In the evaluation of the path integral for the harmonic oscillator, we had left the classical action unevaluated. It's high time to fix that.

(a) From the Lagrangian for the harmonic oscillator

\[\begin{equation*}L(x, \dot{x})=\frac{m}{2} \dot{x}^{2}-\frac{m \omega^{2}}{2} x^{2} \tag{11.154}\end{equation*}\]

show that its classical equation of motion, its Euler-Lagrange equation, is

\[\begin{equation*}m \ddot{x}+m \omega^{2} x=0 . \tag{11.155}\end{equation*}\]

(b) Solve this differential equation for trajectory \(x(t)\), with the boundary conditions that \(x(t=0)=x_{i}\) and \(x(t=T)=x_{f}\), for initial and final positions \(x_{i}\) and \(x_{f}\).

(c) Now, with this trajectory \(x(t)\), calculate the classical action

\[\begin{equation*}S[x]=\int_{0}^{T} d t\left(\frac{m}{2} \dot{x}^{2}-\frac{m \omega^{2}}{2} x^{2}\right) \tag{11.156}\end{equation*}\]