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Question Number 142690 by Rankut last updated on 04/Jun/21

$$\int_{\mathrm{0}\:} ^{\mathrm{1}} \frac{\boldsymbol{{log}}\left(\boldsymbol{{x}}\right)\boldsymbol{{log}}\left(\frac{\boldsymbol{{x}}}{\mathrm{1}−\boldsymbol{{x}}}\right)}{\:\sqrt{\frac{\boldsymbol{{x}}}{\mathrm{1}−\boldsymbol{{x}}}}}\boldsymbol{{dx}} \\$$$$\boldsymbol{\mathrm{Any}}\:\boldsymbol{\mathrm{help}}\: \\$$$$\\$$

Answered by Dwaipayan Shikari last updated on 04/Jun/21

$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}−\mathrm{1}} \left(\frac{{x}}{\mathrm{1}−{x}}\right)^{{b}−\mathrm{1}} {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}+{b}−\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{1}−{b}} {dx}=\frac{\Gamma\left({a}+{b}−\mathrm{1}\right)\Gamma\left(\mathrm{2}−{b}\right)}{\Gamma\left({a}+\mathrm{1}\right)} \\$$$${Here}\:{a}=\mathrm{1}\:\:\: \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left({x}\right){x}^{{a}−\mathrm{1}} \left(\frac{{x}}{\mathrm{1}−{x}}\right)^{{b}−\mathrm{1}} {dx}=\frac{\partial}{\partial{a}}.\frac{\Gamma\left({a}+{b}−\mathrm{1}\right)}{\Gamma\left({a}+\mathrm{1}\right)}\Gamma\left(\mathrm{2}−{b}\right) \\$$$$=\frac{\Gamma\left({a}+\mathrm{1}\right)\Gamma'\left({a}+{b}−\mathrm{1}\right)−\Gamma'\left({a}+\mathrm{1}\right)\Gamma\left({a}+{b}−\mathrm{1}\right)}{\Gamma^{\mathrm{2}} \left({a}+\mathrm{1}\right)}\Gamma\left(\mathrm{2}−{b}\right) \\$$$$=\Gamma'\left({b}\right)\Gamma\left(\mathrm{2}−{b}\right)−\psi\left(\mathrm{2}\right)\Gamma\left({b}\right)\Gamma\left(\mathrm{2}−{b}\right) \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left({x}\right){log}\left(\frac{{x}}{\mathrm{1}−{x}}\right)\left(\frac{{x}}{\mathrm{1}−{x}}\right)^{{b}−\mathrm{1}} {dx} \\$$$$=\Gamma''\left({b}\right)\Gamma\left(\mathrm{2}−{b}\right)−\Gamma'\left(\mathrm{2}−{b}\right)\Gamma'\left({b}\right)−\psi\left(\mathrm{2}\right)\Gamma'\left({b}\right)\Gamma\left(\mathrm{2}−{b}\right)+\psi'\left(\mathrm{2}\right)\Gamma'\left(\mathrm{2}−{b}\right)\Gamma\left({b}\right) \\$$$${Here} \\$$$${put}\:{b}=\frac{\mathrm{1}}{\mathrm{2}\:} \\$$$$\\$$