Question Number 189949 by Universmathematiques last updated on 24/Mar/23
Answered by witcher3 last updated on 25/Mar/23
![f(x)=x((x((x()^(1/4) ))^(1/3) ...))^(1/2) ln(f(x))=ln(x)+(1/2)ln(x)+(1/6)ln(x)..... ln(f(x))=Σ_(k=1) ^∞ ((ln(x))/(k!))=ln(Π_(k≥1) x^(1/(k!)) )=ln(x^(Σ_(m≥1) (1/(m!))) ) =ln(x^(e−1) ) f(x)=x^(e−1) ∫_0 ^1 f(x)dx=∫_0 ^1 x^(e−1) dx=(1/e)[x^e ]_0 ^1 =(1/e)](https://www.tinkutara.com/question/Q189963.png)
$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\sqrt[{\mathrm{2}}]{\mathrm{x}\sqrt[{\mathrm{3}}]{\mathrm{x}\sqrt[{\mathrm{4}}]{}}…} \\ $$$$\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{ln}\left(\mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\left(\mathrm{x}\right)….. \\ $$$$\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{k}!}=\mathrm{ln}\left(\underset{\mathrm{k}\geqslant\mathrm{1}} {\prod}\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{k}!}} \right)=\mathrm{ln}\left(\mathrm{x}^{\underset{\mathrm{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{m}!}} \right) \\ $$$$=\mathrm{ln}\left(\mathrm{x}^{\mathrm{e}−\mathrm{1}} \right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{e}−\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{e}−\mathrm{1}} \mathrm{dx}=\frac{\mathrm{1}}{\mathrm{e}}\left[\mathrm{x}^{\mathrm{e}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{e}} \\ $$