Question Number 100450 by 175 last updated on 26/Jun/20
Answered by maths mind last updated on 26/Jun/20
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{x}\right)^{{a}} }{\sqrt{{x}}.\sqrt{\mathrm{1}−{x}}.\sqrt{\mathrm{1}+{x}}}{dx}={f}\left({a}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{{a}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{a}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} .\left(\mathrm{1}−\left(−\mathrm{1}\right){x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx}={f}\left({a}\right) \\ $$$${f}\left({a}\right)=\beta\left(\frac{\mathrm{1}}{\mathrm{2}},{a}+\frac{\mathrm{1}}{\mathrm{2}}\right)._{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};{a}+\mathrm{1};−\mathrm{1}\right) \\ $$$${we}\:{can}\:{see}\:{f}'\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{x}\right)^{{a}} }{\sqrt{{x}.}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${we}\:{want}\:{f}'\left(\mathrm{0}\right) \\ $$$$\partial_{{a}} \beta\left(\frac{\mathrm{1}}{\mathrm{2}},{a}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\beta\left(\frac{\mathrm{1}}{\mathrm{2}},{a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\Psi\left({a}+\mathrm{1}\right)\right) \\ $$$${too}\:{bee}\:{continued} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$