# Can't solve this integral. Solution involves hypergeometric functions.

I need to re-derive a result from a paper: $$\int^{\infty}_{-\infty}\,{\rm d}x\, \left\vert x\right\vert\,x^{n}\, \Theta\left(\frac{x}{c}\right) \exp\left(% -\,\frac{1}{2}\left\{abc + \frac{\left[bc\right]^{2} + 1}{\beta bc}x \right\}^{2}\,% \right) \tag{1}$$ where $\Theta$ is the step function, $n \in \mathbb{N^+}$ , $a,c \in \mathbb{R}$ , $\beta \in [1,2]$ and $b \in [0,\infty)$.

It says the solution is:

$$2^{n/2} {\Big(\frac{\beta bc}{{(bc)}^2+1}\Big)}^{n}e^{-\frac{{(abc)}^2}{2}} \Big[ \Gamma(1+\frac{n}{2})M(1+\frac{n}{2},\frac{1}{2};\frac{{(abc)}^2}{2}) + \sqrt{2}\:abc\:\Gamma(\frac{3+n}{2})M(\frac{3+n}{2},\frac{3}{2};\frac{{(abc)}^2}{2}) \Big] \tag{2}$$ Where $M$ is the Kummer's confluent hypergeometric function.

I think it can be useful the relation with the Tricomi funcion that you can find in the same wikipedia page:

$$U(\alpha,\gamma;z)=\frac{1}{\Gamma(\alpha)} \int^{\infty}_{0} dt \: e^{-zt} t^{\alpha-1} {(1+t)}^{\gamma-\alpha-1} \tag{3}$$ $$U(\alpha,\gamma;z)=\frac{\Gamma(1-\gamma)}{\Gamma(\alpha-\gamma+1)}M(\alpha,\gamma;z) + \frac{\Gamma(\gamma-1)}{\Gamma(\alpha)}z^{1-\gamma}M(\alpha-\gamma+1,2-\gamma;z) \tag{4}$$

if we put: $$\alpha=1+\frac{n}{2} \\ \gamma=\frac{1}{2} \\ z=\frac{{(abc)}^2}{2} \tag{5}$$ (4) resembles the result that I have to prove. It seems that i have to manipulate the original integral into something like the U integral form (3).

I'm sorry for the cumbersome parameters but I want to preserve the original form of the problem.

UPDATE: I verified numerically that must be an error in (2), there is a minus between the two $M$ functions: $$2^{n/2} {\Big(\frac{\beta bc}{{(bc)}^2+1}\Big)}^{n}e^{-\frac{{(abc)}^2}{2}} \Big[ \Gamma(1+\frac{n}{2})M(1+\frac{n}{2},\frac{1}{2};\frac{{(abc)}^2}{2}) - \sqrt{2}\:abc\:\Gamma(\frac{3+n}{2})M(\frac{3+n}{2},\frac{3}{2};\frac{{(abc)}^2}{2}) \Big] \tag{6}$$

UPDATE 2: I posted a simplified version of the question here, I really hope this is not a problem with double identical questions!

Solved thanks to this question. It is the same integral with the obvious substitution $$y= \frac{{(bc)}^2+1}{\beta bc}x$$