Assume $\mu_0$ is centered. The covariance form of $\mu_0$ is a bilinear form on the dual $X^*$. Regard the continuous linear functionals $\alpha:X\to\bR$ as random variables on the probability space $(X,\mu_0)$. Then $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bE}{\mathbb{E}}$
$$Q: X^*\times X^*\to\bR,\;\; Q (\alpha,\beta)=\bE[\alpha\cdot\beta]=\int_X\alpha(x)\beta(x) \mu_0(dx). $$
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 $v_A=0$.