Performing Wick Rotation to get Euclidean action of scalar field

Question

I'm working with the signature $(+,-,-,-)$ and with a Minkowski space-stime Lagrangian

$$
\mathcal{L}_M = \Psi^\dagger\left(i\partial_0 + \frac{\nabla^2}{2m}\right)\Psi
$$
The Minkowski action is
$$
S_M = \int dt d^3x \mathcal{L}_M
$$
I should obtain the Euclidean action by Wick rotation.

My question is about the way with that I should perform the Wick rotation. 

Since the spacetime interval is defined by $ds^2 = dt^2 - d\vec{x}^2$, If I perform a Wick rotation (just rotating the time axis) I get a negative Euclidean interval. 

1. What's the sense of that? What's the connection between physical actions calculated in two different signature?

2. I can perform the rotation with different signs $t =\pm i\tau$. I know that, if there exist any poles, I must choose the correct sign in order to not cross them. But in this case, apparently I can choose both, and I get

If I choose $ t = i\tau $ I get

\[i\int_{+i\infty}^{-i\infty} d\tau d^3x \Psi^\dagger\left(i\frac{\partial}{\partial i\tau} + \frac{\nabla^2}{2m}\right)\Psi =\\ -i\int_{-i\infty}^{+i\infty} d\tau d^3x \Psi^\dagger(x,i\tau)\left(\frac{\partial}{\partial \tau} + \frac{\nabla^2}{2m}\right)\Psi(x,i\tau) \]

That's different from the standard euclidean action which is with a minus between $\partial_t$ and $\nabla^2$.

Comments

Your action is not Minkowski - the time derivative has first order only but should have second order!