Is it even related to the Gamma Function?

$\begin{aligned} f(x,s) & = \sum_{n=0}^\infty \frac {(-1)^nx^{n+s}}{n+s} \\ \frac \partial {\partial x} f(x) & = \sum_{n=0}^\infty (-1)^nx^{n+s-1} = x^{s-1} - \sum_{n=0}^\infty (-1)^nx^{n+s} = x^s \left(\frac 1x - \frac 1{1+x}\right) = \frac {x^s}{x(1+x)} \end{aligned}$

Therefore,

$\begin{aligned} y & = \int_0^\infty \frac \partial {\partial x} f(x) \ dx \\ & = \int_0^\infty \frac {x^{\color{#3D99F6}\frac 16}}{x(1+x)} \ dx & \small \color{#3D99F6} \text{for }s = \frac 16 \\ & = \int_0^\infty \frac {x^{{\color{#3D99F6}\frac 16}-1}}{(1+x)^{\color{#3D99F6}\frac 16 + \frac 56}} \ dx \\ & = B \left(\frac 16, \frac 56 \right) & \small \color{#3D99F6} B(m,n) \text{ is beta function.} \\ & = \frac {\Gamma \left(\frac 16 \right)\Gamma \left(\frac 56 \right)}{\Gamma \left(\frac 16 + \frac 56 \right)} & \small \color{#3D99F6} \Gamma (z) \text{ is gamma function.} \\ & = \frac {\Gamma \left(\frac 16 \right)\Gamma \left(1-\frac 16 \right)}{0!} \\ & = \frac {\pi}{\sin \frac \pi 6} = 2 \pi \end{aligned}$

$\implies \cos \left(\dfrac {3y}2 \right) = \cos 3\pi = \boxed{-1}$

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References:

• Beta function
• Gamma function