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 $$\int_{periodic\ path} e^{-\frac{1}{2}\int_{0}^1\dot{x}_s^2+x^2_sds}\mathcal{D}x =\int_{periodic\ path} e^{-\frac{1}{2}\int_{0}^1\langle-\Delta +1x,x\rangle} \mathcal{D}x$$ where $$\mathcal{D}x=\mathrm{det}^{1/2}(-\Delta)\frac{dx}{(2\pi)^{\infty/2}}.$$ As the finite cas, $$\int_{periodic\ path} e^{-\frac{1}{2}\int_{0}^1\langle-\Delta +1x,x\rangle} \mathcal{D}x=\frac{\mathrm{det}^{1/2}(-\Delta )}{\mathrm{det}^{1/2}(-\Delta +1)}=\mathrm{det}^{-1/2}(1+(-\Delta)^{-1}).$$ As we know, $Sp(-\Delta)=\{4\pi^2k^2,k\in \mathbb{Z}\}$. We have $$\mathrm{det}^{1/2}(1+(-\Delta)^{-1})=\Pi_{k=1}^{\infty}(1+\frac{1}{4k^2\pi^2})=2\sinh{1/2}$$

It also follows the right answer.

So my question is "How to make it rigorous?"

First, it will need a gaussian measure on the periodic path. But I can not find a natural one.

Edit: Thanks to Alexander Chervov, he give a interesting measure by Fourier Analysis. It is a right one in some sense. But it is not even clear for me that its support is the contious path. And with this measure how can we get the final answer rigorousment.