Given an absolutely integrable function $f:\mathbb R^n \to \mathbb R$, let $R[f]$ be its Radon transform defined for every $(w,b) \in (\mathbb R^n \setminus \{0\}) \times \mathbb R$ by

$$ R[f](w,b) := \int_H f(x)ds(x) = \int_{\mathbb R^n}f(x)\delta(b-x^\top w)\,dx, $$ where $\delta$ is the Dirac distribution and $ds$ is area element on the hyperplane $$ H:=\{x \in \mathbb R^n \mid x^\top w = b\}. $$ In particular, if $P$ is a probability distribution on $\mathbb R^n$ with density $f$ and $z \sim P$, then we can inteprete $R[f](w,b)$ as the density of the random variable $z^\top w$ evaluated at the point $b$.

Now, let $g:\mathbb R^n \to \mathbb R$ be the density of the multivariate Gaussian distribution $N(\mu,\Sigma)$ with mean $\mu \in \mathbb R^n$ and covariance matrix $\Sigma \in \mathbb R^{n \times n}$. Since $z^\top w \sim N(\mu^\top w,w^\top \Sigma w)$ for $z \sim N(\mu,\Sigma)$, it is clear by virtue of the previous remark that, $$ R[g](w,b) = \varphi\left(\frac{b-\mu^\top w}{\|w\|_\Sigma}\right), \tag{1} $$ where $\varphi$ is the density of the standard Gaussian distribution $N(0,1)$, and $\|w\|_\Sigma := (w^\top \Sigma w)^{1/2}$.

Consider the function $h:\mathbb R^n \to \mathbb R$ defined by $h(x)=x_1 g(x)$ for every $x=(x_1,\ldots,x_n) \in \mathbb R^n$.

Question.In the spirit of (1), what is an analytic formula for $R[h](w,b)$ ?

In the special case where $d=1$ so that $\Sigma=\sigma>0$ is just a scalar, a simple computation gives

$$ R[h](w,b) = \int_{-\infty}^{+\infty} x\varphi(x)\delta(b-wx)\,dx = \frac{b}{w}\varphi(\frac{b-w\mu}{2\sigma w}) = \frac{b}{w}R[g](w,b). $$

## Update

The accepted answer shows that

$$ R[h](w,b) = (\alpha(b-\mu_2)+\mu_1)R[g](w,b), $$

where $\mu_1 := e_1^\top \mu$, $e_1=(1,0,\dots,0) \in \mathbb R^n$, $\mu_2 := w^\top \mu$, $\alpha := \dfrac{e_1^\top \Sigma w}{w^\top \Sigma w}$.

Of course, this can be generalized by replacing $e_1$ with any unit-vector $v$.