to prove:

$I=\displaystyle\int_{0}^{1} x \Gamma\left(\dfrac{2+x}{2}\right)\Gamma\left(\dfrac{2-x}{2}\right)dx\leq\dfrac{\sqrt{6\ln2}}{3}$

my attempt:

i tried using cauchy schwarz inequality . therefore

$I^2=\dfrac{1}{3}\times\left[\displaystyle\int_{0}^{1} \Gamma\left(\dfrac{2+x}{2}\right)\Gamma\left(\dfrac{2-x}{2}\right)dx\right]^2$

but i don't know how to proceed further. i am new gamma function please don't used too advanced stuff.

i am thinking of using this $\beta\left[(1+x/2),(1-x/2)\right]=\displaystyle\int_{0}^{1}t^{x/2}(1-t)^{-x/2}dt=\Gamma\left(\dfrac{2+x}{2}\right)\Gamma\left(\dfrac{2-x}{2}\right)$ but couldn't solve

also tried duplication formula .


I think we can slightly improve the upper bound.

Two known identities for the Gamma function are $$\Gamma(1+z)=z\Gamma(z)\,\,\text{ and }\,\,\Gamma(z)\Gamma(1-z)=\pi\csc(\pi z)$$ Combining, we get $$\Gamma(1+z)\Gamma(1-z)=\pi z\csc(\pi z)$$ Therefore $$\Gamma\left(\frac{2+x}{2}\right)\Gamma\left(\frac{2-x}{2}\right)=\frac\pi 2 x \csc\left(\frac \pi 2 x\right)$$ This implies that $$I=\int_0^1x \frac{\frac{\pi x} 2}{\sin\left(\frac{\pi x}2\right)}dx$$ Now, you can verify that for all $u\geq 0$, we have $$\sin(u)\geq u - \frac {u^3}6$$ Therefore, $$I\leq\int_0^1\frac{x}{1-\frac{\pi^2 x}{24}} dx$$ Now, for $a>1$, $$\int_0^1\frac{x}{1-\frac {x^2}{a^2}}dx = \int_0^1\frac{a}{2}\left(\frac 1 {1-\frac x a}-\frac 1 {1+\frac x a}\right)dx=-\frac{a^2}2\ln\left(1-\frac 1 {a^2}\right)$$ Thus, with $a=\frac{2\sqrt{6}}{\pi}$ $$I\leq-\frac{12}{\pi^2}\ln\left(1-\frac{\pi^2}{24}\right)\simeq 0.644...\leq \dfrac{\sqrt{6\ln2}}{3}\simeq 0.679...$$


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