Question Number 219515 by hardmath last updated on 27/Apr/25
$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{x}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\mathrm{dx}\:=\:?\: \\ $$
Answered by SdC355 last updated on 27/Apr/25
$$\mathrm{can}'\mathrm{t}\:\mathrm{Find}\:\mathrm{primitive}\:\mathrm{function}\:\int\:\centerdot\: \\ $$$$\mathrm{but}\:\mathrm{approximatly}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\centerdot\approx\mathrm{0}.\mathrm{8964887819296233}….\: \\ $$
Answered by vnm last updated on 28/Apr/25
![Ξ©=β«_0 ^1 x^x^2 dx=β«_0 ^1 e^(x^2 ln x) dx= β«_0 ^1 Ξ£_(n=0) ^β (((x^2 lnx)^n )/(n!))=Ξ£_(n=0) ^β β«_0 ^(1() ((x^2 lnx)^n )/(n!))dx= Ξ£_(n=0) ^β (1/(n!))β«_0 ^1 x^(2n) (lnx)^n dx=Ξ£_(n=0) ^β (1/(n!))I_n I_n =β«_0 ^1 x^(2n) (lnx)^n dx= [lnx=βat, a>0, x=e^(βat) , dx=βae^(βat) dt, t_1 =+β, t_2 =0] β«_(+β) ^0 (e^(βat) )^(2n) (βat)^n (βae^(βat) )dt= (β1)^(n+1) a^(n+1) β«_(+β) ^0 e^(β(2n+1)at) t^n dt= [(2n+1)a=1, a=(1/(2n+1))] (((β1)^(n+1) )/((2n+1)^(n+1) ))β«_(+β) ^0 e^(βt) t^n dt=(((β1)^n )/((2n+1)^(n+1) ))β«_0 ^(+β) e^(βt) t^n dt= (((β1)^n )/((2n+1)^(n+1) ))βn! Ξ©=Ξ£_(n=0) ^β (1/(n!))β(((β1)^n n!)/((2n+1)^(n+1) ))=Ξ£_(n=1) ^β (((β1)^(nβ1) )/((2nβ1)^n ))β0.896488781930](https://www.tinkutara.com/question/Q219559.png)
$$ \\ $$$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{x}^{\mathrm{2}} } {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}^{\mathrm{2}} \mathrm{ln}\:{x}} {dx}= \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({x}^{\mathrm{2}} \mathrm{ln}{x}\right)^{{n}} }{{n}!}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}\left(\right.} \frac{\left.{x}^{\mathrm{2}} \mathrm{ln}{x}\right)^{{n}} }{{n}!}{dx}= \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}} \left(\mathrm{ln}{x}\right)^{{n}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}{I}_{{n}} \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}} \left(\mathrm{ln}{x}\right)^{{n}} {dx}= \\ $$$$\left[\mathrm{ln}{x}=β{at},\:{a}>\mathrm{0},\:{x}={e}^{β{at}} ,\:{dx}=β{ae}^{β{at}} {dt},\:{t}_{\mathrm{1}} =+\infty,\:{t}_{\mathrm{2}} =\mathrm{0}\right] \\ $$$$\int_{+\infty} ^{\mathrm{0}} \left({e}^{β{at}} \right)^{\mathrm{2}{n}} \left(β{at}\right)^{{n}} \left(β{ae}^{β{at}} \right){dt}= \\ $$$$\left(β\mathrm{1}\right)^{{n}+\mathrm{1}} {a}^{{n}+\mathrm{1}} \int_{+\infty} ^{\mathrm{0}} {e}^{β\left(\mathrm{2}{n}+\mathrm{1}\right){at}} {t}^{{n}} {dt}= \\ $$$$\left[\left(\mathrm{2}{n}+\mathrm{1}\right){a}=\mathrm{1},\:{a}=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right] \\ $$$$\frac{\left(β\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\int_{+\infty} ^{\mathrm{0}} {e}^{β{t}} {t}^{{n}} {dt}=\frac{\left(β\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{+\infty} {e}^{β{t}} {t}^{{n}} {dt}= \\ $$$$\frac{\left(β\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\centerdot{n}! \\ $$$$\Omega=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\centerdot\frac{\left(β\mathrm{1}\right)^{{n}} {n}!}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(β\mathrm{1}\right)^{{n}β\mathrm{1}} }{\left(\mathrm{2}{n}β\mathrm{1}\right)^{{n}} }\approx\mathrm{0}.\mathrm{896488781930} \\ $$
Commented by hardmath last updated on 28/Apr/25
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{darling}\:\mathrm{professor}\:\mathrm{thankyou} \\ $$