Question-195931

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Question Number 195931 by mr W last updated on 13/Aug/23

Commented by mr W last updated on 13/Aug/23

Commented by mr W last updated on 13/Aug/23

$${i}\:{think}\:{this}\:{is}\:{true}. \\ $$$${does}\:{somebody}\:{know}\:{a}\:{proof}\:{of}\:{it}? \\ $$

Answered by witcher3 last updated on 13/Aug/23

∫_0 ^1 x^(−x) dx=∫_0 ^1 e^(−xln(x)) dx =∫_0 ^1 Σ_(n≥0) (((−xln(x))^n )/(n!))dx S=Σ_(n≥0) (1/(n!))∫_0 ^1 (−lne(x))^n x^n dx ∣xln(x)∣≤e^(−1) ⇒Σ(((−xln(x))^n )/(n!))?cv uniformly This show whay we change ∫ and Σ ∫_0 ^1 (−ln(x))^n x^n dx;−ln(x)=y =∫_0 ^∞ y^n e^(−ny) e^(−y) dy=∫_0 ^∞ y^n e^(−(1+n)y) dy =∫_0 ^∞ ((y^n e^(−y) dy)/((n+1)^(n+1) ))=((Γ(n+1))/((n+1)^(n+1) )) S=Σ_(n≥0) (1/(n!))∫_0 ^1 (−xln(x))^n dx=Σ_(n≥0) (1/(n!)).((Γ(n+1))/((n+1)^(n+1) )) =Σ_(n≥0) (1/((n+1)^(n+1) ))=Σ_(n≥1) (1/n^n ) ∫_0 ^1 (dx/x^x )=Σ_(n≥1) (1/n^n )

$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{−\mathrm{x}} \mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\mathrm{xln}\left(\mathrm{x}\right)} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{xln}\left(\mathrm{x}\right)\right)^{\mathrm{n}} }{\mathrm{n}!}\mathrm{dx} \\ $$$$\mathrm{S}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{n}!}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{lne}\left(\mathrm{x}\right)\right)^{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \mathrm{dx} \\ $$$$\mid\mathrm{xln}\left(\mathrm{x}\right)\mid\leqslant\mathrm{e}^{−\mathrm{1}} \Rightarrow\Sigma\frac{\left(−\mathrm{xln}\left(\mathrm{x}\right)\right)^{\mathrm{n}} }{\mathrm{n}!}?\mathrm{cv}\:\mathrm{uniformly} \\ $$$$\mathrm{This}\:\mathrm{show}\:\mathrm{whay}\:\mathrm{we}\:\mathrm{change}\:\int\:\mathrm{and}\:\Sigma \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{ln}\left(\mathrm{x}\right)\right)^{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \mathrm{dx};−\mathrm{ln}\left(\mathrm{x}\right)=\mathrm{y} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \mathrm{y}^{\mathrm{n}} \mathrm{e}^{−\mathrm{ny}} \mathrm{e}^{−\mathrm{y}} \mathrm{dy}=\int_{\mathrm{0}} ^{\infty} \mathrm{y}^{\mathrm{n}} \mathrm{e}^{−\left(\mathrm{1}+\mathrm{n}\right)\mathrm{y}} \mathrm{dy} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{y}^{\mathrm{n}} \mathrm{e}^{−\mathrm{y}} \mathrm{dy}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }=\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} } \\ $$$$\mathrm{S}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{n}!}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{xln}\left(\mathrm{x}\right)\right)^{\mathrm{n}} \mathrm{dx}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{n}!}.\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} } \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{n}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{x}} }=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{n}} } \\ $$

Commented by mr W last updated on 13/Aug/23

$${thanks}\:{sir}! \\ $$

Commented by witcher3 last updated on 13/Aug/23

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcom} \\ $$

Commented by mr W last updated on 14/Aug/23

is there a special name or notation for the value of Σ_(n=1) ^∞ (1/n^n )?

$${is}\:{there}\:{a}\:{special}\:{name}\:{or}\:{notation} \\ $$$${for}\:{the}\:{value}\:{of}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{n}} }? \\ $$

Commented by witcher3 last updated on 14/Aug/23

i serched this one for long i have nothing sorry

$$\mathrm{i}\:\mathrm{serched}\:\mathrm{this}\:\mathrm{one}\:\mathrm{for}\:\mathrm{long}\:\mathrm{i}\:\mathrm{have}\:\mathrm{nothing}\:\mathrm{sorry} \\ $$

Commented by mr W last updated on 14/Aug/23

$${thanks}\:{alot}\:{anyway}! \\ $$

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