In the course of my research, I needed a formula and found it, but I can not understand the derivation process of the formula. How to extract the $t^n$ and get the $\theta(m-p)$ in the last step? Can you explain how to get the last step? Thank you.
The derivation process in the literature is as follows:
The generating function for the Laguerre functions is {$\phi_m(x_3;\alpha)$}
\begin{align}\frac{\alpha ^{1/2}e^{-(1/2)\alpha x_3 (1+s)/(1-s)}}{1-s}= \sum_{m=0}^{\infty} s^m\phi_m(x_3;\alpha)\end{align}
Thus
\begin{align*} &\int_{0}^{\infty}dx_3\frac{\alpha ^{1/2}e^{-(1/2)\alpha x_3 (1+s)/(1-s)}}{1-s}e^{-\beta x_3}\frac{\alpha ^{1/2}e^{-(1/2)\alpha x_3 (1+t)/(1-t)}}{1-t}\\ &\quad=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}s^m t^n\left \langle m|e^{-\beta x_3}|n \right \rangle\\ &\quad=\frac{\alpha}{(\alpha+\beta-\beta t)-(\beta+(\alpha-\beta) t) s}\\ &\quad=\alpha \sum_{m=0}^{\infty} s^{m} \frac{(\beta+(\alpha-\beta) t)^{m}}{(\alpha+\beta-\beta t)^{m+1}}\\ &\quad=\alpha \sum_{m=0}^{\infty} s^{m} \sum_{n=0}^{\infty} t^{n} \sum_{p=0}^{n} \theta(m-p) \frac{(m+n-p) !}{(m-p) !(n-p) ! p !} \frac{(\alpha-\beta)^{p} \beta^{m+n-2 p}}{(\alpha+\beta)^{m+n+1-p}} \end{align*}
where $\theta(n)=1$ for $n=0,1,2, \ldots$ and $\theta(n)=0$ for $n=-1,-2,-3, \ldots$