# 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 $$x$$ .

• Here are some values fro 1 to 50 with step 3 done with Mathematica: ${0.644934066848, 0.0506357428594, 0.0193926015722, 0.00909693982500, \ 0.00575435101450, 0.00367705114317, 0.00271759837650, \ 0.00197640380654, 0.00157689618165, 0.00123156302281, \ 0.00102844646947, 0.000840349805475, 0.000723315503534, \ 0.000609762183667, 0.000536278083883, 0.000462537723680, \ 0.000413413530462}$. May 11 '19 at 8:20
• @user64494 for $n=1$ I get $\int_0^1 \int_0^1 (1-2x)(1-2y)/(1-xy)=5\pi^2/6-8\approx 0.22467$ May 11 '19 at 9:35
• @FedorPetrov I get the same as you (also with Mathematica)
– EGME
May 11 '19 at 9:44
• 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 May 11 '19 at 13:01

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