I am trying to compute an integral that looks like the moments of a Gaussian $\mathcal{N}(\mu, \sigma^2)$, but the main difference is that we only integrate over R+ and not R. I believe we could call this the moments of a Half-Gaussian distribution :
$$I_m = \int_0^{\infty}x^m \exp\left(-\frac{1}{2\sigma^2}(x-\mu)^2\right)\mathrm{d}x, \quad m\in\mathbb{N}~.$$
I was wondering if there is a (recurrence) formula of $I_m$ for any $m\in \mathbb{N}$.
So far I found that:
- $m=0$: $I_0 = \dfrac{\operatorname{erf}\left(\frac{{\mu}}{\sqrt{2}\,{\sigma}}\right)+1}{2}$
- $m=1$: $I_1 = \dfrac{\sqrt{2}\operatorname{\Gamma}\left(1,\frac{{\mu}^2}{2{\sigma}^2}\right)\,{\sigma}+\left(2\sqrt{{\pi}}-\operatorname{\Gamma}\left(\frac{1}{2},\frac{{\mu}^2}{2{\sigma}^2}\right)\right){\mu}}{2\sqrt{{\pi}}}$
- $m=2$: $I_2 = \dfrac{2^\frac{3}{2}\operatorname{\Gamma}\left(1,\frac{{\mu}^2}{2{\sigma}^2}\right)\,{\mu}{\sigma}+\left(2\sqrt{{\pi}}- \operatorname{\Gamma}\left(\frac{1}{2},\frac{{\mu}^2}{2{\sigma}^2}\right) + \left(2\sqrt{{\pi}}-2\operatorname{\Gamma}\left(\frac{3}{2},\frac{{\mu}^2}{2{\sigma}^2}\right)\right){\sigma}^2\right){\mu}^2}{2\sqrt{{\pi}}}$
- ...
with $\Gamma$ the incomplete Gamma function. This led me to the formula for $m>1$: $$I_m = T_{m,\sigma} \mu^{m-1}\sigma + T_{m,\sigma^2} \mu^{m-2}\sigma^2 + \dots + T_{m,\sigma^{m-1}} \mu\sigma^{m-1} + T_{m,\sigma^m}\sigma^m = \sum_{n=1}^m T_{m,\sigma^n} \mu^{m-n}\sigma^n $$
I managed to find a recurrence formula for $T_{m,\sigma^n}$ as a function of $m$ for $n=1,2,3,4$. However, I stopped there because it didn't seem to lead me to a general formula for any $n$. Also, the results I found are assuming that $\mu>0$, which is quite restrictive.
Given the nice form of the integral $I_m$ and its close relationship to the moments of a Gaussian, I was wondering if someone might know how to compute it and help me out...
Thanks in advance!