Question Number 219866 by Nicholas666 last updated on 02/May/25
$$\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\:\underset{\:\mathrm{0}} {\int}^{\:\pi/\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}{x}−\mathrm{1}} \theta\:\mathrm{cos}\:^{\mathrm{2}{y}−\mathrm{1}} \theta\:{d}\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\Gamma\left({x}\right)\Gamma\left({y}\right)}{\Gamma\left({x}\right)+\Gamma\left({y}\right)}\:\:\:\:\: \\ $$$$\: \\ $$
Answered by MrGaster last updated on 03/May/25
$$\:\:\underset{\:\mathrm{0}} {\int}^{\:\pi/\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}{x}−\mathrm{1}} \theta\:\mathrm{cos}\:^{\mathrm{2}{y}−\mathrm{1}} \theta\:{d}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}^{\mathrm{2}{x}−\mathrm{1}} \left(\frac{\phi}{\mathrm{2}}\right)\mathrm{cos}^{\mathrm{2}{y}−\mathrm{1}} \left(\frac{\phi}{\mathrm{2}}\right)\frac{{d}\phi}{\mathrm{2}}\:\left(\mathrm{Let}\:\theta=\phi/\mathrm{2}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{B}\left({x},{y}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\Gamma\left({x}\right)\Gamma\left({y}\right)}{\Gamma\left({x}+{y}\right)}\:\left(\because\:\frac{\Gamma\left({x}\right)\Gamma\left({y}\right)}{\cancel{\Gamma\left({x}\right)+\Gamma\left({y}\right)}\:\left(\mathrm{false}\right)}\right)\:\:\:\: \\ $$
Commented by Nicholas666 last updated on 03/May/25
$${yes} \\ $$