let $\forall n $ integer $p_n(t)=\frac{1}{n!}(t^n(1-t)^n)^{(n)} $ i 'm looking for an explicit constant $0<c<1$ ( very good small c ) independant of $n$, and a constant $b$ ( non explicit) independant of $n$ such that $ I_n:=| \int_{0}^1 \int_{0}^1 \frac{P_n(x)P_n(y)}{1-xy}dxdy| \leq b c^n$ or such that $(I_n)^\frac{1}{n} $ equivalent to $ c $
remarque 0: this integral is similar to Beukers integral
remarque 1: after integrating n times for $ y $ we get $I_n =| \int_{0}^1 \int_{0}^1 \frac{P_n(x)x^ny^n(1-y)^n}{(1-xy)^{n+1}}dxdy| $
remarque 2: $ I_n= \displaystyle \sum_{k=0}^{+\infty} \int_{0}^1 P_n(x)P_n(y)x^ky^k dx dy= \displaystyle \sum_{k=n}^{+\infty} (\int_{0}^1 P_n(x)x^k dx )^2 = \displaystyle \sum_{k=n}^{+\infty} (\int_{0}^1 x^n(1-x)^n \frac{(-1)^n}{n!} k(k-1)..(k-n+1) x^{k-n} dx )^2 $ \ $ =\displaystyle \sum_{k=n}^{+\infty} {{\Big{(}} \frac{k(k-1)..(k-n+1)}{(k+1)(k+2)..(n+k+1)}\Big{)}}^2 $ so we can may be use the hypergeometric term to find c
remarque 3: i'm not sure but i think that $ ( \int_{0}^1 \int_{0}^1 \frac{P_n(x)P_n(y)}{1-xy}dxdy)_{ n \geq 0} $ satisfy a recurrence relation (as Apery number), i don't want that method to find $c$ using poincarre theorem
