An Integration Calculation I'm just having a bit of difficulty understanding the last couple of steps made in the paper Horowitz & Hubeny - Quasinormal Modes of AdS Black Holes and the Approach to Thermal Equilibrium (p.8) which can be found at this link where the following is stated
$$\int_{r_+}^{\infty}dr[f|\psi'|^2+2i\omega\bar{\psi}\psi'+V|\psi|^2]=0$$
and taking the imaginary part gives
$$\int_{r_+}^{\infty}dr[\omega \bar{\psi}\psi'+\bar{\omega}\psi\bar{\psi'}]=0.$$
Integration by parts of the second term yields
$$(\omega - \bar{\omega})\int_{r_+}^{\infty}dr \bar{\psi} \psi'$$
$$=\bar{\omega}|\psi(r_+)|^2$$
given that $\psi(\infty)=0$ and $\psi'$ denotes differentiation w.r.t $r$. Substituting this final result back into the first equation we obtain
$$\int_{r_+}^{\infty}dr[f|\psi'|^2+V|\psi|^2]=\frac{|\omega^2|\psi(r_+)|^2}{Im\omega}.$$
My problem lies in showing these last two results; namely finding $\bar{\omega}|\psi(r_+)|^2$ from the previous equation and then showing the final substitution. 
I've been trying for a very long time with integration by parts, using some complex identities involving the conjugate etc., but I can't arrive at the final result. 
This is simply a case of me trying to fully understand a paper I'm interested in. Any help would be greatly appreciated.

Adding more intermediate steps:
\begin{align*}
\int_{r_+}^{\infty}dr[\omega\bar{\psi}\psi'+\bar{\omega}\bar{\psi}'\psi]&=
\int_{r_+}^{\infty}dr[\omega\bar{\psi}\psi'-\bar{\omega}\bar{\psi}\psi'+\bar{\omega}\bar{\psi}\psi'+\bar{\omega}\bar{\psi}'\psi]=\\
&=\int_{r_+}^{\infty}dr[\color{red}{\omega\bar{\psi}\psi'-\bar{\omega}\bar{\psi}\psi'}+\color{blue}{\bar{\omega}(\bar{\psi}\psi)'}]=\\
&=\color{red}{(\omega-\bar{\omega})\int_{r_+}^{\infty}dr\,\bar{\psi}\psi'}+
\color{blue}{\bar{\omega}|\psi|^2\biggl|_{r_+}^{\infty}}=\\
&=(\omega-\bar{\omega})\int_{r_+}^{\infty}dr\,\bar{\psi}\psi'-\bar{\omega}|\psi(r_+)|^2.
\end{align*}
Since the initial expression was previosly shown to be zero, we obtain the first result.
To show the second result, we substitute $\int_{r_+}^{\infty}dr\,\bar{\psi}\psi'$ by
$\displaystyle \frac{\bar{\omega}|\psi(r_+)|^2}{\omega-\bar{\omega}}$ in the 2nd term of your first equation:
$$2i\omega \int_{r_+}^{\infty}dr\,\bar{\psi}\psi'=2i\omega\cdot \frac{\bar{\omega}|\psi(r_+)|^2}{\omega-\bar{\omega}}=\frac{2i}{\omega-\bar{\omega}}|\omega|^2|\psi(r_+)|^2
=\frac{1}{\Im\omega}|\omega|^2|\psi(r_+)|^2=\frac{|\omega^2||\psi(r_+)|^2}{\Im\omega}.$$
