Question Number 154719 by liberty last updated on 21/Sep/21
![lim_(x→(π/6)) ((2ln [Γ(sin x)]−ln π)/(Γ(sec x)−1)) =?](Q154719.png)
$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2ln}\:\left[\Gamma\left(\mathrm{sin}\:{x}\right)\right]−\mathrm{ln}\:\pi}{\Gamma\left(\mathrm{sec}\:{x}\right)−\mathrm{1}}\:=? \\ $$
Answered by john_santu last updated on 21/Sep/21
$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:{x}\:\Psi^{\mathrm{0}} \left(\mathrm{sin}\:\left({x}\right)\right)}{\mathrm{2tan}\:\left(\mathrm{2}{x}\right)\mathrm{sec}\:\left(\mathrm{2}{x}\right)\Gamma\left(\mathrm{sec}\:\left(\mathrm{2}{x}\right)\right)\Psi^{\mathrm{0}} \left(\mathrm{sec}\:\left(\mathrm{2}{x}\right)\right)}= \\ $$$$\frac{\mathrm{2}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\left(−\gamma−\mathrm{ln}\:\mathrm{4}\right)}{\mathrm{2}.\sqrt{\mathrm{3}}.\mathrm{2}.\mathrm{1}.\left(\mathrm{1}−\gamma\right)}\:=\:\frac{\gamma+\mathrm{ln}\:\mathrm{4}}{\mathrm{4}\left(\gamma−\mathrm{1}\right)} \\ $$$$\gamma\:=\:{Mascherani}\:{constant} \\ $$