Gamma function and sine

Question

I need to prove that the function $$c(\alpha)=\sin\left(\frac{\pi \alpha}{2}\right)\cdot \Gamma(\alpha)$$ is decreasing on $(1,2]$. It is evident if you build the graph of it. E.g. in WolframAlpha you can write "plot sin(pi a/2)*Gamma(a) for a from 1 to 2": Click here to see the graph in WolframAlpha But what if I want to use this in a proof of some theorem?

Answer 1

To prove that $$ c(\alpha) = \sin(\dfrac{\pi\alpha}{2})\cdot \Gamma(\alpha) $$ is decreasing on $(1,2]$ it is sufficient to prove that $$ c'(\alpha) =\dfrac{\pi}{2}\cos(\dfrac{\pi\alpha}{2})\Gamma(\alpha) + \sin(\dfrac{\pi\alpha}{2}) \Gamma'(\alpha) = \cos(\dfrac{\pi\alpha}{2})\Gamma(\alpha)\left(\dfrac{\pi}{2} + \tan(\dfrac{\pi\alpha}{2}) \dfrac{\Gamma'(\alpha)}{\Gamma(\alpha)}\right) < 0 $$ on $(1,2]$. Since $\cos(\dfrac{\pi\alpha}{2})< 0$ and $\Gamma(\alpha)>0$ we try to prove $$ f(\alpha) = \dfrac{\pi}{2} + \tan(\dfrac{\pi\alpha}{2})\cdot \dfrac{\Gamma'(\alpha)}{\Gamma(\alpha)} > 0 $$ on $(1,2]$. The digamma function is defined by $ \psi(\alpha) = \dfrac{\Gamma'(\alpha)}{\Gamma(\alpha)}. $ Since $$ \psi(\alpha) = -\dfrac{1}{\alpha}-\gamma + \sum_{n=1}^{\infty}\left(\dfrac{1}{n} - \dfrac{1}{\alpha+n}\right), $$ where $\gamma$ is Euler's constant, we get $$ \psi'(\alpha) = \sum_{n=0}^{\infty}\dfrac{1}{(\alpha+n)^{2}} > 0. $$ Consequently $\psi$ is strictly increasing. Since $\psi(1) = -\gamma < 0$ (telescoping sum) and $\psi(2) = 1 - \gamma > 0$ there is one and only one point $\alpha_{0}$ in $[1,2]$ such that $\psi(\alpha_{0}) = 0$.

But $\tan(\dfrac{\pi\alpha}{2}) <0$ and $\psi(\alpha) < 0$ on $(1,\alpha_{0}]$. Consequently $f(\alpha) > 0$ on in $(1,\alpha_{0}]$.

According to Gauss' digamma theorem $$ \psi(\frac{4}{3}) = 3 - \gamma - \dfrac{\pi\sqrt{3}}{6} - \dfrac{3}{2}\ln 3 \approx -0.132033782 < 0. $$ Thus $\frac{4}{3} < \alpha_{0}$. Consequently $$-\sqrt{3} = \tan(\dfrac{\pi}{2}\dfrac{4}{3}) < \tan(\dfrac{\pi}{2}\alpha) $$ on $[\alpha_{0},2]$. Finally we conclude $$ f(\alpha) > \dfrac{\pi}{2}+ \tan(\dfrac{\pi}{2}\dfrac{4}{3})\psi(2) = \dfrac{\pi}{2}-\sqrt{3}(1-\gamma) \approx 0.8385123778 >0 $$ if $\alpha \in [\alpha_{0},2]$