Question Number 219554 by Nicholas666 last updated on 28/Apr/25
$$ \\ $$$$\:{I}_{{n}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} …\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left({x}_{\mathrm{1}} {x}_{\mathrm{2}} …{x}_{{n}} \right)^{{a}} }{\left(\mathrm{1}−{x}_{\mathrm{1}} {x}_{\mathrm{2}} …{x}_{{n}} \right)}{ln}\left({x}_{\mathrm{1}\:} \right){ln}\left({x}_{\mathrm{2}} \right)…{ln}\left({x}_{{n}\:} \right){dx}_{\mathrm{1}} {dx}_{\mathrm{2}} …{dx}_{{n}\:} \:\:\:\:\: \\ $$$$ \\ $$
Answered by maths2 last updated on 29/Apr/25
$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} …\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}_{\mathrm{1}} ….{x}_{{n}} \right)^{{a}+{k}} {ln}\left({x}_{\mathrm{1}} \right)….{ln}\left({x}_{{n}} \right){dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({a}+{k}+\mathrm{1}\right)^{\mathrm{2}{n}} }=\left(−\mathrm{1}\right)^{{n}} \underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{1}+{a}+{k}\right)^{\mathrm{2}{n}} }=\left(−\mathrm{1}\right)^{{n}} \zeta\left({a}+\mathrm{1},\mathrm{2}{n}\right) \\ $$$$\zeta\left({x},{n}\right)=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({x}+{m}\right)^{{n}} }\:\:{zeta}\:{Hurwitz}\:{function} \\ $$