Understanding how to derive the Feynman rules out of the path integral formalism I am studying interacting scalar fields (from Osborn) using the path integral approach.
We define the functional integral
\begin{equation*}
Z[J] := \int d[\phi] e^{iS[\phi] + i\int d^d x J(x) \phi(x)} \tag{1}
\end{equation*}
The idea is to differentiate $Z[J]$ with respect to $J$ and end up defining correlation functions. We can define this integral by a perturbation expansion. This can be expressed in terms of Feynman diagrams, and for each diagram there is an amplitude given by the Feynman rules.
I see that, formally, $(1)$ takes the form
\begin{align*} Z[J] &= \exp\left(\frac{i}{2} \int d^d x d^d y \frac{\delta}{\delta \phi(x)} \Delta_F (x-y) \frac{\delta}{\delta \phi(y)}\right) \times \\ &\times \exp\left( i \int d^d x (-V(\phi(x))+J(x)\phi(x))\right) \Big|_{\phi=0} \end{align*}
Then Osborn states to "expand this integral to get the perturbation expansion" and then he goes straight to explain the Feynman rules.
My issue is that I do not see how this perturbative expansion leads to write down the Feynman rules.
Could you please explain me the easiest case I could find, $V(\phi(x))=\lambda \frac{\phi^3}{3!}$? In other words, how to expand the following integral perturbatively
\begin{align*} Z[J] &= \exp\left(\frac{i}{2} \int d^d x d^d y \frac{\delta}{\delta \phi(x)} \Delta_F (x-y) \frac{\delta}{\delta \phi(y)}\right) \times \\ &\times \exp\left( i \int d^d x (-\lambda \frac{\phi^3}{3!}+J(x)\phi(x))\right) \Big|_{\phi=0} \tag{2} \end{align*}
So that I see how to establish the Feynman rules (i.e. to find out what the factor of (…) for a $\phi^3$ is etc), what kind of integrals (at least 1-loop) are associated to which diagrams and find all connected one- and two-loop graphs
which contribute to $\langle \phi(x_1) \phi(x_2) \rangle$ and $\langle \phi(x_1) \phi(x_2) \phi(x_3)\rangle$
Please note that , to do so, we should only do an inspection of $(2)$ and not the explicit full expansion.
Please note that this is not a homework exercise: I am just looking for a particular solved example so that I can understand how the whole machinery works. You could explain it based on another potential if you wish.
Source: Osborn notes, section 2.2. Interacting Scalar Field Theories
EDIT 0
By expanding $(2)$ as $e^x e^y = 1 + xy + (xy)^2/4 + ...$ i.e.
\begin{equation*} Z[J] = 1+ \left(-\frac{1}{2} \int d^d x d^d y \frac{\delta}{\delta \phi(x)} \Delta_F (x-y) \frac{\delta}{\delta \phi(y)}\right) \times \left( \int d^d x (-\lambda \frac{\phi^3}{3!}+J(x)\phi(x))\right) + ...\end{equation*}
I get what to me looks like a messy expression. What am I missing?
EDIT 1
Alright, so based on your comment I would say we get
\begin{align*} Z[J] &= 1+ \left(-\frac{1}{2} \int d^d x d^d y \frac{\delta}{\delta \phi(x)} \Delta_F (x-y) \frac{\delta}{\delta \phi(y)}\right) \times \\ &\times \left( \int d^d x (-\lambda \frac{\phi^3}{3!}+J(x)\phi(x))\right) + ... \\ &= 1+ \left(- \int d^d y \frac{\delta}{\delta \phi(y)} \Delta_F (x-y) \right) \times \\ &\times \left( (-\lambda \phi^2/2+J(x)\right) + ...
&= \end{align*}
 A: So, this is not a full answer but I'll do the calculations up to the first-order in the perturbative expansion (so no loops, see the comments) :
\begin{align*}
&\left.e^{\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}}e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\\
&=\left[1+\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}+\cdots \right]\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\\
&=\left[1+\frac{i}{2}\int d^d x\int d^d y \frac{\delta}{\delta\phi(x)}\Delta_F(x-y)\left( -\frac{\lambda}{2}i\phi^2(y)+iJ(y)\right) +\cdots\right]\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\\
&=\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}+\frac{i}{2}\int d^d x\int d^d y\,\Delta_F(x-y)\left( -\lambda i\phi(y)\delta(x-y)\right)\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\\
&\hphantom{=}+\frac{i}{2}\int d^d x \int d^d y\,\Delta_F (x-y)\left( -\frac{\lambda}{2}i\phi^2(y)+iJ(y)\right)\left( -\frac{\lambda}{2}i\phi^2(x)+iJ(x)\right)\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\\
&\hphantom{=}+\cdots
\\
&=1+\frac{i}{2}\times 0+\frac{i}{2}\int d^d x \int d^d y\,(i(J(x))\Delta_F(x-y)(iJ(y))+\cdots
\end{align*}
So at the first order the propagator term is given by $\frac{i}{2}\int d^d x \int d^d y\,(i(J(x))\Delta_F(x-y)(iJ(y))$, more precisely the propagator is $\Delta(x-y)$ and the external lignes are $iJ$.
Hope this helps a bit
