Path Integral Derivation of the QED free field Generating Functional
This is following a derivation found in Mandl & Shaw's QFT.
The free field generating functional for QED is $Z_0 [J_{\kappa}, \sigma, \bar{\sigma}] = Z_0 [J_{\kappa}]Z_0 [\sigma, \bar{\sigma}]$. It is separated into two parts, one with the free electromagnetic field and one with the free spinor fields. They are derived in a similar way, for instance for the electromagnetic field
$$ Z_0 [J_{\kappa}] = \frac{1}{N} \int \mathcal{D}A \text{exp}\big{iX[J_{\kappa}]\big},$$
where $N$ ensures normalization and $X[J_{\kappa}]$ is the action without any spinor fields defined by
$$ X[J_{\kappa}] = \int d^4x\big{\frac{-1}{2}(\partial_{\nu}A_{\mu})(\partial^{\nu}A^{\mu}) + J_{\kappa}(x)A^{\kappa}(x)\big}.$$
In Mandl & Shaw the generating functional is found by first manipulating the expression for the action. 818
By defining a wave operator
$$ K^{\mu\nu}(x,x') = -g^{\mu\nu}\delta^{(4)}(x-x')\Box_{x'} $$
and its inverse by
$$ \int d^4x' K^{\mu\nu}(x,x') K_{\nu\tau}^{-1}(x',x'') = g^{\mu}_{\tau}\delta^{(4)}(x-x''), $$
the action may be expressed as
$$ X[J_{\kappa}] = \frac{-1}{2}\int d^4xd^4x' A_{\mu}(x)K^{\mu\nu}(x,y)A_{\nu}(y) + \int d^4xJ_{\kappa}(x)A^{\kappa}(x). $$
By introducing a shifted electromagnetic field $A_{\mu}'(x)$ according to
$$ A_{\mu}'(x) = A_{\mu}(x) - \int d^4x'K^{-1}_{\mu\nu}(x,x')J^{\nu}(x')$$
and substituting it into the definition of the action, I obtain the following expression
$$ X[J_{\kappa}] = \frac{-1}{2} \int d^4x d^4y \big[A_{\mu}(x) - \int d^4x' K_{\mu\tau}^{-1}(x,x')J^{\tau}(x')\big] K^{\mu\nu}(x,y) \big[A_{\nu}(y) - \int d^4y' K_{\nu\sigma}^{-1}(y,y')J^{\sigma}(y')\big] \ + \int d^4xJ_{\kappa}g^{\kappa\mu} \big[A_{\mu}(x) - \int d^4x' K_{\mu\nu}^{-1}(x,x')J^{\nu}(x')\big]. $$
Apparently it should be possible, by only using the definition of the inverse wave operator above, to express this action in terms of the shifted potential as
$$ X[J_{\kappa}] = \frac{1}{2} \big[J^{\kappa}K^{-1}{\kappa\lambda}J^{\lambda}\big] - \frac{1}{2}\big[A{\mu}'K^{\mu\nu}A_{\nu}'\big].$$
I can handle some of the terms and sort of see how this might be possible. 818