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 $n=-1,-2,-3, \ldots$
Best Answer
We obtain \begin{align*} &\frac{\left(\beta+(\alpha-\beta)t\right)^m}{(\alpha+\beta-\beta t)^{m+1}}\\ &\quad=\frac{\beta^m}{\left(\alpha+\beta\right)^{m+1}}\, \frac{\left(1+\frac{\alpha-\beta}{\beta}t\right)^m}{\left(1-\frac{\beta}{\alpha+\beta}t\right)^{m+1}}\\ &\quad=\frac{\beta^m}{\left(\alpha+\beta\right)^{m+1}} \sum_{k=0}^\infty\binom{-m+1}{k}\left(-\frac{\beta}{\alpha+\beta}\right)^kt^k\left(1+\frac{\alpha-\beta}{\beta}\right)^m\tag{1}\\ &\quad=\frac{\beta^m}{\left(\alpha+\beta\right)^{m+1}} \sum_{k=0}^\infty\binom{m+k}{k}\left(\frac{\beta}{\alpha+\beta}\right)^kt^k\sum_{p=0}^m\binom{m}{p}\left(\frac{\alpha-\beta}{\beta}\right)^pt^p \tag{2}\\ &\quad=\frac{\beta^m}{\left(\alpha+\beta\right)^{m+1}}\sum_{n=0}^\infty \left(\sum_{{k+p=n}\atop{k\geq 0,0\leq p\leq m}}\binom{m+k}{k}\binom{m}{p} \left(\frac{\beta}{\alpha+\beta}\right)^k\left(\frac{\alpha-\beta}{\beta}\right)^p\right)t^n\tag{3}\\ &\quad=\sum_{n=0}^\infty \left(\sum_{{p=0}\atop{0\leq p\leq m}}^n\binom{m}{p}\binom{m+n-p}{n-p}\frac{(\alpha-\beta)^p\beta^{m+n-2p}} {(\alpha+\beta)^{m+n+1-p}}\right)t^n\tag{4}\\ &\quad\,\,\color{blue}{=\sum_{n=0}^\infty\left(\sum_{p=0}^{\min\{m,n\}}\frac{(m+n-p)!}{(n-p)!p!(m-p)!}\,\frac{(\alpha-\beta)^p\beta^{m+n-2p}} {(\alpha+\beta)^{m+n+1-p}}\right)t^n}\tag{5} \end{align*} and the claim follows.
Comment:
In (1) we make a binomial series expansion.
In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ and expand the binomial.
In (3) we calculate the Cauchy product.
In (4) we substitute $k=n-p$ and collect terms.
In (5) we use $\binom{n}{k}=\frac{n!}{k!(n-k)!}$.