OPE in linear dilaton theory I'm trying to do the following question from David Tong's problem sheets on string theory:

A theory of a free scalar field has OPE $$\partial X(z)\partial X(w) = \frac{\alpha'}{2}\frac{1}{(z-w)^2}+...$$ Consider the putative candidate for the stress energy tensor $$T(z) = \frac{1}{\alpha '}: \partial X (z) \partial X (z) : -Q \partial^2 X(z).$$ Use $TX$ OPE to determine the transformation of $X$ under conformal transformations $\delta z = \epsilon(z)$

Now to determine $T(z)X(w)$, I thought I would contract the normally ordered field with $X(w)$. So to get:
$$2\langle \partial X (z) \partial X (w) \rangle -Q^2 \partial^3 X(z).$$
Is that the correct way of proceeding? I'm not sure if I need to contract the not-normally ordered fields $\partial^2 X(z)$ with $X(w)$ as well?
Also, I don't quite understand how would I continue with this question after I worked out the OPE of $TX$.
 A: So, from p.73 of Tong's notes you can see that an operator $O(w)$ should transform under $\delta z =\epsilon(z)$ by equation 4.12:
$$\delta O(w) = -Res[\epsilon(z)T(z)O(w)]$$
meaning that knowing an operator's OPE with the stress tensor is knowing how it transforms under $\delta z$. So first we must calculate the OPE $T(z)O(w,\bar{w})$.
The way we do OPEs is we sum over $\textbf{all}$ allowed contractions between the operators (but not between normal ordered operators). The OPE you're trying to do is
$$ T(z)X(w) = \left(\frac{1}{a'}:\partial X(z)\partial X(z): - Q\partial^2 X(z)\right)X(w) $$ so in the first term you can either contract $X(w)$ with the first $\partial X(z)$ or with the second, but not $\partial X(z)$ with $\partial X(z)$, so that will give you the same term (as I think you have already ghessed) therefore you can write the first term as $(2/a'):\partial X(z):\langle \partial X(z)X(w)\rangle$ where the normal ordering is now redundant. The second term only has one possible contraction and that is $-Q\langle\partial^2 X(z)X(w)\rangle$.
To evaluate these contractions you need $\langle X(z) X(w)\rangle = \frac{a'}{2}\ln(z-w)$ which can be differentiated to give
$$\langle \partial X(z)X(w)\rangle = \partial_z\langle X(z) X(w)\rangle = \partial_z \frac{a'}{2}\ln(z-w) = \frac{a'}{2}\frac{1}{z-w}$$
$$\langle\partial^2 X(z)X(w)\rangle = \partial_z^2\frac{a'}{2}\ln(z-w) = -\frac{a'}{2}\frac{1}{(z-w)^2}.$$
The last ingredient you need to get only $(w)$ dependance on the RHS of $T(z)O(w,\bar{w})$ is to Laurent expand $\partial X(z)$ around $w$ by $\partial X(z) = \partial X(w) + (z-w)\partial^2X(w)+O(z-w)^2$. Sometimes this expansion will give you more singular terms but in this case we only need it to first order.
Putting everything together we see that
$$T(z)X(w) = \frac{Qa'/2}{(z-w)^2}+\frac{\partial X(w)}{z-w}+n.s$$ meaning that $X(w)$ does not transform like a primary operator (because there is no $X(w)$ in the numerator of the first term, as you will probably see later there is a better candidate for a primary operator).
All that is left now is to calculate the residue of this OPE with $\epsilon(z) = \epsilon(w) + (z-w)\partial \epsilon(w)+... $
$$\delta X(w) = -Res\Big[(\epsilon(w) + (z-w)\partial \epsilon(w)+...)\left( \frac{Qa'/2}{(z-w)^2}+\frac{\partial X(w)}{z-w}+n.s \right)\Big] $$
$$=-Res\Big[ \frac{\partial\epsilon(w)Qa'/2+\epsilon(w)\partial X(w)}{z-w} \Big]$$
$$=-\frac{Qa'}{2}\partial\epsilon(w)-\epsilon(w)\partial X(w)$$
