Question Number 161407 by mathlove last updated on 17/Dec/21
Answered by puissant last updated on 18/Dec/21
![Ω=∫_0 ^1 (√(x((x((x((...))^(1/5) ))^(1/4) ))^(1/3) )) dx = ∫_0 ^1 x^((1/2)+(1/6)+(1/(24))...) dx = ∫_0 ^1 x^(Σ_(n=2) ^∞ (1/(n!))) dx = ∫_0 ^1 x^(e−1) dx = (1/e)[x^e ]_0 ^1 =(1/e).. ..........Le puissant...........](https://www.tinkutara.com/question/Q161458.png)
$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}\sqrt[{\mathrm{3}}]{{x}\sqrt[{\mathrm{4}}]{{x}\sqrt[{\mathrm{5}}]{…}}}}\:{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{24}}…} {dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}} {dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{e}−\mathrm{1}} {dx}\:=\:\frac{\mathrm{1}}{{e}}\left[{x}^{{e}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{{e}}.. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:……….\mathscr{L}{e}\:{puissant}……….. \\ $$
Commented by Tawa11 last updated on 18/Dec/21
$$\mathrm{Great}\:\mathrm{sir} \\ $$