Let us write down how the muon coordinates $(t,vt,b,0)$ transform from the LF to the FM. The FM moves with velocity $v$ relative to the LF along the $x$-axis:
\[
\begin{cases}
ct_1' = ct \cosh \theta - vt \sinh \theta = \gamma c t (1 - v^2/c^2) = \frac{ct}{\gamma}\\
x_1' = vt \cosh \theta - ct \sinh \theta = \gamma ct ( v/c - v/c) = 0 \\
y_1' = b
\end{cases}
\]

Now let us write down how the particle coordinates $(t,0,0,0)$ transform from the LF to the FM:
\[
\begin{cases}
ct_2' = ct \cosh \theta = \gamma ct \\
x_2' = -ct \sinh \theta = -\gamma vt\\
y_2' = 0\\
\end{cases}
\]

In FM the EM filed produced by the muon is electrostatic:
\[
\vec{E}' = \frac{-e}{4 \pi \varepsilon_0}
\begin{pmatrix}
\frac{-\gamma vt}{(b^2 + \gamma^2 v^2 t^2)^{3/2}} \\ \frac{-b}{(b^2 + \gamma^2 v^2 t^2)^{3/2}} \\ 0
\end{pmatrix}
, \quad \vec{B}' = 0\]

To calculate fields $\vec{E}$ and $\vec{B}$ we should apply the boost of velocity $v$ in positive direction of $x$-axis.
\[
\vec{E} = \frac{-e}{4 \pi \varepsilon_0}
\begin{pmatrix}
\frac{-\gamma vt}{(b^2 + \gamma^2 v^2 t^2)^{3/2}} \\
\frac{-\gamma b}{(b^2 + \gamma^2 v^2 t^2)^{3/2}} \\
0
\end{pmatrix}
, \quad
\vec{B} = \frac{\gamma v}{c^2} \frac{-e}{4 \pi \varepsilon_0}
\begin{pmatrix}
0 \\
0 \\
\frac{- b}{(b^2 + \gamma^2 v^2 t^2)^{3/2}} \\
\end{pmatrix}
\]

The interaction with the magnetic field is negligible because the particle velocity is small compared to the speed of light. From the symmetry of $E_x$ under $t \to -t$ , it is clear that the total momentum acquired along the $x$-axis is zero.

\[ |\Delta p| = \left|\int\limits_{-\infty}^{+\infty} Ze E_y \, dt \right| = \frac{|Z|e^2}{4 \pi \varepsilon_0} \int\limits_{-\infty}^{+\infty} \frac{\gamma b }{(b^2 + \gamma^2 v^2 t^2)^{3/2}} dt \]Using the substitution $x=\gamma v t /b$ we obtain
\[|\Delta p | = \frac{|Z|e^2}{4 \pi \varepsilon_0} \frac{1}{bv} \int\limits_{-\infty}^{+\infty} \frac{dx}{(1+x^2)^{3/2}}\]

Using the substitution $x = \sinh u$ we can calculate the integral:
\[ I = \int\limits_{-\infty}^{+\infty} \frac{\cosh u }{\cosh^3 u} du = \int\limits_{-1}^{+1} d ( \tanh u ) = 2\]

Let us consider a thin-walled cylinder of length $dL$, radius $b$, and wall thickness $db$ in the medium. There are $Z_0 n_0 \cdot 2\pi b \, db \, dL$ electrons inside this cylinder. Thus, the energy loss is
\[dE = Z_0 n_0 \, dL \int\limits_{b_\text{min}}^{b_\text{max}} \frac{e^4}{8 \pi^2 \varepsilon_0^2} \frac{1}{m_ev^2b^2} \cdot 2 \pi b \, db = Z_0 n_0 \, dL \, \frac{e^4}{4\pi \varepsilon_0^2} \frac{1}{m_e v^2} \ln \frac{b_\text{max}}{b_\text{min}} \]