Question Number 135382 by Bird last updated on 12/Mar/21
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {ln}\left(\mathrm{1}−{x}^{\mathrm{4}} \right){dx}\:{with}\:{n} \\ $$$${integr}\:{natural} \\ $$
Answered by Dwaipayan Shikari last updated on 12/Mar/21
$${x}^{\mathrm{4}} ={t} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{{n}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}} {log}\left(\mathrm{1}−{t}\right){dt} \\ $$$${I}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{a}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{b}−\mathrm{1}} {dt}=\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)} \\ $$$$\frac{\partial}{\partial{b}}{I}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{a}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{b}−\mathrm{1}} {log}\left(\mathrm{1}−{t}\right)=\frac{\left.\Gamma\left({a}+{b}\right)\Gamma'\left({b}\right)\Gamma\left({a}\right)−\Gamma'\left({a}+{b}\right)\Gamma\left({a}\right)\Gamma\left({b}\right)\right)}{\Gamma^{\mathrm{2}} \left({a}+{b}\right)} \\ $$$$=\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)}\left(\psi\left({b}\right)−\psi\left({a}+{b}\right)\right) \\ $$$${b}=\mathrm{1} \\ $$$${a}=\frac{{n}+\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{I}\left(\frac{{n}+\mathrm{1}}{\mathrm{4}},\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{{n}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}} {log}\left(\mathrm{1}−{t}\right){dt}=\frac{\Gamma\left(\frac{{n}+\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}\right)}\left(−\gamma−\psi\left(\frac{{n}+\mathrm{5}}{\mathrm{4}}\right)\right) \\ $$$$=−\frac{\mathrm{4}}{\mathrm{4}{n}+\mathrm{1}}\left(\gamma+\psi\left(\frac{{n}+\mathrm{5}}{\mathrm{4}}\right)\right) \\ $$
Commented by mathmax by abdo last updated on 12/Mar/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$