I am studying the phase distribution for coherent states, as is defined in quantum optics. (See, for example, Introductory Quantum Optics by Gerry and Knight, pages 46–48).
In this situation, we seek the phase probability distribution $P(\varphi)$ for a coherent state; then, we take
$$P(\varphi)=\frac{1}{2\pi} \left| \left\langle \varphi \right| \left. \alpha \right\rangle \right|^2 \tag{1}$$
where the factor $(1/2\pi)$ is to guarantee the normalization of the probability distribution, the state $\left|\alpha\right\rangle = e^{-\frac{1}{2}\left|\alpha \right|^2}\sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n!}} \left|n \right\rangle$ is the coherent state, and $\left|\varphi \right\rangle = \sum_{n=0}^{\infty} e^{i\varphi n} \left| n\right\rangle$ is the state of phase as is defined by Gerry and Knight, Pag. 38; besides, $\left|n\right\rangle$ is the number state. Then, by utilizing these definitions in Eq. (1), we have
$$P(\varphi)=\frac{1}{2\pi} \left| \left(\sum_{n'=0}^{\infty} e^{-i\varphi n} \left\langle n'\right| \right)\cdot \left(e^{-\frac{1}{2}\left|\alpha \right|^2}\sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n!}} \left|n \right\rangle\right) \right|^2 \\ = \frac{1}{2\pi} \left| \left(\sum_{n'=0}^{\infty} \sum_{n=0}^{\infty} e^{-i\varphi n} e^{-\frac{1}{2}\left|\alpha \right|^2} \frac{\alpha^{n}}{\sqrt{n!}} \delta_{n,n'}\right) \right|^2 $$ $$=\frac{1}{2\pi} \left| \left( \sum_{n=0}^{\infty} e^{-i\varphi n} e^{-\frac{1}{2}\left|\alpha \right|^2} \frac{\alpha^{n}}{\sqrt{n!}}\right) \right|^2,\tag{2}$$
Now, we use the polar form of $\alpha$, that is $\alpha=\left| \alpha\right|e^{i\theta}$; therefore, we can express the Eq. (2) as
$$P(\varphi)=\frac{e^{-\left|\alpha \right|^2}}{2\pi}\left| \sum_{n=0}^{\infty} e^{-in (\varphi-\theta)} \frac{\left| \alpha\right|^{n}}{\sqrt{n!}}\right|^2. \tag{3}$$
Now comes the important step. Gerry and Knight invoke the fact that the Poisson distribution can be approximated by a Gaussian for large $\left|\alpha\right|^2$; then, they use the following approximation
$$e^{-\left|\alpha \right|^2/2} \frac{\left| \alpha\right|^{2n}}{n!} e^{-\left|\alpha \right|^2/2} e^{-\left|\alpha \right|^2} \approx \left(2\pi \left|\alpha \right|^2\right)^{-\frac{1}{2}} \exp\left[-\frac{(n-\left|\alpha \right|^2)^2}{2\left|\alpha \right|^2}\right]\tag{4} $$
in the sum of the Eq. (3) to evaluate it; then, they obtain
$$P(\varphi) \approx \left(\frac{2 \left|\alpha \right|^{2}}{\pi} \right)^{\frac{1}{2}} \exp \left[-2\left| \alpha \right|^2 (\varphi - \theta)^2 \right], \tag{5}$$
which represents a Gaussian centered at $\varphi=\theta$. Then, my question is, how to use Eq. (4) in Eq. (3) to obtain the Eq. (5)?
