The Laguerre Polynomials have the following integral representations $$L_{n}^{\alpha} (x) = x^{-\alpha} e^x \frac{1}{2\pi i } \oint_c \frac{e^{-z} z^{n+\alpha}}{(z-x)^{n+1}} dz$$ where $c$ is an anticlockwise contour around $z=x$. Use these to show that the generating function defined as $$F^{\alpha} (t,x) = \sum_{n=0}^{\infty} t^n L_n^{\alpha} (x),$$ is given by $$F^{\alpha}(t,x) = \frac{1}{(1-t)^{\alpha+1}}e^{-\frac{xt}{1-t}}$$ Assume that for small enough $t$, the order of integration and summation may be swapped and any resulting poles lie within the contour.
Attempt: Swapping the order of integration as hinted at gives $$F = \frac{1}{2\pi i } \oint_c \sum_{n=0}^{\infty} \left(\frac{zt}{(z-x)}\right)^n \frac{e^{-z+x} z^{\alpha} x^{-\alpha}}{(z-x)}dz$$ Now for $t$ small, I can write the sum as a geometric series (and in fact there is a hint in the question that a geometric series should be obtained), so I can write the sum of the series as $1/(1- (zt/(z-x))$. Now inputting, rearranging gives the formula $$\frac{1}{2\pi i} e^x x^{-\alpha} \oint \frac{e^{-z} z^{\alpha}}{z(1-t)-x}dz$$ Per the question, we assume the pole $z = x/(1-t)$ lies within the contour and evaluating by residue theorem gives the result $$\frac{1}{(1-t)^{\alpha}}e^{\frac{-xt}{(1-t)}}$$ so I am off by a factor of $1/(1-t)$ just. I can't see it, any help would be great. I hope it is something obvious and my method is fine. Many thanks.
First of all, good job on this computation. Everything looks correct, except for your calculation of the residue. Note that \begin{align*} \text{Res}\left(\frac{1}{z(1-t) - x}, \frac{x}{1-t}\right) = \lim_{z \to \frac{x}{1-t}} \left(z - \frac{x}{1-t}\right) \frac{1}{z(1-t) - x} = \frac{1}{1-t} \end{align*} which accounts for the missing factor of $\frac{1}{1-t}$.