Question Number 142553 by 777316 last updated on 02/Jun/21
$$ \\ $$$$\:\:\:\:{Evaluate}\::\: \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{log}\left({x}\right){log}\left(\frac{{x}}{\mathrm{1}−{x}}\right)}{\:\sqrt{\frac{{x}}{\mathrm{1}−{x}}}}\:{dx} \\ $$
Answered by Dwaipayan Shikari last updated on 02/Jun/21
$$\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left({x}\right){x}^{−\mathrm{1}/\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{1}/\mathrm{2}} −{log}\left({x}\right){log}\left(\mathrm{1}−{x}\right){x}^{−\mathrm{1}/\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{1}/\mathrm{2}} {dx} \\ $$$$=\frac{\partial^{\mathrm{2}} }{\partial{a}^{\mathrm{2}} }\mid_{{a}=\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\Gamma\left({a}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left({a}+\frac{\mathrm{3}}{\mathrm{2}}\right)}−\frac{\partial^{\mathrm{2}} }{\partial{a}\partial{b}}.\mid_{{a}=\frac{\mathrm{1}}{\mathrm{2}}\:{b}=\frac{\mathrm{3}}{\mathrm{2}}} \frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)} \\ $$
Answered by mnjuly1970 last updated on 02/Jun/21
$$\:\:\:{solution}\left({gercekboss}\right) \\ $$$$\:\:\:\Theta\left({a},{b}\right):=\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{a}} \left(\frac{{x}}{\mathrm{1}−{x}}\right)^{{b}−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$\:\:\:\Theta:=\frac{\partial^{\mathrm{2}} \:\Theta\left({a},{b}\right)}{\partial{a}.\partial{b}}\mid_{{a}\:,{b}=\mathrm{0}} \\ $$$$\:\:\:\: \\ $$