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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?

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\begin{align} \mathcal{L}&= -\frac{1}{4} F^{\mu\nu}F_{\mu\nu} -\frac{\zeta}{2}\left( \partial\cdot A\right)^2 \\ &=-\frac{1}{2} \partial _\mu A_\nu\partial^\mu A^\nu +\frac{1}{2} \partial _\mu A_\nu\partial^\nu A^\mu -\frac{\zeta}{2}\left( \partial\cdot A\right)^2 \\ &=-\frac{1}{2} \partial _\mu A_\nu\partial^\mu A^\nu +\frac{1}{2} \partial _\mu A_\mu\partial^\nu A^\nu -\frac{\zeta}{2}\left( \partial\cdot A\right)^2 \\ &=-\frac{1}{2} \partial _\mu A_\nu\partial^\mu A^\nu +\frac{1}{2} \left( \partial\cdot A\right)^2 -\frac{\zeta}{2}\left( \partial\cdot A\right)^2 \\ &=-\frac{1}{2} \partial _\mu A_\nu\partial^\mu A^\nu -\frac{\zeta-1}{2}\left( \partial\cdot A\right)^2 \\ \end{align} where in the second line we just use the definition of $F^{\mu \nu}$, and in the third line we did integration by parts twice to swap places for the two derivatives

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