For the choice $\zeta = 1$ the Lagrangian can be brought into a particularly simple form upon integration by parts in the action integral. Equation$$\mathcal{L}' = -{1\over4}F_{\mu\nu}F^{\mu\nu} - {1\over2}\zeta(\partial_\sigma A^\sigma)^2$$with $\zeta = 1$ can be transformed into$$\mathcal{L}' = -{1\over2}\partial_\mu A_\nu \partial^\mu A^\nu + {1\over2}\partial_\mu A_\nu \partial^\nu A^\mu - {1\over2}\partial_\mu \partial_\nu A^\nu$$$$= -{1\over2}\partial_\mu A_\nu \partial^\mu A^\nu + {1\over2} \partial_\mu [A_\nu(\partial^\nu A^\mu) - (\partial_\nu A^\nu)A^\mu].$$The last term is a four-divergence which has no influence on the field equations. Thus the dynamics of the electromagnetic field (in the Lorentz gauge) can be described by the simple Lagrangian$$\mathcal{L}'' = -{1\over2}\partial_\mu A_\nu \partial^\mu A^\nu.\tag*{$(*)$}$$
Later in the book I am reading, we have the following, where it's worked out for the case of arbitrary $\zeta$:
If the gauge-fixing parameter is $\zeta \neq 1$ the Lagrangian $(*)$ is changed to$$\mathcal{L}'' = -{1\over2} \partial_\mu A_\nu \partial^\mu A^\nu - {{\zeta - 1}\over2}(\partial_\nu A^\nu)^2.$$
To me, though, it is not so clear how this formula for $\mathcal{L}''$ comes from here in the case of arbitrary $\zeta$. Could anyone help explain?
