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I'm trying to solve the following definite integral:

$\mathcal{I} = \int_0^1dx\ x^{P+k/2-m}(1-x)^me^{-\sqrt{x}}, $

where $P\in\mathcal{N}$ (whole positive numbers and zero), $m\in\mathcal{N}$, $k\in\mathcal{N}$, $0\leq k\leq P$ and $0\leq m\leq P$.

It's similar to the Beta function except for the exponential factor. I've been trying different approaches and looking around with no luck.

Mathematica gives the answer as

$0.443113\ m! \left[2 \Gamma \left(\frac{k}{2}-m+\text{P}+1\right) \, _1\tilde{F}_2\left(\frac{k}{2}-m+\text{P}+1;\frac{1}{2},\frac{k}{2}+\text{P}+2;\frac{1}{4}\right)+\Gamma \left(\frac{1}{2} (k-2 m+3)+\text{P}\right) \, _1\tilde{F}_2\left(\frac{1}{2} (k-2 m+3)+\text{P};\frac{3}{2},\frac{k+5}{2}+\text{P};\frac{1}{4}\right)\right]$

where $\ _p\tilde{F}_q\Big(\{a_1,\ldots,a_p\};\{b_q,\ldots,b_q\};z\Big)\ $ is the regularized generalized hypergeometric function.

I'm hoping that there is a simpler solution that becomes apparent when solving the integral (if so that's probably faster than double checking Mathematica's answer and start going through identities to rewrite it).

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    $\begingroup$ This can be integrated to $p(\sqrt{x})\, e^{-\sqrt x}$ with polynomial $p$ (make the change of variables $t=\sqrt{x}$). $\endgroup$ Oct 17, 2014 at 16:43
  • $\begingroup$ I don't understand exactly, do you mean so as to end up with a sum from $0$ to $m$ of incomplete Gamma functions? $\endgroup$
    – jorgen
    Oct 17, 2014 at 17:01
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    $\begingroup$ No. The integral breaks into a sum of pieces like $\int x^k e^{-\sqrt x}dx$ with $k$ positive integer or half-integer. After the change of variables $t=\sqrt x$ this becomes something like $\int t^{2k+1}e^{-t}dt$ which can be easily integrated by parts. $\endgroup$ Oct 17, 2014 at 17:08
  • $\begingroup$ Ok! When you say 'sum of pieces' this includes a sum over $m$ from the binomial theorem applied to $(1-x)^m$, right? $\endgroup$
    – jorgen
    Oct 17, 2014 at 17:11
  • $\begingroup$ Yes. So one ends with a double sum in fact. $\endgroup$ Oct 17, 2014 at 17:12

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