on the velocity of the shell. So we now calculate the coefficients $\beta^1_{\omega'\omega}$. We expand the outgoing modes into plane waves. The outgoing modes are functions that depend on $u$ only, and we obtain $$\Psi \simeq \frac{1}{4\pi \sqrt{2\omega}} e^{-\mathrm{i}\omega\xi_0(\eta_1(u))}
= \int_{-\infty}^{\infty} A^1_{\omega}(\omega')\frac{1}{4 \pi \sqrt{2\omega'}}
e^{-{\mathrm{i}}\omega' u} \; {\mathrm{d}}\omega' \; \; \mbox{for} \; \; u>0\ ,$$ with $$\begin{aligned}
\label{DefA1}
A^1_{\omega}(\omega') = \left\{ \begin{array}{rr}
\alpha_{\omega'\omega}^{1*} \; \; \omega' > 0 \\
{\mathrm{i}} \beta_{- \omega'\omega}^1 \; \; \omega' < 0
\end{array} \right. \;\;.\end{aligned}$$ This is a Fourier transformation, which allows us to compute the Bogoliubov coefficients without evaluatiing the scalar product (\[skpr\]). A Fourier transformation of the outgoing modes gives $$I := \sqrt{\frac{\omega}{\omega'}}A^1_{\omega}(\omega')
     = \frac{1}{2\pi} \int_{0}^{\infty} 
        e^{-{\mathrm{i}}\omega \xi_0(\eta_1(u))} 
        e^{{\mathrm{i}}\omega' u} \; {\mathrm{d}}u \ .$$ Using the approximations for $\xi_0(V)$ and $\eta_1(u)$, and introducing the notation $\lambda := d_0'2M(1+\nu)/(d_0^2\nu)$, we find $$I = \frac{1}{2\pi} \int_{0}^{\infty}
      e^{-{\mathrm{i}}(\omega\frac{c_1}{d_0}-\omega')u} \left[ 
      \frac{(R_0 - d_0'2M/d_0)(1+\nu)/\nu + c_1 u}
           {(R_0 - d_0'2M/d_0)(1+\nu)/\nu}
      \right]^{-\mathrm{i}\omega\lambda} \; \mathrm{d}u \ .$$ We further define $x-x_0 := c_1 u / d_0 $ and $x_0:=(R_0 -
d_0'2M/d_0)(1+\nu)/(\nu d_0)$. The transition coefficients then become $$\label{beta1}
\beta^1_{\omega'\omega} = 
-\frac{1}{2\pi}\sqrt{\frac{\omega'}{\omega}}
e^{\mathrm{i}\omega_+ x_0} x_0^{\mathrm{i}\omega\lambda} \frac{d_0}{c_1}
\int_{x_0}^{\infty} x^{-\mathrm{i}\omega\lambda} e^{-\mathrm{i}\omega_+ x} \; 
{\mathrm{d}}x \ ,$$ where $\omega_+ := \omega + \omega' \frac{d_0}{c_1}$.

