Rewrite position-space Feynman diagram in momentum-space I'm following Peskin, but don't understand how to rewrite position-space feynman diagram in momenum-space.
Suppose we are in $\phi^4$-theory and $\phi$ is a real scalar field. It would be instructive with an example. I suppose its possible to:


*

*go from the analytical expression in position space to the analytical expression in momentum space

*go from the feynman diagram in position space to the analytical expression in momentum space.


Can you show how to do this step by step? I can offer a feynman diagram to save time:

And here's the analyitical expression
$$
-\frac{(i \lambda)^2}{2 (4!)^2} \int d ^4 z\, \int d^4 w\, D_F(x-z) D_F(y-w) D_F(z-w)^3
$$
 A: When you're proving it for the first time you apply the following identities by substitution: $$\begin{align}D_F(x-y) &= \int \operatorname{d}^4p \frac{\operatorname{e}^{-i p (x - y)}}{4\pi^2} \left(\frac{1}{p^2-m^2 + i\epsilon}\right)\\
\delta^4(x-y) &= \int \operatorname{d}^4p \frac{\operatorname{e}^{-i p (x - y)}}{4\pi^2}. \end{align}$$ The $i\epsilon$ is somewhat arbitrary, and gives the Feynman propagator.
Note that you can do this process at the level of the action/Lagrangian density integral, before you derive Feynman diagrams, short circuiting a lot of re-proving the same identities for different Feynman diagrams. Doing that requires using the substitution:
$$\phi(p) \equiv \int \operatorname{d}^4 x \frac{\operatorname{e}^{ipx}}{4\pi^2} \phi(x),$$ in addition to the delta function one above. 
When you get further along in your career you go straight to the momentum space propagators where each line gives a momentum space propagator (the quantity in the parentheses above), and the net 4-momentum flowing into each vertex is zero (conservation of momentum delta functions that arise from the space integrals of the exponentials).
