term can be written as the integral from $0$ to $\infty$ minus the integral from $0$ to $\omega$. The latter integration is finite as well, since the difference of the $\Gamma$-functions vanishes as $\omega_+ \to 0$. We are left with $$\begin{aligned}
&&\int_0^{\infty} \vert \beta^1_{\omega'\omega} \vert^2 \; 
\mathrm{d}\omega' -
\lim_{x_0 \to 0}\int_0^{\infty} \vert\beta^1_{\omega'\omega}\vert^2 \;
\mathrm{d}\omega'
= 
\mbox{finite} +
\frac{1}{4\pi^2 \omega} e^{-\pi\omega\lambda}\times \nonumber \\
&& \int_0^{\infty} \omega_+^{- 1} 
\left[\left|\int_{\mathrm{i}\omega_+ x_0}^{\mathrm{i}\omega_+ x_0 + \infty} 
y^{-\mathrm{i}\omega\lambda} e^{- y} \mbox{d}y \; \right|^2 - 
\vert \Gamma(1-\mathrm{i}\omega\lambda) \vert^2 \right] \; 
\mbox{d}\omega_+ \ .\end{aligned}$$ Now we neglect all finite terms and use the fact that the second term inside the square bracket is again (up to finite terms) the result of the $x_0 \to 0$ limit (\[x0lim\]). We thus see that the divergent part of (\[beforesub\]) can be written as $$\begin{aligned}
&&\int_0^{\infty} \vert \beta^1_{\omega'\omega} \vert^2 \; \mathrm{d}\omega'=
\lim_{\epsilon \to 0 \atop \varepsilon \to 0} 
\frac{1}{4\pi^2 \omega} e^{-\pi\omega\lambda}\times \nonumber \\
&&\int_{0-\mathrm{i}\epsilon}^{\infty-\mathrm{i}\epsilon}
\omega_+^{-\mathrm{i}\lambda\varepsilon-1}
\int_{\mathrm{i}\omega_+ x_0}^{\mathrm{i}\omega_+ x_0 + \infty} e^{-y} 
y^{-\mathrm{i}\omega \lambda} \; {\mathrm{d}}y \; 
\int_{-\mathrm{i}\omega_+^* x_0}^{-\mathrm{i}\omega_+^* x_0 + \infty} 
e^{-\tilde{y}} 
\tilde{y}^{-\mathrm{i} \omega \lambda} \; {\mathrm{d}}\tilde{y} \;
{\mathrm{d}}\omega_+ \ . 
\label{zurueck}\end{aligned}$$ For positive $\varepsilon$ we can replace the contour parallel to the positive real axis by an integration along the negative 