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0-1-x-x-x-x-x-1-5-1-4-1-3-dx-




Question Number 216486 by Tawa11 last updated on 08/Feb/25
∫_( 0) ^( 1)  x(√(x  ((x  ((x  ((x ...))^(1/5) ))^(1/4) ))^(1/3) ))  dx
$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{x}\sqrt{\mathrm{x}\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:\sqrt[{\mathrm{4}}]{\mathrm{x}\:\:\sqrt[{\mathrm{5}}]{\mathrm{x}\:…}}}}\:\:\mathrm{dx} \\ $$
Answered by mehdee7396 last updated on 09/Feb/25
x×x^(1/2) ×x^(1/6) ×x^(1/(24)) ×x^(1/(120)) ×...  =x^(Σ_1 ^∞ (1/(n!)).) =x^(e−1)   ⇒I=∫_0 ^1 x^(e−1) dx=(1/(e−1))(x^(e−1) )]_0 ^1   =(1/(e−1))
$${x}×{x}^{\frac{\mathrm{1}}{\mathrm{2}}} ×{x}^{\frac{\mathrm{1}}{\mathrm{6}}} ×{x}^{\frac{\mathrm{1}}{\mathrm{24}}} ×{x}^{\frac{\mathrm{1}}{\mathrm{120}}} ×… \\ $$$$={x}^{\underset{\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}.} ={x}^{{e}−\mathrm{1}} \\ $$$$\left.\Rightarrow{I}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{e}−\mathrm{1}} {dx}=\frac{\mathrm{1}}{{e}−\mathrm{1}}\left({x}^{{e}−\mathrm{1}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{e}−\mathrm{1}}\: \\ $$$$ \\ $$
Commented by Tawa11 last updated on 09/Feb/25
Thanks sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

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