaning of this property?
 A: Here is the procedure that you should follow to get the Wightman functions. First, lets agree that we want to compute
$$
\langle0|\mathcal O_L(t,0)\mathcal O_L(0)|0\rangle,
$$
where $t$ is Lorentzian time (i.e. set $x=0$ for now). We start with the Euclidean correlator, and we consider it in the quantization by flat time slices. Euclidean correlators are always (Euclidean time)-ordered, and we have for $\tau>0$
$$
\langle0|\mathcal{O}_E(\tau,0)\mathcal O_E(0)|0\rangle=\tau^{-2\Delta}.
$$
We now add a Lorentzian component $t$ to $\tau$, $\tau\to\tau+it$. We first take $t$ small, and then this is unambiguous. We have then
$$
(\tau^2+2it\tau-t^2)^{-\Delta}.
$$
Now we take $\tau$ to zero while keeping it positive. The intuition here is that positivity of $\tau$ keeps the correlator ordered in the same way as we started with. (If we change the sign of $\tau$ in Euclidean, this flips the ordering -- remember that Euclidean correlators always correspond to time-ordered vevs in a given quantization.) Then you can see that this limiting procedure gives two different answers depending on the sign of $t$. We obtain
$$
e^{-i\pi\Delta}|t|^{-2\Delta},\quad t>0,\\
e^{+i\pi\Delta}|t|^{-2\Delta},\quad t<0.\\
$$
Covariantizing, we get for time-like separations
$$
\langle0|\mathcal O_L(t,x)\mathcal O_L(0)|0\rangle=e^{-i\pi\Delta\,\mathrm{sign}\,t}|x^2-t^2|^{-\Delta},\quad x^2-t^2<0.
$$
You can by the way notice that the answer is $PT$-invariant, if you remember that $T$ is anti-unitary and thus you need to accompany it with a conjugation.
You can compare this with $\pm i0$ prescriptions and see that your guess about (anti-)time-ordered correlator was correct. If you have more operators in the correlation function, you follow the same procedure, keeping the Euclidean times ordered as needed when you simultaneously take them to zero. I am not sure how you should interpret integral $\Delta$s though.
