Why the action is taking phase in considering Huygens principle in matter waves? From Dirac's  remarks 
$$\langle x_2,t_2|x_1,t_1\rangle=\exp\left[ \frac{i\int_{t_1}^{t_2}\mathrm dt\, L_{\text{classical}}{\left(\dot{x},x\right)}}{\hbar}\right].$$
How can I conclude from Huygens principle a space time trajectory is formed by a particle overall contribution of all equal smaller classical path contribution with different phase?
 A: Let $\lambda \in \mathbb{R}$, $\lambda \gg 1$, let $f,g$ be analytic real/complex functions near $c \in [a,b] \subseteq \mathbb{R}$. Let $g^\prime(c)=0$ for some $c \in (a,b)$ and $g^\prime(t)\neq 0$ for all $t \in [a,b] \setminus \{c\}$. Assume also that $g^{\prime \prime}(c)\neq 0$, $f(c)\neq 0$, let $\mu$ denote the sign of $g^{\prime \prime}(c)$. Then it holds that: 
$$I(\lambda) = \int^b_a f(t)e^{i\lambda g(t)}dt \approx f(c)e^{i\pi \mu/4} e^{i\lambda g(c)}\sqrt{\frac{2\pi}{\lambda|g^{\prime \prime}(c)|}}.$$ 
Proof:
$$I(\lambda) = \int^b_a f(t)e^{i\lambda g(t)}dt = e^{i\lambda g(c)}\int^b_a f(t)e^{i\lambda(g(t)-g(c))}dt.$$
$\exp(i\lambda(g(t)-g(c)))$ is highly oscillatory for $t\neq c$ and $\lambda \gg 1$, hence for some $\epsilon \ll 1$ we have that:
$$I(\lambda) = e^{i\lambda g(c)}\int^{c+\epsilon}_{c-\epsilon}f(t)e^{i\lambda(g(t)-g(c))}dt$$
$$\approx f(c)e^{i \lambda g(c)}\int^{c+\epsilon}_{c-\epsilon} e^{i\lambda(g(t)-g(c))}.$$ 
We can Taylor expand $g$ around $t=c$ to second order: $g(t) \approx g(c)+g^{\prime}(c)(t-c)+\frac{1}{2}g^{\prime \prime}(c)(t-c)^2$. $g^{\prime}(c)=0$ by assumption so it follows that:
$$I(\lambda) = f(c)e^{i\lambda g(c)}\int^{c+\epsilon}_{c-\epsilon} e^{i\lambda g^{\prime \prime}(c)(t-c)^2/2}dt \approx f(c)e^{i\lambda g(c)}\int^{\infty}_{-\infty} e^{i\lambda g^{\prime \prime}(c)(t-c)^2/2}dt$$
$$ = f(c)e^{i\lambda g(c)}\int^{\infty}_{-\infty} e^{i\lambda g^{\prime \prime}(c)s^2/2}ds.$$
Using the standard Gaussian integral formula it follows that
$$I(\lambda) = f(c)e^{i\lambda g(c)}\sqrt{\frac{2\pi}{\lambda|g^{\prime \prime}(c)|}}\sqrt{i\mu},$$
where we used that $\mu g^{\prime \prime}(c)=|g^{\prime \prime}(c)|$. Notice that since $\mu \in \{1,-1\}$ and $\sqrt{i}=(e^{i\pi/2})^{1/2} = e^{i\pi/4}$ it holds that $\sqrt{\mu i} = e^{i\pi/4}$. Therefore indeed it follows that 
$$I(\lambda) \approx f(c)e^{i\pi \mu/4}e^{i\lambda g(c)}\sqrt{\frac{2\pi}{\lambda|g^{\prime \prime}(c)|}}.$$
Here in our case the function $g$ is replaced by the functional $S[x(t)]$. The idea is similar. $\delta S = 0$ happens for $x=x_{\mathrm{clas}}$ and hence you get that equation. Notice here that $1/\hbar \rightarrow \infty$ and that the stationary phase approximation corresponds to the Huygens principle in this way. The fact that $$\exp(i/\hbar (S[x]-S[x_{\mathrm{clas}}]))$$ is highly oscillatory corresponds to what you want to prove.
I have the idea of this proof from somewhere online and I don't know the reference exactly.
