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](https://www.wolframalpha.com/input?i=integrate%20x%20*%20exp%28-%28x-%5Cmu%29%5E2%2F%282*sigma%5E2%29%29*delta%28b-w*x%29%2Fsqrt%282*pi*sigma%5E2%29%20dx%20from%20-inf%20to%20inf) 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 $v$.



Radon transform of the function $h(x_1,\ldots,x_n) = x_1 g(x_1,\ldots,x_n)$, where $g$ is the density of multivariate Gaussian $N(\mu,\Sigma)$
$\newcommand{\si}{\sigma}\newcommand{\Si}{\Sigma}\newcommand{\ep}{\varepsilon}\newcommand{\vpi}{\varphi}\newcommand{\R}{\mathbb R}$In this "Gaussian" setting especially, it is convenient to approximate the delta function by the normal distribution $N(0,\ep^2)$ with $\ep\downarrow0$, so that  

\begin{equation\*}
 R[f](w,b)=\lim\_{\ep\downarrow0}R\_\ep[f](w,b), \tag{1}\label{1}
\end{equation\*}
where
\begin{equation\*}
 R\_\ep[f](w,b):=\int\_{\R^n}dx\,\vpi\_\ep(w^\top x-b)f(x), 
\end{equation\*}
\begin{equation\*}
 \vpi\_\ep(t):=\frac1\ep 
 \vpi\Big(\frac t\ep\Big), 
\end{equation\*}
and $\vpi$ is the standard normal density.


Now, for $h(x)\equiv x\_1 g(x)$ and $g$ the density of $N(\mu,\Si)$, we can write
\begin{equation\*}
 R\_\ep[h](w,b)
 =\int\_{\R^n}dx\,g(x) x\_1 \vpi\_\ep(w^\top x-b)
 =E e\_1^\top X\, \vpi\_\ep(w^\top X-b), 
\end{equation\*}
where $e\_1:=[1,0,\dots,0]^\top\in\R^{n\times1}=\R^n$ and $X\sim N(\mu,\Si)$ .


Note that the joint distribution of $e\_1^\top X$ and $w^\top X$ is bivariate normal with respective means
\begin{equation\*}
\mu\_1:=e\_1^\top\mu\quad\text{and}\quad\mu\_2:=w^\top\mu, \tag{2}\label{2} 
\end{equation\*}
respective standard deviations
\begin{equation\*}
\si\_1:=\sqrt{e\_1^\top\Si e\_1} \quad\text{and}\quad \si\_2:=\sqrt{w^\top\Si w}, \tag{3}\label{3}
\end{equation\*}
and correlation
\begin{equation\*}
 \rho:=\frac{e\_1^\top\Si w}{\si\_1\si\_2}. \tag{4}\label{4}
\end{equation\*}


So, straightforward calculations yield
\begin{equation\*}
 R\_\ep[h](w,b)
 =
 \frac{ \rho \si \_1 \si \_2 (b-\mu \_2)+\mu \_1 (\si
 \_2^2+\ep ^2)}{\sqrt{2 \pi } (\si \_2^2+\ep ^2){}^{3/2}}\,
 \exp\Big\{-\frac{(b-\mu \_2){}^2}{2 (\si \_2^2+\ep ^2)}\Big\}. 
\end{equation\*}
Finally, by \eqref{1},
\begin{equation\*}
 R[h](w,b)
 =
 \frac{\rho\si\_1(b-\mu\_2)+\mu\_1\si\_2}{\sqrt{2\pi}\,\si \_2^2}\,
 \exp\Big\{-\frac{(b-\mu\_2)^2}{2 \si \_2^2}\Big\}, 
\end{equation\*}
with $\mu\_1,\mu\_2,\si\_1,\si\_2,\rho$ given by \eqref{2}--\eqref{4}.


