Closed form $\int_{0}^{\frac{r}{2}} {\binom{n}{p} \binom{n-p}{r-2p} 2^{r-2p}}{\binom{2n}{r}^{-1}} \ \text{d}p$ Note: This is exact copy of my Math.SE question, which I am reposting here, as despite bounty it did not receive any answers.

Let there be $n$ pairs of shoes in a box.
The the probability that from the $r \le n$ shoes I am taking out of the box there are exactly $p$ pairs is given by
\begin{equation*}
        \mathbb{P}_{n}^{(r)}(p)
        = \frac{\binom{n}{p} \binom{n-p}{r-2p} 2^{r-2p}}{\binom{2n}{r}}.
\end{equation*}
For $n = 15$ and $r \in \{6,8,10\}$. The function (assuming the continuous factorial equivalents) looks like this:


I am interested in finding the area under that curve, namely
  $$\int_{0}^{\frac{r}{2}} \frac{\binom{n}{p} \binom{n-p}{r-2p} 2^{r-2p}}{\binom{2n}{r}} \ \text{d}p$$

I consulted this question but could derive how that would help me.
I also thought about writing the first product of binomial coefficients as $$
\binom{n}{n - p}\binom{n - p}{r - 2p}
$$
which is similar to the form $\binom{f(x)}{f(y)} \binom{f(y)}{f(x)}$ mentioned in this question.
 A: I suspect some form of CLT for large $n, r$, which may possibly be proved by adopting Laplace's method of approximating the integrand by an appropriate gaussian kernel.
Interestingly enough, we can prove that:
$$ \int_{-\infty}^{\infty} \frac{\binom{n}{p}\binom{n-p}{r-2p}2^{r-2p}}{\binom{2n}{r}}\,\mathrm{d}p = 1. $$
To show this, we may apply the Legendre duplication formula to write
$$ \frac{\binom{n}{p}\binom{n-p}{r-2p}2^{r-2p}}{\binom{2n}{r}} = \frac{n!}{\binom{2n}{r}} \frac{\sqrt{\pi}}{\Gamma(1+p)\Gamma(n-r+1+p)\Gamma(\frac{r}{2}+\frac{1}{2}-p)\Gamma(\frac{r}{2}+1-p)}$$
and then apply the Ramanujan's beta integral (see the formula (5.3.14) of DLMF: 5.13). 
A: Following on from Carlo's answer, someone noted in response to one of my questions once that for a unimodal nonnegative function $f$, we have 
$$ \biggl| \sum_{i=0}^k f(i) - \int_0^k f(x)\,dx \biggr|
  \le 2\max_{0\le x\le k} f(x).$$
The proof is by drawing a picture.
So to prove closeness of the sum and integral it suffices to prove that the function is unimodal and has a decreasing maximum. That shouldn't be too hard.
A: As I mentioned in a comment, I would expect the large $n,r$
 limit to be unity for any ratio $f=r/n\in(0,1]$, since the difference between $\sum_p$ and $\int dp$ should vanish in that limit (and $\sum_p \mathbb{P}_{n}^{(r)}(p)=1$ by normalization). Proving this might be cumbersome, but the numerical evidence is clear:
 
Plot of $\int_0^{fn/2} \mathbb{P}_{n}^{(fn)}(p)\,dp$ versus $n$ for $f=1/4,1/2,3/4,1$. The convergence to unity is slower for smaller $f$, but it's there, as expected.
