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.


**Edit2 and Answer** Thanks to A.J. Tolland and Glimm & Jaffe's book. I just complete the answer to my question. 


Let $P$ is the measure of the brownian bridge. 


Consider the operator forme $C^{-\infty}(S^1)$ to $C^{\infty}(S^1)$,
$$-\Delta+1, $$
There is a unique gaussian mesure $Q$ on $C^{-\infty}(S^1)$(in fact its support is $C^0(S^1)$) whose matrix of covariance is 
$$(-\Delta+1)^{-1}$$


By the uniqueness, we have
$$\frac{dx}{\sqrt{2\pi}}e^{-\frac{1}{2}\int\_{0}^1(x\_s+x)^2ds}dP=\frac{\mathrm{det}^{1/2}(-\Delta)}{\mathrm{det}^{1/2}(-\Delta+1)}dQ$$


After integral, we have get the resultat.


Remark, the existance and the uniqueness of the gaussian measure is the big theorem in the Appendix A.4 of Glimm & Jaffe's book.



Path integral and harmonic oscillator
Take a look at Appendix A of the 2nd edition of Glimm & Jaffe's book. They give a rigorous construction of the measure you're after aka, the Ornstein-Uhlenbeck measure, which is the cylinder measure you get by taking the continuum limit of the Euclidean signature harmonic oscillator). The key point is that the cylinder measure defined using the momentum basis (or the lattice approximation) 'vanishes at infinity', hence is an actual measure.


They also prove that this measure is supported on (distributions almost everywhere equal to) continuous functions.


