Integral involving Laguerre polynomials I am trying to solve
$$ \int_{-\infty}^{\infty} dz  \exp\left[-\left(z - \tfrac{1}{2}r\right)^2 \right] L_n(z^2),$$
where
$L_n$ denotes the n-th Laguerre polynomial such that 
$$ L_n(x) = \sum_{k=0}^n \frac{(-1)^k}{k!} \binom{n}{k}x^k.$$
Background info: In quantum mechnanics in phase space, this integral arises in the derivation of Husimi function of the n-th Hermite function. The Husimi function is obtained by the convolution of the Wigner function by a Gaussian. The Wigner function of the n-th Hermite function reads
$$W_n(r) = \frac{(-1)^n}{\pi\hslash} \exp(-\tfrac{1}{2} r^2 )L_n(r^2)$$
where $r^2 = 2 \left[ \left(\frac{p}{\hslash \kappa} \right)^2 + \left(\kappa x \right)^2 \right]$. See, for instance, p.106 of Schleich's Quantum Optics in Phase Space.
My ansatz is simply $Q_n(r) = (W_n \ast W_0)(r)$, where $Q_n$ denotes the Husimi function. It is of course well-known, and reads  $Q_n(r) = \frac{1}{2\pi \hslash n!}\left(\tfrac{1}{4}r^2 \right)^n \exp( - \tfrac{1}{4}r^2)$.
I am aware that one can reduce the integral to a Gaussian integral by doing the substitution $\beta = z - \tfrac{1}{2} r$, applying the definition of Laguerre polynomials, using the binomial theorem and so on. However, in the end I get an expression with sums, binomial coefficients and a double factorial, which is a bit more complex than the actual solution:
 $$Q_n(r) = (W_n \ast W_0) (r) = \frac{(-1)^n}{(\pi\hslash)^2} \exp\left(-\frac{1}{4}r^2 \right)\sum_{k=0}^n \frac{(-1)^k}{k!}\binom{n}{k}\sum_{j=0}^{k} \binom{2k}{2j} \left(\frac{1}{2}r \right)^{2k-2j} \\
\times \int d\beta \exp (-\beta^2) \beta^{2j},$$
with $\int d\beta \exp (-\beta^2) \beta^{2j} = \sqrt{\pi} \frac{(2j-1)!!}{2^j}$ from Wikipedia.
I looked at some handbooks for integration and summation formulas, however I was not able to find something with which I could make progress on this problem. 

Let $a=r/2$, so that your integral has the form
$$f_n = \int_{-\infty}^\infty dx \, e^{-(x-a)^2} L^{(-1/2)}_n(x^2) \; .$$ 
The evaluation will be based on the generating function
$$\frac{1}{\sqrt{1-z}} \exp\left(-\frac{x^2 z}{1-z}\right) = \sum_n z^n L_n^{(-1/2)}(x^2) \; .$$
Multiply the generating function by $e^{-(x-a)^2}$ to obtain
$$\int_{-\infty}^\infty dx \, e^{-(x-a)^2} \frac{1}{\sqrt{1-z}} \exp\left(-\frac{x^2 z}{1-z}\right) = \sum_n z^n f_n \; .$$
For this particular generalized Laguerre polynomial ($\alpha=-1/2$), the integral of the generating function has the convenient form
$$\sqrt{\pi} e^{-a^2 z} = \sqrt{\pi} \sum_n \frac{(-a^2 z)^n}{n!} = \sum_n z^n f_n \; .$$
Thus, we have shown
$$f_n = \frac{\sqrt{\pi}}{n!} {(-a^2)}^n \; .$$
So, I suspect your Husimi functions (never heard of them) are written in terms of the Hermite polynomials
$$H_{2n}(x) = {(-4)}^n n! \, L_n^{(-1/2)}(x^2) \; .$$
