Gamma multiplied by Gamma

Relevant wiki: Gamma Function

$\begin{aligned} \displaystyle I &= \int_0^\infty \sqrt[3] {\dfrac{8 x^{11}}{3^{18} e^{{x^4}/{3^5}}}} \, dx \\ &= \displaystyle \int_0^\infty \dfrac{2 x^{{11}/{3}}}{3^6 e^{{x^4}/{3^6}}} \, dx \\ &= \displaystyle \dfrac 32 \int_0^\infty t^{{1}/{6}} e^{-t} \, dt & \small {\color{#3D99F6} \dfrac{x^4}{3^6} = t \implies \dfrac{4x^3}{3^6} dx = dt \text{ and } \displaystyle \int_0^\infty \mapsto \int_0^\infty} \\ &= \displaystyle \dfrac 32 \Gamma \left( \dfrac 76 \right) \\ &= \displaystyle \dfrac 14 \Gamma \left( \dfrac 16 \right) & \small {\color{#3D99F6} \Gamma (s+1) = s \Gamma (s) } \end{aligned}$

Thus $a=16$ and $b=6$ .

Now

$\begin{aligned} \displaystyle \sin \left( \dfrac{2 \Gamma \left( \dfrac 36 \right) \Gamma \left( \dfrac 96 \right)}{4} \right) &= \displaystyle \sin \left( \dfrac{ \Gamma \left( \dfrac 12 \right) \cdot \dfrac 12 \cdot \Gamma \left( \dfrac 12 \right)}{2} \right) \\ &= \displaystyle \sin \left( \dfrac{\pi}{4} \right) & \small {\color{#3D99F6} \Gamma \left( \dfrac 12 \right) = \sqrt{\pi} } \\ &= \displaystyle \dfrac{1}{\sqrt 2} \approx \boxed{0.71} \end{aligned}$