Distribution change of variables I'm working on distributions for a physics problem and I am quite stumped on how to proceed properly.
The problem is as follows: I start at the point $(0,0)$ and go a distance $\eta$ in x direction $(\eta, 0)$, from there I go in a random direction $\phi$ (uniformly distributed) a distance $x$ that is gaussian distributed. I need the distribution of $y$, with $y$ being the distance of the resulting point from $(0,0)$
To formalize that:
$f(x)=\frac{1}{b\sqrt{\pi}}\exp(-x^2/b^2)$
(is the distribution of the random x direction) but I also know that
$y = \sqrt{2 x \eta  \cos (\phi )+\eta
   ^2 + x^2}$
with $\phi$ being uniformly randomly distributed between $0, 2\pi$
Is it correct to say (with performing change of variables between x to y, and phi:
$g(y, \phi)\propto f(x)|\frac{dx}{dy}||\frac{dx}{d\phi}|$?
I'm pretty sure this is wrong (I suppose I need the Jacobian in some way), but I can't find good references for it

$$
\begin{aligned}
g_Y(y) &=\int_0^{2\pi} \frac{d\phi}{2\pi} \int_{-\infty}^{\infty} dx \frac{1}{b\sqrt{\pi}} \exp(\frac{-x^2}{b^2}) \times \\
& ~~~~~~~~~~~~~~~~~~~~~\times \delta \bigg( y-\sqrt{2x\eta\cos{\phi}+\eta^2+x^2} 
~~\bigg) \\
&=\int_0^{2\pi} \frac{d\phi}{2\pi} \frac{1}{b\sqrt{\pi}}\exp{\frac{-\tilde{x}^2}{b^2}}\frac{1}
{| \frac{\partial}{\partial x }\sqrt{2x\eta\cos{\phi}+\eta^2+x^2}|_{x=\tilde{x}}|}\\
&\text{where }\tilde{x} \text{ solves } \\
y^2 &= 2\tilde{x}\eta \cos{\phi}+\tilde{x}^2+\eta^2
~ \\
\end{aligned}
$$
Complete the square and solve for $\tilde{x}: y^2=2\tilde{x}\eta\cos{\phi}+\eta^2+\tilde{x}^2$ for $\tilde{x}$ in terms of $\phi, y,$ and $\eta$. And now the work begins . . .
But the integral looks vaguely familiar. $\sqrt{2x\eta\cos{\phi}+\eta^2+x^2} $ is screaming out to be $Y=|X+\eta|_2$, or the distribution of the Euclidean length of $X+\eta$ where $X$ and $\eta$ are in $\mathbb{R}^d$ and $\phi$ is simply the angle between them. It suggests that there is an earlier statement of this problem that might yield geometrical transformation and a much easier problem to solve. Maybe not, but you never know.
