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Question Number 189949 by Universmathematiques last updated on 24/Mar/23

Answered by witcher3 last updated on 25/Mar/23

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\sqrt[2]{\mathrm{x}\sqrt[3]{\mathrm{x}\sqrt[4]{}}...}$$$$\ln \left(\mathrm{f}\left(\mathrm{x}\right)\right)=\ln \left(\mathrm{x}\right)+\frac{1}{2}\ln \left(\mathrm{x}\right)+\frac{1}{6}\ln \left(\mathrm{x}\right).....$$$$\ln \left(\mathrm{f}\left(\mathrm{x}\right)\right)=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{\ln \left(\mathrm{x}\right)}{\mathrm{k}!}=\ln \left(\prod _{\mathrm{k}⩾1}{\mathrm{x}}^{\frac{1}{\mathrm{k}!}}\right)=\ln \left({\mathrm{x}}^{\sum _{\mathrm{m}⩾1}\frac{1}{\mathrm{m}!}}\right)$$$$=\ln \left({\mathrm{x}}^{\mathrm{e}-1}\right)$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{e}-1}$$$${\int }_0^1\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^1{\mathrm{x}}^{\mathrm{e}-1}\mathrm{dx}=\frac{1}{\mathrm{e}}{\left[{\mathrm{x}}^{\mathrm{e}}\right]}_0^1=\frac{1}{\mathrm{e}}$$

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