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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)) =?

$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2ln}\:\left[\Gamma\left(\mathrm{sin}\:{x}\right)\right]−\mathrm{ln}\:\pi}{\Gamma\left(\mathrm{sec}\:{x}\right)−\mathrm{1}}\:=? \\ $$

Answered by john_santu last updated on 21/Sep/21

 lim_(x→(π/6))  ((2cos x Ψ^0 (sin (x)))/(2tan (2x)sec (2x)Γ(sec (2x))Ψ^0 (sec (2x))))=  ((2.((√3)/2).(−γ−ln 4))/(2.(√3).2.1.(1−γ))) = ((γ+ln 4)/(4(γ−1)))  γ = Mascherani constant

$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:{x}\:\Psi^{\mathrm{0}} \left(\mathrm{sin}\:\left({x}\right)\right)}{\mathrm{2tan}\:\left(\mathrm{2}{x}\right)\mathrm{sec}\:\left(\mathrm{2}{x}\right)\Gamma\left(\mathrm{sec}\:\left(\mathrm{2}{x}\right)\right)\Psi^{\mathrm{0}} \left(\mathrm{sec}\:\left(\mathrm{2}{x}\right)\right)}= \\ $$$$\frac{\mathrm{2}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\left(−\gamma−\mathrm{ln}\:\mathrm{4}\right)}{\mathrm{2}.\sqrt{\mathrm{3}}.\mathrm{2}.\mathrm{1}.\left(\mathrm{1}−\gamma\right)}\:=\:\frac{\gamma+\mathrm{ln}\:\mathrm{4}}{\mathrm{4}\left(\gamma−\mathrm{1}\right)} \\ $$$$\gamma\:=\:{Mascherani}\:{constant} \\ $$

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