Improper integral of the product of exponential function and Laguerre polynomial I saw this integral in the book [Gerry C.C.,Knight P.L.] Introductory quantum optics:
$$\frac{1}{\pi^2}\int_{-\infty}^{\infty}L_n(\lvert\lambda\rvert^2)e^{\lambda^*\alpha-\lambda\alpha^*-\frac{1}{2}\lvert\lambda\rvert^2}d^2\lambda$$
where $L_n(\lvert\lambda\rvert^2)$ is Laguerre polynomial and $\lambda$ and $\alpha$ complex
It is related to the Wigner distribution in Quantum mechanics. How can this be integrated?
 A: To evaluate this integral, first separate it into its radial and angular dependences, by setting $\lambda = r e^{i\theta}$ and $\alpha=s e^{i\varphi}$, which produce
\begin{align}
I & :=
\frac{1}{\pi^2}\int_{-\infty}^{\infty}L_n(\lvert\lambda\rvert^2)e^{\lambda^*\alpha-\lambda\alpha^*-\frac{1}{2}\lvert\lambda\rvert^2}\mathrm d^2\lambda
%
\\& = 
%
\frac{1}{\pi^2}
\int_{0}^{\infty}\!\!\!
\int_{0}^{2\pi}
L_n(r^2)
e^{rs e^{-i(\theta-\varphi)}-rs e^{i(\theta-\varphi)}-\frac{1}{2}r^2}
r\,\mathrm d\theta\,\mathrm dr
%
\\& = 
%
\frac{1}{\pi^2}
\int_{0}^{\infty}\!\!\!
\int_{0}^{2\pi}
e^{-2i\,rs \sin(\theta-\varphi)}
\mathrm d\theta
\,
L_n(r^2)
e^{-\frac{1}{2}r^2}
r\,\mathrm dr
%
\\& = 
%
\frac{2}{\pi}
\int_{0}^{\infty}
J_0(2rs)
L_n(r^2)
e^{-\frac{1}{2}r^2}
r\,\mathrm dr
%
\\& = 
%
(-1)^n
\frac{2}{\pi}
e^{-2s^2} 
L_n\mathopen{}\left(4s^2\right)\mathclose{},
%
\\& = 
%
(-1)^n
\frac{2}{\pi}
e^{-2|\alpha|^2} 
L_n\mathopen{}\left(4|\alpha|^2\right)\mathclose{},
%
\end{align}
where the last step comes from standard integral tables, specifically integral (7.421.1) in Gradshteyn & Ryzhik,
$$
\int_0^\infty x e^{-\frac12\alpha x^2} L_n(\tfrac12\beta x^2)J_0(xy) \mathrm dx 
= 
\frac{(\alpha-\beta)^n}{\alpha^{n+1}}
e^{-\frac{1}{2\alpha}y^2} 
L_n\mathopen{}\left(\frac{\beta y^2}{2\alpha(\beta-\alpha)}\right)\mathclose{},
$$
which is referenced to Erdelyi's Tables of Integral Transforms vol II, p. 13 eq. (4), in case you don't have access to G&R.
Obviously, if you're going to depend on any of this, you should re-calculate everything to double-check I haven't messed up any details.
