Two-point correlation function in path integral formulation

[besprnt]

Suppose that I have already calculated the two-point correlation function for a Lagrangian with no interations using the path integral formulation.
\langle \Omega | T[\phi(x)\phi(y)] | \Omega \rangle = \frac{ \int \mathcal{D}\phi \phi(x)\phi(y) \exp[iS_0] }{ \int \mathcal{D}\phi \exp[iS_0] }.
If I now add an interaction, such that the new action may be written as S = S_0 + S_I, the new two-point correlation function is obviously
\langle \Omega | T[\phi(x)\phi(y)] | \Omega \rangle = \frac{ \int \mathcal{D}\phi \phi(x)\phi(y) \exp[i(S_0+S_I)] }{ \int \mathcal{D}\phi \exp[i(S_0+S_I)] }.
My question is:
Would I have to do the calculations again for the new expression? Or is there some short cut, such as factoring out \exp[iS_I] so that the new expression is a product of the old expression and \exp[iS_I]?

 

[dextercioby]

Sure, the new 2-point correlation functions will be some nonlinear function of the 'free one'. Therefore you need a series expansion for it, in the same way you need it for the generating functional(s).