3
$\begingroup$

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!

Thank you for your patience.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.