Question Number 172359 by mnjuly1970 last updated on 25/Jun/22
Answered by Mathspace last updated on 26/Jun/22
$${J}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {ln}^{\mathrm{2}} {xdx} \\ $$$${let}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$${we}\:{have}\:{f}^{''} \left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {ln}^{\mathrm{2}} {xdx} \\ $$$$\Rightarrow{f}^{''} \left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {ln}^{\mathrm{2}} {x}\:{dx} \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dx} \\ $$$$={B}\left({a}+\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({a}+\mathrm{1}\right)} \\ $$$$=\sqrt{\pi}×\frac{\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({a}+\mathrm{1}\right)} \\ $$$$\Rightarrow{f}^{'} \left({a}\right)=\sqrt{\pi}×\frac{\Gamma^{'} \left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({a}+\mathrm{1}\right)−\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma^{'} \left({a}+\mathrm{1}\right)}{\Gamma^{\mathrm{2}} \left({a}+\mathrm{1}\right)} \\ $$$${we}/{have}\:\psi\left({x}\right)=\frac{\Gamma^{'} \left({x}\right)}{\Gamma\left({x}\right)}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)=\sqrt{\pi}×\frac{\psi\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({a}+\mathrm{1}\right)−\Gamma\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\psi\left({a}+\mathrm{1}\right)\Gamma\left({a}+\mathrm{1}\right)}{\Gamma^{\mathrm{2}} \left({a}+\mathrm{1}\right)} \\ $$