Obtaining Euclidean action from Minkowski action

[spaghetti3451]

The Euclidean classical action $S_{\text{cl}}[\phi]$ for a scalar field $\phi$ is given by

\begin{equation}
S_{\text{cl}}[\phi]=\int d^{4}x\ \bigg(\frac{1}{2}(\partial_{\mu}\phi)^{2}+U(\phi)\bigg).
\end{equation}

This can be obtained from the action $S_{\text{Mn}}[\phi]$ in Minkowski space for the same scalar field $\phi$ as given by

\begin{equation}
S_{\text{Mn}}[\phi]=\int d^{4}x\ \bigg(\frac{1}{2}(\partial_{\mu}\phi)^{2}-U(\phi)\bigg).
\end{equation}

To transform the action $S_{\text{Mn}}[\phi]$ in Minkowski space to the Euclidean classical action $S_{\text{cl}}[\phi]$, one needs to perform a Wick rotation. I need someone to show the derivation.

 

[samalkhaiat]

Continue from real x^{0} to x^{0} = - i x_{4}, and from S_{Min} so obtained get S_{Euc} using S_{Euc} = - i S_{Min}:
S_{M} = \int dx^0 \ d^3x \ \left( (1/2) (\frac{\partial \phi}{\partial x^{0}})^{2} - (1/2) (\nabla \phi)^{2} - U(\phi) \right)
S_{M} = -i \int dx_{4} \ d^3x \ \left( - (1/2) (\frac{\partial \phi}{\partial x_{4}})^{2} - (1/2) (\nabla \phi)^{2} - U(\phi) \right)
S_{M} = i \int dx_{4} \ d^3x \ \left( (1/2) (\frac{\partial \phi}{\partial x_{4}})^{2} + (1/2) (\nabla \phi)^{2} + U(\phi) \right) = i S_{E}
S_{E} = \int d^{4}x_{E} \ \left( (1/2) (\frac{\partial \phi}{\partial x_{\mu}})^{2} + U(\phi) \right)