# 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: $$2i\omega \int_{r_+}^{\infty}dr\,\bar{\psi}\psi'=2i\omega\cdot \frac{\bar{\omega}|\psi(r_+)|^2}{\omega-\bar{\omega}}=\frac{2i}{\omega-\bar{\omega}}|\omega|^2|\psi(r_+)|^2 =\frac{1}{\Im\omega}|\omega|^2|\psi(r_+)|^2=\frac{|\omega^2||\psi(r_+)|^2}{\Im\omega}.$$