Path integral derivation of exact identity for bosonic field

Question

Let $\eta(t)$ be a non-dynamical Euclidean Gaussian bosonic field with partition function $$ Z=\int D[\eta]\exp\left(-\frac{1}{2\sigma^2}\int_0^t \mathrm{d}\tau\,\eta(\tau)^2\right) $$ so that $\left\langle\eta(\tau)\eta(\tau')\right\rangle = \sigma^2\delta(\tau-\tau')$ for $0<\tau, \tau'<t$. Such fields can represent e.g. white noise in Brownian motion. Using Wick's theorem I have found a very simple identity, namely $$ \left\langle\exp\left(a\int_0^t\mathrm{d}\tau \int_0^t\mathrm{d}\tau'\, \eta(\tau)\eta(\tau')\right)\right\rangle = \frac{1}{\sqrt{1-2\sigma^2at}}. $$ I feel like it should be possible to derive this directly from the path integral in a similar manner to the well-known identity discussed here, but I have not yet been able to come up with such a proof. Can anyone enlighten me?

I would also be happy with references which discuss it.

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Here is my messy proof. \begin{align} \left\langle\exp\left(a\int_0^t\mathrm{d}\tau \int_0^t\mathrm{d}\tau'\, \eta(\tau)\eta(\tau')\right)\right\rangle ={}& \sum_{n=0}^{\infty} \frac{a^n}{n!}\int_0^t\mathrm{d}\tau_1\cdots\int_0^t\mathrm{d}\tau_{2n}\left\langle\eta(\tau_1)\cdots\eta(\tau_{2n})\right\rangle \\ ={}& \sum_{n=0}^{\infty} \frac{a^n}{n!}\int_0^t\mathrm{d}\tau_1\cdots\int_0^t\mathrm{d}\tau_{2n}\left[\delta(\tau_1-\tau_2)\cdots\delta(\tau_{2n-1}-\tau_{2n})+\text{other contractions}\right]\\ ={}& \sum_{n=0}^{\infty} \frac{\sigma^{2n}a^n}{n!}\frac{(2n)!}{2^n n!} \bigg[\underbrace{\int_0^t\mathrm{d}\tau_1\int_0^t\mathrm{d}\tau_2\,\delta(\tau_1-\tau_2)}_t\bigg]^n\\ ={}&\sum_{n=0}^\infty \frac{(2n)!}{(-4)^n (n!)^2}(-2\sigma^2 at)^n\\ ={}&\frac{1}{\sqrt{1-2\sigma^2at}} \end{align} In the third line I have used that the total number of terms resulting from the wick contraction is $$ \frac{1}{2^n} {n \choose 2}{n-2 \choose 2}\cdots{2 \choose 2} =\frac{(2n)!}{2^n n!}, $$ and that each contraction may be factorised into a product of $n$ identical terms, and in the final line I have used the identity $$ \sum_{n=0}^{\infty} \frac{(2n)!}{(-4)^n (n!)^2}x^n = [1+x]^{-1/2}. $$

Answer 1

I found one way to do it, but there may well be more elegant solutions so I will leave the question open for the time being in case anyone else has something to offer.

Essentially, the identity is a result of the fact that $x(t)=\int_0^t\mathrm{d}\tau\,\eta(\tau)$ is a common-or-garden Gaussian variable. Firstly, we use \begin{align} F(t)=\left\langle\exp\left(a\int_0^t\mathrm{d}\tau \int_0^t\mathrm{d}\tau'\, \eta(\tau)\eta(\tau')\right)\right\rangle ={}&\left\langle\int\mathrm{d}x (t)\,\delta\left(x(t)-\int_0^t\mathrm{d}\tau\, \eta(\tau)\right)\exp\left(ax(t)^2\right)\right\rangle \\ ={}& \int\mathrm{d}x(t)\left\langle\int\frac{\mathrm{d}k}{2\pi}\exp\left(\mathrm{i} k\left[x(t)-\int_0^t\mathrm{d}\tau\, \eta(\tau)\right]\right)\exp\left(ax(t)^2\right)\right\rangle \\ ={}& \int\mathrm{d}x(t)\int\frac{\mathrm{d}k}{2\pi}\exp\left(ax(t)^2+\mathrm{i}k x(t)\right)\left\langle\exp\left(-\mathrm{i} k\int_0^t\mathrm{d}\tau\, \eta(\tau)\right)\right\rangle \end{align} and then note that \begin{align} \left\langle\exp\left(-\mathrm{i} k\int_0^t\mathrm{d}\tau\, \eta(\tau)\right)\right\rangle={}&\int D[\eta]\exp\left(-\frac{1}{2\sigma^2}\int_0^t\mathrm{d}\tau\,\eta(\tau)^2-\mathrm{i} k\int_0^t\mathrm{d}\tau\, \eta(\tau)\right)\\ ={}&\int D[\eta]\exp\left(-\frac{1}{2\sigma^2}\int_0^t\mathrm{d}\tau\,\left[\eta(\tau)^2+2\sigma^2\mathrm{i} k\eta(\tau)\right]\right)\\ ={}&\int D[\eta]\exp\left(-\frac{1}{2\sigma^2}\int_0^t\mathrm{d}\tau\,\left[\left(\eta(\tau)+\mathrm{i}k\sigma^2\right)^2+(k\sigma^2)^2\right]\right)\\ ={}&\exp\left(-\frac{1}{2}k^2 \sigma^2t\right). \end{align} Thus \begin{align} F(t)={}& \int\mathrm{d}x\int\frac{\mathrm{d}k}{2\pi}\exp\left(ax^2+\mathrm{i}k x-\frac{k^2\sigma^2 t}{2}\right) = \int\mathrm{d}x\,\frac{1}{\sqrt{2\pi\sigma^2 t}}\exp\left(-\frac{x^2}{2\sigma^2 t}+a x^2 \right) = \frac{1}{\sqrt{1-2at\sigma^2}} \end{align} as before.