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 $v_A=0$.