imaginary axis. Then we substitute $e^z=\mathrm{i} \frac{\omega_+}{\omega}$ and have $$\begin{aligned}
&&\int_0^{\infty} \vert \beta^1_{\omega'\omega} \vert^2 \; \mathrm{d}\omega' =
\lim_{\epsilon \to 0 \atop \varepsilon \to 0} 
\frac{1}{4\pi^2 \omega} e^{-\pi\omega\lambda}\times \nonumber \\
&&\int_{\ln \left(\frac{\epsilon}{\omega}\right)}^{\infty} e^{-\mathrm{i}\lambda \varepsilon z}
\int_{x_0 \omega e^z}^{x_0 \omega e^z+\infty} y^{- \mathrm{i}\omega\lambda} e^{- y} \mbox{d}y 
\; \int_{x_0 \omega e^{z*}}^{x_0 \omega e^{z*}+\infty} \tilde{y}^{\mathrm{i}\omega\lambda} 
e^{- \tilde{y}} 
\mbox{d}\tilde{y} \; \mbox{d} z \ .
\label{almostdone}\end{aligned}$$ We can now take the limit $\epsilon \to 0$ and see that, since the integration is along the real axis, our integral can be written as $$\int_0^{\infty} \vert \beta^1_{\omega'\omega} \vert^2 \; \mathrm{d}\omega' =
\lim_{\varepsilon \to 0} 
\int_{- \infty}^{\infty} e^{-\mathrm{i}\lambda \varepsilon z} h(z)\;
\mbox{d}z\ , \label{defh}$$ with $$\begin{aligned}
h(z):=\frac{e^{-\pi\omega\lambda}}{4\pi^2 \omega} \int_{x_0 \omega e^z}^{\infty} 
y^{-\mathrm{i}\omega\lambda} e^{- y} \mbox{d}y \; 
\int_{x_0 \omega e^z}^{\infty} \tilde{y}^{\mathrm{i}\omega\lambda} e^{- \tilde{y}} 
\mbox{d}\tilde{y} \ ,\end{aligned}$$ for any complex $z$. For real $z$, as we integrate over on the r.h.s. of Eq. (\[defh\]), this yields the quantity we search for; $h(z)$ is identical to the product of the inner integrations in (\[almostdone\]) for real $z$. This function is holomorphic in $z$. It has an upper bound $B$ for any real $z$ in $[-R,R]$. Thus $$\begin{aligned}
\left\vert \int_{-R}^R h(z) \mbox{d}z \right\vert \leq 2 R B \; \; ,\end{aligned}$$ and, because the 