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?
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.
