How can I show that $$S_E[x]=\int_{t_i}^{t_f} dt \left(\frac{m}{2}\dot{x}^2+V(x)\right),$$ starting from the definition of transition amplitude $$A=\langle x_f\,|\,e^{-\frac{i}{\hbar}(t_f-t_i)\hat{H}}\,|\, x_i\rangle?$$
Here you are my try. I am completely following the Feynman derivation for the phase space path integral. Starting with Wick rotation $t \rightarrow -i\tau$ I got $$A=\langle x_f\,|\,e^{-\frac{1}{\hbar}(\tau_f-\tau_i)\hat{H}}\,|\, x_i\rangle=\langle x_f\,|\,e^{\left(-\frac{1}{\hbar}\frac{(\tau_f-\tau_i)}{N}\hat{H}\right)^N}\,|\, x_i\rangle$$
and using $\epsilon=\frac{\tau_f-\tau_i}{N}$ I got $$\int\left(\prod^{N-1}_{k=1}dx_k\right)\langle x_k|e^{-\frac{\epsilon}{\hbar}\hat{H}}|x_{k-1}\rangle=\int\left(\prod^{N-1}_{k=1}dx_k\right)\left(\prod^{N}_{j=1}\frac{dp_j}{2\pi\hbar}\right)\langle x_k|p_j\rangle \langle p_j|e^{-\frac{\epsilon}{\hbar}\hat{H}}|x_{k-1}\rangle$$ then I use the approximation $e^x=1+x+O(x^2)$ neglecting $O(x^2)$ so it becomes $$\int\left(\prod^{N-1}_{k=1}dx_k\right)\left(\prod^{N}_{j=1}\frac{dp_j}{2\pi\hbar}\right)\langle x_k|p_j\rangle \langle p_j|x_{k-1}\rangle\left(1-\frac{\epsilon}{\hbar}H(x_{k-1},p_j)\right)\approx\int\left(\prod^{N-1}_{k=1}dx_k\right)\left(\prod^{N}_{j=1}\frac{dp_j}{2\pi\hbar}\right)e^{\frac{i}{\hbar}p_jx_{k}}e^{-\frac{i}{\hbar}p_jx_{k-1}}e^{-\frac{\epsilon}{\hbar}H(x_{k-1},p_j)}=\int\left(\prod^{N-1}_{k=1}dx_k\right)\left(\prod^{N}_{j=1}\frac{dp_j}{2\pi\hbar}\right)e^{\frac{i}{\hbar}p_j(x_{k}-x_{k-1})-\frac{\epsilon}{\hbar}H(x_{k-1},p_j)}$$ and so I have the exponential in the form: $$e^{\frac{i}{\hbar}p_j(x_{k}-x_{k-1})-\frac{\epsilon}{\hbar}H(x_{k-1},p_j)}=e^{-\frac{1}{\hbar}\epsilon\left(p_j\frac{(x_{k}-x_{k-1})}{-i\epsilon}+H(x_{k-1},p_j)\right)}$$
EDIT: I just noticed that using the Wick rotation the lagrangian becomes exactly what I was looking for since $\partial_t\rightarrow i\partial_{\tau}$ and this means $$\frac{m}{2}{\partial_t x}^2-V(x)\rightarrow -\left(\frac{m}{2}{\partial_{\tau} x}^2+V(x)\right)$$ but how could I handle the minus sign?