Another way to test $(9.39)$ is to substitute it in (9.2) and see if the resulting intracavity electric field agrees with (9.24). Substituting (9.39) in (9.2), we find that $\hat{E}_{\text {in }}(x)=\hat{E}_{\text {in }, \mathrm{D}}(x)+\hat{E}_{\text {in }, \mathrm{C}}(x)$, where
$$
\begin{aligned}
& \hat{E}_{\text {in, } \mathrm{D}}(x) \equiv-i 2 \sum_{n=1}^{\infty} \hat{a}_n \\
& \times \underbrace{\int_0^{\infty} d k \mathcal{E}_{\mathrm{vac}}(k) \sin ([x+L] k)\left\{\mathcal{L}(k) e^{i k L} \alpha_{n 1}(k)-\mathcal{L}^*(k) e^{-i k L} \alpha_{n 2}(k)\right\}}_{D_n(x)}+\text { H.c. }
\end{aligned}
$$
and
$$
\hat{E}_{\text {in, } \mathrm{C}}(x) \equiv-i 2 \int_0^{\infty} d k^{\prime} \hat{b}\left(k^{\prime}\right)
$$
$$
\times \underbrace{\int_0^{\infty} d k \mathcal{E}_{\mathrm{vac}}(k) \sin ([x+L] k)\left\{\mathcal{L}(k) e^{i k L} \beta_1\left(k, k^{\prime}\right)-\mathcal{L}^*(k) e^{-i k L} \beta_2^*\left(k, k^{\prime}\right)\right\}}_{I\left(k^{\prime}, k\right)}
$$
1. Using $(9.40)$ and $(9.41)$, show that $D_n(x)$ is given by
$$
D_n(x)=\sqrt{\frac{\hbar c}{\pi L k_n}} \int_{-\infty}^{\infty} d k k|\mathcal{L}(k)|^2 \sin ([x+L] k) \frac{\sin \left(\left[k-k_n\right] L\right)}{k-k_n}
$$
and that the integral above can be rewritten as
$$
\frac{(-1)^n}{2} \operatorname{Re} \underbrace{\int_{-\infty}^{\infty} d k k|\mathcal{L}(k)|^2 \mathcal{P} \frac{1}{k-k_n}\left\{e^{i k x}-e^{i(x+2 L) k}\right\}}_{I_n(x)} .
$$
Now noticing that inside the cavity $x$ varies from $-L$ to 0 only, do the integral $I_n(x)$ by contour integration and verify that it is given by ${ }^{20}$
$$
\begin{aligned}
& I_n(x)=-i 2 \pi e^{i k_n x} k_n\left|\mathcal{L}\left(k_n\right)\right|^2 \\
+ & \frac{\pi}{L} \underbrace{\sum_{m=-\infty}^{\infty}\left\{\frac{k_m-i \gamma}{k_m-k_n-i \gamma} e^{i\left(k_m-i \gamma\right) x}-\frac{k_m+i \gamma}{k_m-k_n+i \gamma} e^{i(x+2 L)\left(k_m+i \gamma\right)}\right\}}_{S_n(x)} .
\end{aligned}
$$
Show that $S_n(x)$ can be rewritten as $S_n(x)=S_{1 n}(x)+k_n S_{2 n}(x)$, where
$$
\begin{aligned}
& S_{1 n}(x)=\sum_{m=-\infty}^{\infty}\left\{e^{i\left(k_m-i \gamma\right) x}-e^{i\left(k_m+i \gamma\right)(x+2 L)}\right\}, \\
& S_{2 n}(x)=\sum_{m=-\infty}^{\infty}\left\{\frac{e^{i k_m x}}{k_m-k_n-i \gamma} e^{\gamma x}-\frac{e^{i(x+2 L) k_m}}{k_m-k_n+i \gamma} e^{-(x+2 L) \gamma}\right\} .
\end{aligned}
$$
Now, using
$$
\sum_{m=-\infty}^{\infty} e^{i k_m u}=2 L \sum_{m^{\prime}=-\infty}^{\infty} \delta\left(u-2 L m^{\prime}\right)
$$
and noticing that $-L<x<0$, show that
$$
S_{1 n}(x)=2 L\left\{1-e^{-2 L \gamma}\right\} \delta(x) .
$$
To sum $S_{2 n}(x)$, use the Fourier series for $\cos ([x+\pi] a)$ and its derivative with respect to $x$. Verify that
$$
\sum_{m=-\infty}^{\infty} \frac{e^{i k_m x}}{k_m-k_n+i \gamma}=-i 2 L \frac{e^{-L \gamma}}{e^{L \gamma}-e^{-L \gamma}} e^{\gamma u} e^{i k_n u}
$$
where $0<u<2 L$. Now using that ${ }^{21}\left|\mathcal{L}\left(k_n\right)\right|^2=-\operatorname{Re}\{i \operatorname{cotan}(-i \gamma L)\}=$ $[\exp (L \gamma)+\exp (-L \gamma)] /[\exp (L \gamma)-\exp (-L \gamma)]$, finally show that
$$
\begin{aligned}
\int_{-\infty}^{\infty} d k & k|\mathcal{L}(k)|^2 \sin ([x+L] k) \frac{\sin \left(\left[k-k_n\right] L\right)}{k-k_n} \\
= & (-1)^n \pi\left\{1-e^{-2 L \gamma}\right\} \delta(x)+\pi k_n \sin \left([x+L] k_n\right) .
\end{aligned}
$$
2. Analogously, show that
$$
\begin{aligned}
I\left(k^{\prime}, x\right)= & \frac{1}{2} \sqrt{\frac{\hbar c}{\pi k^{\prime}}} \delta(x)\left\{1-e^{-2 L \gamma}\right\} e^{i k^{\prime} L} \mathcal{L}\left(k^{\prime}\right)\left\{\cos k^{\prime} L-\frac{1}{k^{\prime} L} \sin k^{\prime} L\right\} \\
& +\frac{1}{4 \pi} \sqrt{\frac{\hbar c}{\pi k^{\prime}}}\left|\frac{1+r}{t}\right| \delta(x)\left\{1-e^{-2 L \gamma}\right\} \int_0^{\infty} d k^{\prime \prime}\left|\mathcal{L}\left(k^{\prime \prime}\right)\right|^2 \cos k^{\prime \prime} L \\
& \times \lim _{\delta \rightarrow 0^{+}}\left[\frac{1}{k^{\prime \prime}+k^{\prime}+i \delta}\left\{\cos k^{\prime \prime} L-\frac{1}{k^{\prime \prime} L} \sin k^{\prime \prime} L\right\}\right. \\
& \left.+\frac{1}{k^{\prime \prime}-k^{\prime}-i \delta}\left\{\cos k^{\prime \prime} L-\frac{1}{k^{\prime \prime} L} \sin k^{\prime \prime} L\right\}\right] .
\end{aligned}
$$