Exotic distribution During the development of a variational mean field algorithm I have found a distribution with the form:
$q(x) \propto x e^{-ax^2 +bx}$
with $x\in(0,+\infty)$
Does such a distribution have a name? What is the normalizing factor? And the expected moments? I always can compute them, but I'm wondering if that's already done. 
 A: Writing $$x\exp(-ax^2+bx) =  \exp\left(\frac{b^2}{4a}\right)x^{2-1}\exp\left(-\left(\frac{x-b/(2a)}{1/\sqrt{a}}\right)^2\right)$$ exhibits this distribution as a Generalized Gamma  with scale parameter $1/\sqrt{a}$ and shape parameters $d=2, p=2$ that has been shifted by $\mu=b/(2a)$ and truncated at the left at $b/(2a)$.  Because the power of $x$ in the exponential is $p=2$ it can also be called a shifted left-truncated Rayleigh distribution.
The raw moments are relatively easy to find using the moment generating function (MGF) $\mathbb{E}(\exp(-tX))$ because the scaling and shifting tell us to change the variable from $x$ to $y=\sqrt{a}(x - b/(2a))$, after which the integral splits into two that are readily evaluated in terms of the standard Normal distribution function $\Phi$:
$$\phi(t) \propto 2\sqrt{a} + 2\sqrt{\pi}(b-t)\exp\left(\frac{(b-t)^2}{4a}\right)\Phi\left(\frac{b-t}{\sqrt{2a}}\right).$$
Because the MGF is directly proportional to the density function, the constant of proportionality (which so far has been ignored) can now be found by evaluating $\phi(0)$, because the value of the MGF at zero must equal $1$. In other words, the MGF is $\phi(t)/\phi(0)$.
