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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 $x$ .

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  • $\begingroup$ Here are some values fro 1 to 50 with step 3 done with Mathematica: ${0.644934066848, 0.0506357428594, 0.0193926015722, 0.00909693982500, \ 0.00575435101450, 0.00367705114317, 0.00271759837650, \ 0.00197640380654, 0.00157689618165, 0.00123156302281, \ 0.00102844646947, 0.000840349805475, 0.000723315503534, \ 0.000609762183667, 0.000536278083883, 0.000462537723680, \ 0.000413413530462}$. $\endgroup$
    – user64494
    May 11 '19 at 8:20
  • $\begingroup$ @user64494 for $n=1$ I get $\int_0^1 \int_0^1 (1-2x)(1-2y)/(1-xy)=5\pi^2/6-8\approx 0.22467$ $\endgroup$ May 11 '19 at 9:35
  • $\begingroup$ @FedorPetrov I get the same as you (also with Mathematica) $\endgroup$
    – EGME
    May 11 '19 at 9:44
  • $\begingroup$ thanks mr Fedor petrov for your quick answer, i'm sorry because my question wasn't clear, in fact i'm looking to constant $ 0<c<1$ independent of $n$ such that $I_n<b*c^n$ whith $b$ is a constant independant of $n$ . an equivalent to $(I_n)^(1/n)$ will be perfect for me $\endgroup$
    – mamiladi
    May 11 '19 at 13:01
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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.

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  • $\begingroup$ Calculations confirm it. $\endgroup$
    – user64494
    May 11 '19 at 9:58
  • $\begingroup$ @user64494 I get about $1/(2n^2)$, and your calculations show something like $1/n^2$. And we get smth different already for $n=1$. $\endgroup$ May 11 '19 at 11:00

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