I'm studying the Lagrangian $$ \mathcal{L} = \frac{1}{2}\partial_\mu\phi \partial^\mu\phi+\lambda\phi\partial_\mu\phi\partial^\mu\phi~=~\frac{1}{2}(1+2\lambda \phi)\partial_\mu\phi \partial^\mu\phi.\tag{1} $$ We can take the field redefinition to make it looks like free field theory: $$ \frac{1}{2}(\partial\phi')^2 = \frac{1}{2}(\partial\phi)^2+\lambda\phi(\partial\phi)^2\tag{2} $$ and we can solve and obtain that $$ \phi' = \frac{1}{3\lambda}(1+2\lambda\phi)^{3/2}.\tag{3} $$ My question is since this redefinition is non-linear, how can we see $$ \int\mathcal{D}\phi\rightarrow\int\mathcal{D}\phi'\tag{4} $$ produce the same theory? I heard about this would give an additional Jacobi factor, which vanishes in dimensional regularization because it’s scaleless, but I don't know how to make sense of this. Also, how can we tell if a field could be redefined?
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Yes, OP's field redefinition (3) can be made perturbative by a simple shift: $$ \phi^{\prime}~=~\frac{1}{3\lambda}(1+2\lambda\phi)^{3/2}-\frac{1}{3\lambda}~=~\phi+{\cal O}(\phi^2).\tag{3'}$$
Moreover, in dimensional regularization the Jacobian is 1 due to the presence of $\delta^d(0)$, cf. e.g. my Phys.SE answer here.
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