Issue in asymptotic expansion of Bogoliubov coefficients in Hawking 1975 paper

$p_{omega}^{(2)}$ is given by $$frac{1}{sqrt{2piomega}}frac{1}{r}P_{omega}^-exp{Bigg(-ifrac{omega}{kappa}}bigg(logBig(frac{v_0-v}{CD}Big)bigg)Bigg)$$
this expression is valid for $v<v_0$ and $v_0-vapprox epsilon$ where $epsilon$ is a small number such that our ray optics analysis is valid. Also $p_{omega}^{(2)}=0$ for $v>v_0$. To calculate $alpha_{omegaomega’}^{(2)}$ we need to take fourier transform of $p_{omega}^{(2)}$ w.r.t $v$
$$alpha_{omegaomega’}^{(2)}=frac{rsqrt{omega’}}{sqrt{2pi}}int_{-infty}^{infty}p_{omega}^{(2)}e^{-iomega’v}dv$$
$$=frac{rsqrt{omega’}}{sqrt{2pi}}int_{-infty}^{v_0}p_{omega}^{(2)}e^{-iomega’v}dv$$
$$*=frac{rsqrt{omega’}}{sqrt{2pi}}{int_{-infty}^{v_0}dv e^{-iomega’v}frac{1}{sqrt{2piomega}}frac{1}{r}P_{omega}^-exp{Bigg(-ifrac{omega}{kappa}}bigg(logBig(frac{v_0-v}{CD}Big)bigg)Bigg)}$$
$$frac{P_{omega}^-}{2pi}sqrt{frac{omega’}{omega}}(CD)^{frac{iomega}{kappa}}int_{-infty}^{v_0}dv e^{-iomega’v}(v_0-v)^{frac{-iomega}{kappa}}$$
doing the substitution $z=v_0-v$
$$frac{P_{omega}^-}{2pi}sqrt{frac{omega’}{omega}}(CD)^{frac{iomega}{kappa}}e^{-iomega’v_0}int_{0}^{infty}dzexp{(izomega’)}z^{-frac{iomega}{kappa}}$$
$$becauseint_{0}^{infty}t^{b}e^{-at}=frac{Gamma(b+1)}{a^{b+1}}$$
$$alpha_{omegaomega’}^{(2)}=frac{P_{omega}^-}{2pi}sqrt{frac{omega’}{omega}}(CD)^{frac{iomega}{kappa}}e^{-iomega’v_0}Gammabigg(1-frac{iomega}{kappa}bigg)(-iomega’)^{-1+frac{iomega}{kappa}}$$
This is the expression that is given in the paper (2.19). My doubt lies in the expression marked $*$, I understand since we are calculating asymptotic expansion we can let $omega’$ quite large so $v$ term in $e^{iomega’v}$ can be made quite small but the problem lies in $p_{omega}^2$ term since our first expression is only valid for $v$ close to $v_0$}. Since I’m not well versed with asymptotic expansion it may actually be the right way of doing expansion. Can somebody help me with this expansion?