# Definite integral similar to beta function but with exponential negative square root

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.

$0.443113\ m! \left[2 \Gamma \left(\frac{k}{2}-m+\text{P}+1\right) \, _1\tilde{F}_2\left(\frac{k}{2}-m+\text{P}+1;\frac{1}{2},\frac{k}{2}+\text{P}+2;\frac{1}{4}\right)+\Gamma \left(\frac{1}{2} (k-2 m+3)+\text{P}\right) \, _1\tilde{F}_2\left(\frac{1}{2} (k-2 m+3)+\text{P};\frac{3}{2},\frac{k+5}{2}+\text{P};\frac{1}{4}\right)\right]$
where $\ _p\tilde{F}_q\Big(\{a_1,\ldots,a_p\};\{b_q,\ldots,b_q\};z\Big)\$ is the regularized generalized hypergeometric function.
• This can be integrated to $p(\sqrt{x})\, e^{-\sqrt x}$ with polynomial $p$ (make the change of variables $t=\sqrt{x}$). Oct 17, 2014 at 16:43
• I don't understand exactly, do you mean so as to end up with a sum from $0$ to $m$ of incomplete Gamma functions? Oct 17, 2014 at 17:01
• 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. Oct 17, 2014 at 17:08
• Ok! When you say 'sum of pieces' this includes a sum over $m$ from the binomial theorem applied to $(1-x)^m$, right? Oct 17, 2014 at 17:11