Convergence Property of Path-Integral

Question

Let the action be $$S= \int \bigg\{ \frac{1}{2} \big(\frac{dX}{dt}\big)^2 - V(X) \bigg\} d\tau$$

and the corresponding Path-Integral

$$Z= \int DX(t) e^{iS}.$$

Since the convergence is not clear we Euclideanize the time coordinate $t$ by the Wick rotation

$$ t \rightarrow -i \tau$$

and get the Path-Integral

$$Z_E=\int DX(\tau) e^{-S_E},$$

with $$S_E= \int \bigg\{ \frac{1}{2} \big(\frac{dX}{d\tau}\big)^2 + V(X) \bigg\} d\tau.$$

And now my question - the Euclideanized path-Integral allegedly has a better convergence property, but i do not quite see why this is the case?

Answer 1

In a nutshell, the Euclidean Boltzmann factor is exponentially suppressed because the Euclidean action is bounded from below (assuming the potential $V$ is bounded from below), while the Minkowskian Boltzmann factor has modulus 1 and is oscillatory.

Answer 2

You just have to look at the exponential. So assuming your Lagrangian is bounded from below when written in its euclidean version, you can see that the Wick rotation is taking an imaginary exponential which is oscillatory and turning it into a decaying exponential which can be approximated more easily, for example summing over its extremal points.