Does a distribution on $\mathbb{R}$ exists such that for a random variable $X$ on $\mathbb{R}$ with said distribution we have $$ \mathbb{E}[X^{2l}] = l! \ \ \text{ and } \ \ \mathbb{E}[X^{2l+1}] = 0 $$ for all $l \in \mathbb{N}_0$?
The equality $\mathbb{E}[X^{2l+1}] = 0$ clearly holds for every symmetric distribution (where the moments exist) but I don't know of a distribution with such even moments.
Any help is much apprechiated.
On
Here's a solution relying of generating functions and some simple guesswork.
We are looking for a function $f(x)$ obeying the equation
$$\int_{-\infty }^{\infty } f(x) x^{2 k} \, dx=k!\tag{1}$$
Multiplying by $(-z)^k$ and trying to sum over $k$ needs a convergence forcing factor. The most natural choice is $\frac{1}{k!}$.
Since on the l.h.s. under the integral we have
$$\sum _{k=0}^{\infty } \frac{\left(x^2 (-z)\right)^k}{k!}=e^{-x^2 z}\tag{2}$$
and on the r.h.s.
$$\sum _{k=0}^{\infty }(-z)^k=\frac{1}{1+z}$$
We find for $f(x)$ in place of $(1)$
$$\int_{-\infty }^{\infty } f(x)e^{-x^2 z} \, dx=\frac{1}{1+z}\tag{3}$$
In order to generate somehow $1+z$ in the denominator we let
$$f(x) = e^{-x^2} g(x)$$
with $g(x) = g(-x)$ so that
$$2 \int_{0 }^{\infty } g(x)e^{-x^2 (1+z)} \, dx=\frac{1}{1+z}$$
Now since $d(x^2) = 2 x dx$ we find that $g(x) = x$ gives the desired solution for $x\gt 0$. Extending this to negative $x$ gives $g(x) = |x|$, and the final result becomes
$$f(x) = |x|e^{-x^2}\tag{4}$$
This coincides with the result derived in a different manner by other users.
From comments:
You could look for a $Y$ with $\mathbb E\left[Y^n\right]=n!$ and then let $X=\pm\sqrt{Y}$ with equal probability. A little bit of extra work would let you specify the distribution more simply.
Such a $Y$ is well known: it is the exponential distribution with parameter $1$.
This leads to $X$ having density $$f(x)=|x|\, e^{-x^2}$$ over the reals, as illustrated below. This is clearly symmetric about $0$, so the odd moments are $0$. A relatively simple integration shows the even moments are the desired factorials.