Question 9.33: The “inversion theorem” for Fourier transforms states that φ...

$\int_{\infty}^{-\infty} \tilde{A}(-l)^{*} e^{i l z}(-d l)=\int_{-\infty}^{\infty} \tilde{A}(-l)^{*} e^{i l z} d l=\int_{-\infty}^{\infty} \tilde{A}(-k)^{*} e^{i k z} d k \text { (renaming the dummy variable } l \rightarrow k$ ).

$f(z, 0)=\operatorname{Re}[\tilde{f}(z, 0)]=\frac{1}{2}\left[\tilde{f}(z, 0)+\tilde{f}(z, 0)^{*}\right]=\int_{-\infty}^{\infty} \frac{1}{2}\left[\tilde{A}(k)+\tilde{A}(-k)^{*}\right] e^{i k z} d k$ . Therefore

$\frac{1}{2}\left[\tilde{A}(k)+\tilde{A}(-k)^{*}\right]=\frac{1}{2 \pi} \int_{-\infty}^{\infty} f(z, 0) e^{-i k z} d z$.

$\text { Meanwhile, } \dot{\tilde{f}}(z, t)=\int_{-\infty}^{\infty} \tilde{A}(k)(-i \omega) e^{i(k z-\omega t)} d k \Rightarrow \dot{\tilde{f}}(z, 0)=\int_{-\infty}^{\infty}[-i \omega \tilde{A}(k)] e^{i k z} d k$.

(Note that $\omega=|k| v$, here, so it does not come outside the integral.)

$=\int_{-\infty}^{\infty}\left[i|k| v \tilde{A}(-k)^{*}\right] e^{i k z} d k=\int_{-\infty}^{\infty}\left[i \omega \tilde{A}(-k)^{*}\right] e^{i k z} d k$ .

$\dot{f}(z, 0)=\operatorname{Re}[\dot{\tilde{f}}(z, 0)]=\frac{1}{2}\left[\dot{\tilde{f}}(z, 0)+\dot{\tilde{f}}(z, 0)^{*}\right]=\int_{-\infty}^{\infty} \frac{1}{2}\left[-i \omega \tilde{A}(k)+i \omega \tilde{A}(-k)^{*}\right] e^{i k z} d k$ .

$\frac{-i \omega}{2}\left[\tilde{A}(k)-\tilde{A}(-k)^{*}\right]=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \dot{f}(z, 0) e^{-i k z} d z, \text { or } \frac{1}{2}\left[\tilde{A}(k)-\tilde{A}(-k)^{*}\right]=\frac{1}{2 \pi} \int_{-\infty}^{\infty}\left[\frac{i}{\omega} \dot{f}(z, 0)\right] e^{-i k z} d z$.

Adding these two results, we get $\tilde{A}(k)=\frac{1}{2 \pi} \int_{-\infty}^{\infty}\left[f(z, 0)+\frac{i}{\omega} \dot{f}(z, 0)\right] e^{-i k z} d z$  .   qed