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The problem is as follows:

Use the path-integral formulation of stochastic dynamics for a particle in a harmonic potential $U(r)= \frac{1}{2}kr^2$ to show that $$P(x,t|x_0,t_0)=(\frac{\beta k}{2\pi (1-e^{-2k(t-t_0)/\zeta})})^{\frac{3}{2}}exp(-\frac{\beta k}{2}\frac{(x-x_0 e^{-k(t-t_0)/\zeta})^2}{1-e^{-2k(t-t_0)/\zeta}}).$$

I have gotten as far as expressing the probability as: $P(x,t|x_0,t_0)=C e^{-\frac{\beta}{2}(U(x)-U(x_0))}\int_{r(t_0)}^{r(t)}Dr(\tau)e^{-S}$ where C is a normalising constant and S is the action given by: $S=\int_{t_0}^{t}d\tau (\frac{\dot r^2}{4D}+\frac{1}{4}D(\beta\nabla U)^2-\Theta(0)D\beta\nabla^2U)$

I'm unsure how to proceed and whether to use Ito's or Stratanovich's values for $\Theta(0)$

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  • $\begingroup$ This is not off topic here, but for questions of stochastic calculus, you are probably going to get better answers on a math site. $\endgroup$ – Buzz Feb 22 at 0:06

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