The massless scalar field action on Minkowski background is given by \begin{equation} S[\phi]=\int_{\mathbb{R}^D}d^Dx~\eta^{\mu\nu}\partial_\mu\phi\partial_\nu\phi. \end{equation} This action is often referred as one of the easiest CFT one can have so I would like to show that is it conformally invariant. If we only consider dilatations $x'^\mu\mapsto x^\mu+\epsilon x^\mu$ for example, the scalar field transforms as \begin{equation} \delta\phi(x)=\epsilon x^\mu\partial_\mu\phi(x). \end{equation} I can't manage to show that the action only changes up to a boundary term. How would one show that ?
[EDIT]
I already saw a lot of proofs of this (e.g. here) but they always transform the metric or the partial derivatives. This is equivalent of considering the trivial action of a diffeomoprhism that acts on each object in the integral and therefore trivially leaves it invariant. Am I wrong ?
