From Minkowski to Euclidean Time in Path Integrals I'm trying to prove the following equality: $$ <x_{f},\, it_{f}|x_{i},\, it_{i}>=\mathcal{N}\int_{\left\{ x\in\mathbb{R}^{\mathbb{R}}:\, x\left(t_{f}\right)=x_{f}\wedge x\left(t_{i}\right)=x_{i}\right\} }\mathcal{D}x\exp\left\{ -\frac{1}{\hbar}\int_{t_{i}}^{t_{f}}dt\left\{ \frac{1}{2}m\left[x'\right]^{2}-\left(-V\left[x\right]\right)\right\} \right\} $$ where the definition of $ <x_f, t_f|x_i, t_i>$ is given by: 
$$ <x_{f},\, t_{f}|x_{i},\, t_{i}>\equiv\mathcal{N}\int_{\left\{ x\in\mathbb{R}^{\mathbb{R}}:\, x\left(t_{f}\right)=x_{f}\wedge x\left(t_{i}\right)=x_{i}\right\} }\mathcal{D}x\exp\left\{ \frac{i}{\hbar}\int_{t_{i}}^{t_{f}}dt\left\{ \frac{1}{2}m\left[x'\right]^{2}-\left(V\left[x\right]\right)\right\}\right\}  $$
What I have done so far:


*

*Assume that the domain of integration, that is, $\left\{ x\in\mathbb{R}^{\mathbb{R}}:\, x\left(t_{f}\right)=x_{f}\wedge x\left(t_{i}\right)=x_{i}\right\}$ is such that all the functions in this set can be analytically continued to a new domain of integration $ \left\{ x\in\mathbb{C}^{\mathbb{C}}:\, x\left(it_{f}\right)=x_{f}\wedge x\left(it_{i}\right)=x_{i}\right\} $. Is this valid?

*Plug in the definitions: 
$<x_{f},\, it_{f}|x_{i},\, it_{i}>=\mathcal{N}\int_{\left\{ x\in\mathbb{C}^{\mathbb{C}}:\, x\left(it_{f}\right)=x_{f}\wedge x\left(it_{i}\right)=x_{i}\right\} }\mathcal{D}x\exp\left\{ \frac{i}{\hbar}\int_{it_{i}}^{it_{f}}dt\left\{ \frac{1}{2}m\left[x'\right]^{2}-\left(V\left[x\right]\right)\right\}\right\}$

*Now to compute $ \int_{it_{i}}^{it_{f}}dt\left\{ \frac{1}{2}m\left[x'\right]^{2}-\left(V\left[x\right]\right)\right\} $ make a change of variable (is this valid?? Don't you need Cauchy's theorem and also to assume that the time boundaries go to infinity?) $t \mapsto -it$ to get: $i \int_{t_{i}}^{t_{f}}dt\left\{ \frac{1}{2}m\left[x'\right]^{2}+V\left[x\right]\right\} $ so you get the correct exponent. 

*But how do you prove that $\mathcal{N}\int_{\left\{ x\in\mathbb{C}^{\mathbb{C}}:\, x\left(it_{f}\right)=x_{f}\wedge x\left(it_{i}\right)=x_{i}\right\} }\mathcal{D}x = \mathcal{N}\int_{\left\{ x\in\mathbb{R}^{\mathbb{R}}:\, x\left(t_{f}\right)=x_{f}\wedge x\left(t_{i}\right)=x_{i}\right\} }\mathcal{D}x$? Is it even the same $\mathcal{N}$?

 A: Carefully  following Feynman's procedure one actually finds:
$$\langle x''|e^{-i\frac{(z''-z')}{\hbar}H}| x' \rangle $$ $$=\langle x'', z''|x', z'\rangle  =\lim_{N\to \infty\: \epsilon \to 0} \left[\frac{m}{2\pi i \hbar \epsilon} \right]^{N/2}\int_{-\infty}^{+\infty}\cdots \int_{-\infty}^{+\infty} \left(\prod_{i=1}^{N-1} dx_i \right) \exp\left\{\frac{i\epsilon}{\hbar} \sum_{j=0}^{N-1} \left[ \frac{m}{2}\left(\frac{x_{j+1}-x_j}{\epsilon} \right)^2-V(x_j)\right]\right\}\quad (1)$$
where $z',z'' \in \mathbb C$, and $\epsilon = \frac{z''-z'}{N}$. 
Therefore,


*

*the procedure encompasses the general case of complex time lapse $z''-z'$;

*$\epsilon$ is a complex number of the same nature as that of $z''-z'$ and this fact is responsible for both the change of sign in front of the kinetic energy, passing from Lorentzian to Euclidean formalism, and the disappearance of the overall factor $i$ in front of the action in the same situation.
As the formula (1) holds for generally complex time $z$, the "Wick rotation" is automatic: It is nothing but the specification of the nature of $z$, real or imaginary.
In (1), there are no true paths parametrized by the parameter $z$, however you are free to interpret $x_{j}$  as a possible  position at complex time $z_j = z'+ j\epsilon$. Actually an effective (powerful I might say!) interpretation is that the sum is computed along the class of all such "broken" paths, in view of the integrations in $dx_i$ connecting in all possible ways $x_j$ and $x_{j+1}$. 
In the limit as $N\to +\infty$ one expects that these paths become smooth (actually the story is different, since the set of smooth paths has zero measure...) and, formally, one writes down the said limit as
$$ <x'',\, z''|x',\, z'>=\mathcal{N}\int_{\left\{ x\in\mathbb{R}^{\mathbb{R}}:\, x\left(z''\right)=x''\wedge x\left(z'\right)=x'\right\} }\mathcal{D}x\exp\left\{ \frac{i}{\hbar}\int_{z'}^{z''}dz\left[ \frac{1}{2}m\left(\frac{dx}{dz}\right)^{2}-V\left(x\right)\right] \right\}\:.$$
You see, in particular, that the Euclidean factor $\mathcal{N}$ has to be interpreted as the analytic continuation of the Lorentzian one since
$$\mathcal{N} = \lim_{\epsilon \to 0} \left[\frac{m}{2\pi i \hbar \epsilon} \right]^{N/2}$$
and $\epsilon$ depends on the nature of the considered notion of time.
