Question about an OPE for the free massless scalar CFT

In page 78 of David Tong's notes on CFT https://www.damtp.cam.ac.uk/user/tong/string/four.pdf, he finds that the propagator for a theory of free massless scalars is

$$\langle X(\sigma)X(\sigma')\rangle=-\frac{\alpha'}{2}\text{log}(\sigma-\sigma')^2$$

Then he goes on saying that in equation (4.22) the OPE of $$X(\sigma)X(\sigma')$$ is

$$X(\sigma)X(\sigma')=-\frac{\alpha'}{2}\text{log}(\sigma-\sigma')^2 + ... \label{eq}$$

My question is: what are the $$...$$ in the above equation?

Certainly they are there because he is saying that we could consider the path integral with other isertions away from $$\sigma$$ and $$\sigma'$$, but then equation (4.20) would turn to

$$\langle \partial^2 X(\sigma)X(\sigma')...\rangle=-2 \pi \alpha' \langle \delta^2(\sigma,\sigma') ... \rangle$$

Implying that

$$\langle X(\sigma)X(\sigma')...\rangle =-\frac{\alpha'}{2}\langle \text{log}(\sigma-\sigma')^2... \rangle \\ \hspace{2.8cm}=-\frac{\alpha'}{2}\text{log}(\sigma-\sigma')^2\langle ... \rangle$$

Because all insertions are away from $$\sigma$$ and $$\sigma'$$.

In the free theory the answer is really simple, $$X(\sigma)X(\sigma') = - \frac{\alpha'}{2} \log(\sigma - \sigma') ~ + : X(\sigma)X(\sigma'):$$ where $$:~:$$ denotes normal ordering.