Question Number 64528 by mathmax by abdo last updated on 19/Jul/19
$${find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{−{x}} {dx}\:\:\:{study}\:{first}\:{the}\:{convergence}. \\ $$
Commented by mathmax by abdo last updated on 19/Jul/19
![we have x^(−x) =e^(−xln(x)) let A =∫_0 ^1 x^(−x) dx ⇒A =∫_0 ^1 e^(−xlnx) dx we have lim_(x→0^+ ) xln(x) =0 ⇒lim_(x→0^ ) e^(−xlnx) =1 so the function x→e^(−xlnx) is integrable on [0,1] we have A =∫_0 ^1 Σ_(n=0) ^∞ (((−xlnx)^n )/(n!))dx =Σ_(n=0) ^∞ (((−1)^n )/(n!)) ∫_0 ^1 x^n (lnx)^n dx =Σ_(n=0) ^∞ (((−1)^n )/(n!))w_n w_n =∫_0 ^1 x^n (lnx)^n dx =_(lnx =−t) ∫_(+∞) ^0 e^(−nt) (−t)^n (−e^(−t) )dt = ∫_0 ^∞ (−1)^n t^n e^(−(n+1)t) dt =_((n+1)t =u) (−1)^n ∫_0 ^∞ (u^n /((n+1)^n )) e^(−u) (du/((n+1))) =(((−1)^n )/((n+1)^(n+1) )) ∫_0 ^∞ u^n e^(−u) du we know that Γ(x) =∫_0 ^∞ t^(x−1) e^(−t) dt ⇒ ∫_0 ^∞ u^n e^(−u) du =Γ(n+1)=n! ⇒w_n =(((−1)^n n!)/((n+1)^(n+1) )) ⇒ A =Σ_(n=0) ^∞ (((−1)^n )/(n!)) (((−1)^n n!)/((n+1)^(n+1) )) =Σ_(n=0) ^∞ (1/((n+1)^(n+1) )) =Σ_(n=1) ^∞ (1/n^n ) =1 +(1/2^2 ) +(1/3^3 ) +(1/4^4 ) +....](https://www.tinkutara.com/question/Q64565.png)
$${we}\:{have}\:\:{x}^{−{x}} \:={e}^{−{xln}\left({x}\right)} \:{let}\:\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{−{x}} {dx}\:\Rightarrow{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{xlnx}} {dx} \\ $$$${we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{xln}\left({x}\right)\:=\mathrm{0}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}^{} } \:\:{e}^{−{xlnx}} \:=\mathrm{1}\:\:{so} \\ $$$${the}\:{function}\:{x}\rightarrow{e}^{−{xlnx}} \:{is}\:{integrable}\:{on}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$${we}\:{have}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−{xlnx}\right)^{{n}} }{{n}!}{dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \left({lnx}\right)^{{n}} \:{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}{w}_{{n}} \\ $$$${w}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \left({lnx}\right)^{{n}} {dx}\:=_{{lnx}\:=−{t}} \:\:\:\int_{+\infty} ^{\mathrm{0}} \:{e}^{−{nt}} \:\left(−{t}\right)^{{n}} \:\left(−{e}^{−{t}} \right){dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}} \:{t}^{{n}} \:{e}^{−\left({n}+\mathrm{1}\right){t}} \:{dt}\:=_{\left({n}+\mathrm{1}\right){t}\:={u}} \:\:\:\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}} }\:{e}^{−{u}} \:\frac{{du}}{\left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\int_{\mathrm{0}} ^{\infty} \:\:{u}^{{n}} \:{e}^{−{u}} \:{du}\:\:\:{we}\:{know}\:{that}\:\Gamma\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{u}^{{n}} \:{e}^{−{u}} \:{du}\:=\Gamma\left({n}+\mathrm{1}\right)={n}!\:\Rightarrow{w}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\Rightarrow \\ $$$${A}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{{n}} } \\ $$$$=\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{4}} }\:+…. \\ $$$$ \\ $$