special relativity - Performing Wick Rotation to get Euclidean action of a scalar field $Psi$

I'm working with the signature $(+,-,-,-)$ and with a Minkowski space-time 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. The topic is something has been already asked (look here). The problem is that the answer to the question is not satisfactory for me and for this special case.

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 always the same negative euclidean interval but different results.

   
   

If I choose $ t = i\tau $ I get $$ S_M ~=~i\int_{+i\infty}^{-i\infty} d\tau d^3x \Psi^\dagger\left(i\frac{\partial}{\partial i\tau} + \frac{\nabla^2}{2m}\right)\Psi ~=~ -i\int_{-i\infty}^{+i\infty} d\tau d^3x \Psi^\dagger\left(\frac{\partial}{\partial \tau} + \frac{\nabla^2}{2m}\right)\Psi $$

If I choose $ t = -i\tau $ I get $$ S_M ~=~-i\int_{-i\infty}^{+i\infty} d\tau d^3x \Psi^\dagger\left(-i\frac{\partial}{\partial i\tau} + \frac{\nabla^2}{2m}\right)\Psi ~=~ -i\int_{-i\infty}^{+i\infty} d\tau d^3x \Psi^\dagger\left(-\frac{\partial}{\partial \tau} + \frac{\nabla^2}{2m}\right)\Psi $$

I think there is a slight difference that I don't understand

3. Is the Euclidean action defined by $S_M = i S_E$ or by $S_M = -iS_E$?