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mamiladi
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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
$$I_n:=\left|\int_{0}^1 \int_{0}^1 \frac{p_n(x) x^n y^n(1-y)^n}{(1-xy)^{(n+1)}} dx dy \right| ,$$$$I_n:=\left|\int_{0}^1 \int_{0}^1 \frac{p_n(x) x^n y^n(1-y)^n}{(1-xy)^{n+1}} dx dy \right| ,$$
I don't know what to do after this since I can't have interesting expression inside the integral when I derive n times $x$ .

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 $x$ .

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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Martin Sleziak
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Looking for bound in intgeralintegral involving Legendre polynomial

I'm looking for an apperupper 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 iI can't have interesting expression inside the integral when iI derive n times $x$ .

Looking for bound in intgeral involving Legendre polynomial

I'm looking for an apper 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 $x$ .

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 $x$ .

looking Looking for bound in intgeral involving Legendre polynomial

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András Bátkai
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mamiladi
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