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I have something like

$F(t)=\int_0^{t} e^{-kx} x^\alpha (1+x)^\beta dx$

Is this a form of some known distribution (more specifically density)?

EDIT:

I was asked where I encountered the problem.

Its is as follows...

$\theta|\lambda \sim\mathcal{N}_p(0,\frac{1-\lambda}{\lambda}I)$ and $\lambda$ follows density $f(x)\sim(1-b)x^{-b}$ for $(b<1)$.

I calculated the density of $\theta$ (unconditional).

(While fiddling about 5.4 of this paper)

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    $\begingroup$ Are you giving us a density $f$? $\endgroup$ Oct 27, 2018 at 10:30
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    $\begingroup$ @StubbornAtom Sorry for the cross post. Can you please provide a link to the relevant Mathematica page that has the closed form? $\endgroup$
    – Qwerty
    Oct 27, 2018 at 10:42
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    $\begingroup$ No link, I used my copy of the software. Please decide where you want to ask the question. Cross-posting is against SE rules in general. $\endgroup$ Oct 27, 2018 at 10:58
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    $\begingroup$ There is a fundamental problem with this question: lhs depends on $x$, rhs integrates in $x$... $\endgroup$
    – Xi'an
    Oct 27, 2018 at 12:24
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    $\begingroup$ Would you perhaps intend to write $$F(x;k,\alpha,\beta)=\int_0^x e^{-kx} x^\alpha (1+x)^\beta dx$$ for $x\gt 0$? Regardless, please edit your post to remove the ambiguity arising from equating a function of $x$ with a function of $k,\alpha,\beta$ only. $\endgroup$
    – whuber
    Oct 27, 2018 at 16:57

1 Answer 1

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I will interpret your question as if the positive function given by $$ f(x)= e^{-kx} x^\alpha (1+x)^\beta $$ (removing the superfluous integral) is proportional to some known (named) distribution.

First, I will rewrite (and rename, reparametrize, so note that $\alpha$ below has new meaning) this function to show that it is a multiplicative modification of a gamma density: $$ f(x) \propto \frac{k^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-kx} \cdot (1+x)^\beta \quad \text{for $x>0$} $$ which has a finite integral for $k>0, \alpha>0, \beta\in\mathbb{R}$. For $\beta=0$ it is a gamma density.

Attempting to find a closed form for the normalization constant (with the help of maple) result in an expression to large to be very useful, including Laguerre polynomials or Kummer functions. Part of the nonusefulness of the resulting expression is because it includes, for some parameter values, gamma functions evaluated at negative integer arguments.

But numerical integration works without problems. I don't know any standard name for this distribution. Where did you encounter it, why are you interested?

EDIT

After some more work with maple (doing first some by hand), I get this expression for the $\int_0^\infty \cdot \; dx$: $$ {\frac {{\beta\choose -\alpha} {\mbox{$_1$F$_1$}(\alpha;\,\alpha+\beta+1;\,k)}\pi\,\sin \left( \pi\, \beta \right) {k}^{\alpha}}{\sin \left( \pi\,\alpha+\pi\,\beta \right) \sin \left( \pi\,\alpha \right) \Gamma \left( \alpha \right) }} $$ which is much simpler than what I got before, but still needs more work to simplify for some difficult integer parameter cases. (for instance, for $\beta=0$ plugging directly into that expression gives 0)

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    $\begingroup$ There are some typographical errors in your second expression for $f$: you need to replace $\alpha$ by $\alpha+1$ or else you need to include a factor of $x.$ The problem with Gamma evaluated at negative integers lies with Maple, not with $f$ itself, because such values can always be eliminated (they represent limits of diverging quantities and therefore will be canceled by other quantities). It should give you a multiple of a confluent hypergeometric function. $\endgroup$
    – whuber
    Oct 28, 2018 at 18:21
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    $\begingroup$ Thanks for this. I think eq 2 is correct (note the range for k, $\alpha$), but $\alpha$ changed meaning from eq 1. About Maple, I suspected so much, really needs to learn more about special functions ... I will write a report to Maple ... $\endgroup$ Oct 28, 2018 at 18:44
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    $\begingroup$ +1. For small positive integral $\beta,$ simply expand $(1+x)^\beta,$ yielding a mixture of Gamma distributions with Binomial weights. To deal with integral arguments to $_1F_1,$ take limits using L'Hopital's Rule. $\endgroup$
    – whuber
    Oct 28, 2018 at 21:23
  • $\begingroup$ @kjetilbhalvorsen I gave the source of doubt $\endgroup$
    – Qwerty
    Oct 28, 2018 at 22:35

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