Maybe this is not a research level question. I post it because I heard that the path integral can be rigorous by Brownian motion. But my knowledge of probability is so limited.

If $$L=\frac{1}{2}(-\frac{d^2}{dx^2}+x^2),$$
we know that $Sp(L)=\{1/2,3/2,5/2,...\}$. So we get 
$$\mathrm{Tr}[e^{-L}]=\frac{1}{2\sinh1/2}.$$

I would like to recover it by following "method".

If $E_x$ denote the expectation of the Brownian motion $x_.$ start from x. By Feymann-Kac formula, we have
$$e^{-L}f(x)=E_x[e^{-\frac{1}{2}\int_0^1x_s^2ds}f(x_1)].$$

If $p(x,y)$ denote the kernel of $e^{-L}$, we get
$$p(x,y)=E_x[e^{-\frac{1}{2}\int_0^1x_s^2ds};x_1=y]\frac{e^{-\frac{1}{2}|x-y|^2}}{\sqrt{2\pi}}$$
where $E_x[...;x_1=y]$ is the conditional expectation.

So we get
$$\mathrm{Tr}e^{-L}=\int_{x\in \mathbb{R}}\frac{dx}{\sqrt{2\pi}}E_x[e^{-\frac{1}{2}\int_0^1x_s^2ds};x_1=x]$$

All the thing is rigorous until now. But in some physics book, it follows that the right side is
$$\int_{periodic\ path} e^{-\frac{1}{2}\int_{0}^1\dot{x}_s^2+x^2_sds}Dx
=\int_{periodic\ path} e^{-\frac{1}{2}\int_{0}^1\langle-\Delta +1x,x\rangle} Dx
=\mathrm{det}^{-1/2}(-\Delta +1)$$
where $\mathrm{det}$ is defined the Zeta fonction method. This will also follows the right answer.

So my question is "How to make it rigorous?" This will need a gaussian measure on the periodic path. But I　can not find a natural one.