Looking for bound in integral involving Legendre polynomial I'm looking for an upper bound to the following integral or equivalent when $n$ leads to $ +\infty $  to the following expression
$$I_n:=\left|\int_{0}^1 \int_{0}^1 \frac{p_n(x) p_n(y)}{(1-xy)} dx dy \right| $$
with 
$$ p_n(t):=\frac{1}{n!}(t^n(1-t)^n)^{(n)}.$$
This integral is similar to Beukers integral; after integrating $n$ times to $y$ ,I obtain
$$I_n:=\left|\int_{0}^1 \int_{0}^1 \frac{p_n(x) x^n y^n(1-y)^n}{(1-xy)^{n+1}} dx dy \right| ,$$
I don't know what to do after this since I can't have interesting expression inside the integral when I derive n times  $x$ .  
 A: We have 
$$
\int_{0}^1 \int_{0}^1 \frac{p_n(x) p_n(y)}{1-xy} dx dy=
\int_{0}^1 \int_{0}^1 p_n(x) p_n(y)\sum_{k=0}^\infty(xy)^k dx dy=
\sum_{k=0}^\infty \left(\int_0^1 p_n(x)x^kdx\right)^2.
$$
Next, integrating by parts we have 
$$
(-1)^n\int_0^1 p_n(x)x^kdx={k\choose n}\int_0^1x^k(1-x)^n dx=\\={k\choose n}\cdot \frac{k! n!}{(k+n+1)!}=\frac{{k\choose n}}{{k+n\choose n}}\cdot \frac1{k+n+1}.
$$
We have to estimate the sum of squares of these guys over $k=0,1,\ldots$. For $k\leqslant n-1$ they are just zeros. For $k\geqslant n$ we have
$$
\frac{{k\choose n}}{{k+n\choose n}}=\prod_{i=0}^{n-1}\left(1-\frac{n}{k-i+n}\right)\leqslant \prod_{i=0}^{n-1}\exp\left(-\frac{n}{k-i+n}\right)\leqslant \exp(-n^2/(n+k)).
$$
The function $f(x)=\exp(-2n^2/(x+n))(n+x+1)^{-2}$ for $x\in [n,\infty)$ has the unique maximum point. Indeed, $f$ tends to 0 at infinity and $d(\log f)/dx=2n^2/(x+n)^2-2/(n+x+1)$, it equals to 0 when $n^2(1/(x+n)+1/(x+n)^2)=1$, by monotonicity it has unique positive root $x_0$ for which $n^2/(x_0+n)<1$, $x_0>n^2-n$. The value $f(x_0)$ is at most $n^{-4}$. The sum $\sum_{k\geqslant n} f(k)$ is therefore $O(n^{-4})+\int_n^\infty f(x)dx$. For the integral we have
$$
\int_n^\infty \exp(-2n^2/(x+n))(n+x+1)^{-2}dx\leqslant \int_n^\infty \exp(-2n^2/(x+n))(x+n)^{-2}dx=\frac{1}{2n^2}\left(1-e^{-n}\right).
$$
So $I_n=O(n^{-2})$, and up to multiplicative factor this is sharp.
