I would like to compute
$$\int\_X \exp\left(-\frac{1}{2}(Au)^2\right)\mathrm d\mu\_0(u)$$
with a linear and continuous operator on a Banach space $A:X\to \mathbb R$ (in my case $X=C([0,1])$) and $\mu\_0$ a centred Gaussian measure with covariance operator $Q$ on $X$. Is there any hope this can be calculated explicitly? It looks like this should be feasible as this is the normalization of a measure which is "Gaussian w.r.t another Gaussian".
I have found something similar in Proposition 2 of this paper: <http://projecteuclid.org/download/pdfview_1/euclid.ba/1440594948>


But they need a Hilbert setting, which I don't have here.



Normalization of Gaussian w.r.t. Gaussian in a Banach space
Assume $\mu\_0$ is centered. The covariance **form** (not operator) of $\mu\_0$ is a bilinear form $Q$ on the dual $X^\*$. $\newcommand{\bR}{\mathbb{R}}$ Regard the continuous linear functionals $\alpha:X\to\bR$ as random variables on the probability space $(X,\mu\_0)$. Then
 $\newcommand{\bE}{\mathbb{E}}$


$$Q: X^\*\times X^\*\to\bR,\;\; Q (\alpha,\beta)=\bE[\alpha\cdot\beta]=\int\_X\alpha(x)\beta(x) \mu\_0(dx). $$


A continuous linear functional $A: X\to\bR$ is a Gaussian random variable with mean $0$ and variance $v\_A:=Q(A,A)$. Assume first that $v\_A\neq 0$. Then, for any $t\in\bR$, we have


$$ \int\_X e^{- \frac{1}{2}(tA(x))^2} \mu\_0(dx)=\bE\Big[e^{-\frac{1}{2}(tA)^2}\Big] $$


$$= \frac{1}{\sqrt{2\pi v\_A}}\int\_{\bR} e^{-\frac{1}{2}t^2a^2} e^{-\frac{a^2}{2v\_A}} da =\frac{1}{\sqrt{2\pi v\_A}}\int\_{\bR}e^{-\frac{a^2}{2cA(t)}},$$


where 


$$ \frac{a^2}{2c\_A(t)}= \frac{a^2}{2}\Big( t^2+\frac{1}{v\_A}\Big)\Rightarrow c\_A(t)=\frac{v\_A}{v\_A t^2+1}. $$


Now observe that


$$ \int\_{\bR}e^{-\frac{a^2}{2cA(t)}}=\sqrt{2\pi c\_A(t)}, $$


so 


$$ \int\_X e^{- \frac{1}{2}(tA(x))^2} \mu\_0(dx)=\frac{1}{v\_At^2+1}. $$


This equality holds also when $v\_A=0$.


