Question Number 137717 by mnjuly1970 last updated on 05/Apr/21
Commented by Dwaipayan Shikari last updated on 05/Apr/21
$${f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}−{a}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−{a}}{\mathrm{2}}\right)\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}−{a}}{\mathrm{2}}\right)=\frac{\mathrm{1}−{a}}{\mathrm{4}}.\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}{a}\right)} \\ $$$$=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}−{a}\right){sec}\left(\frac{\pi}{\mathrm{2}}{a}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}{f}'\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{12}} \\ $$