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Question-143046

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Question Number 143046 by 0731619 last updated on 09/Jun/21
Answered by mindispower last updated on 09/Jun/21
x^x^x =f(x)⇒ln(fx)=f(x)ln(x) ln(f(x))e^(−ln(f(x))) =ln(x) ⇒−ln(f(x))=W(−ln(x)) f(x)=e^(−W(−ln(x))) =((W(−ln(x)))/(−ln(x))) ∫(e^(−2W(−ln(x))) /(x(1−ln(x)e^(−W(−ln(x))) ))) ln(x)=y =∫(e^(−2W(−y)) /((1−ye^(−W(−y)) )))dy ∫(e^(−W(−y)) /(e^(W(−y)) −y))dy W(y)e^(W(y)) =y ⇒W′(y)(1+W(y))e^(W(y)) =1 W′(y)=(e^(−W(y)) /(1+W(y))) e^(W(−y)) −y=f(y),f′(y) let g(y)=e^(W(−y)) g′(y)=−W′(−y)e^(W(−y)) =((−1)/(1+W(−y))) =((−e^(W(−y)) )/(e^(W(−y)) −y))⇒dg(y)e^(W(−y)) =(dy/(e^(W(−y)) −y)) ∫(1/g^2 )dg=−(1/g)+c =−e^(−W(−y)) +c =−e^(−W(−ln(x))) +c=∫(x^x^(...^2 ) /(x(1−x^x^x^. ln(x)))dx
$${x}^{{x}^{{x}} } ={f}\left({x}\right)\Rightarrow{ln}\left({fx}\right)={f}\left({x}\right){ln}\left({x}\right) \\ $$$${ln}\left({f}\left({x}\right)\right){e}^{−{ln}\left({f}\left({x}\right)\right)} ={ln}\left({x}\right) \\ $$$$\Rightarrow−{ln}\left({f}\left({x}\right)\right)={W}\left(−{ln}\left({x}\right)\right) \\ $$$${f}\left({x}\right)={e}^{−{W}\left(−{ln}\left({x}\right)\right)} =\frac{{W}\left(−\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)\right)}{−\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)} \\ $$$$\int\frac{{e}^{−\mathrm{2}\boldsymbol{{W}}\left(−{ln}\left({x}\right)\right)} }{{x}\left(\mathrm{1}−{ln}\left({x}\right){e}^{−{W}\left(−{ln}\left({x}\right)\right)} \right)} \\ $$$${ln}\left({x}\right)={y} \\ $$$$=\int\frac{{e}^{−\mathrm{2}{W}\left(−{y}\right)} }{\left(\mathrm{1}−{ye}^{−{W}\left(−{y}\right)} \right)}{dy} \\ $$$$\int\frac{{e}^{−{W}\left(−{y}\right)} }{{e}^{{W}\left(−{y}\right)} −{y}}{dy} \\ $$$${W}\left({y}\right){e}^{{W}\left({y}\right)} ={y} \\ $$$$\Rightarrow{W}'\left({y}\right)\left(\mathrm{1}+{W}\left({y}\right)\right){e}^{{W}\left({y}\right)} =\mathrm{1} \\ $$$${W}'\left({y}\right)=\frac{{e}^{−{W}\left({y}\right)} }{\mathrm{1}+{W}\left({y}\right)} \\ $$$${e}^{{W}\left(−{y}\right)} −{y}={f}\left({y}\right),{f}'\left({y}\right) \\ $$$${let}\:{g}\left({y}\right)={e}^{{W}\left(−{y}\right)} \\ $$$${g}'\left({y}\right)=−{W}'\left(−{y}\right){e}^{{W}\left(−{y}\right)} =\frac{−\mathrm{1}}{\mathrm{1}+{W}\left(−{y}\right)} \\ $$$$=\frac{−{e}^{{W}\left(−{y}\right)} }{{e}^{{W}\left(−{y}\right)} −{y}}\Rightarrow{dg}\left({y}\right){e}^{{W}\left(−{y}\right)} =\frac{{dy}}{{e}^{{W}\left(−{y}\right)} −{y}} \\ $$$$\int\frac{\mathrm{1}}{{g}^{\mathrm{2}} }{dg}=−\frac{\mathrm{1}}{{g}}+{c} \\ $$$$=−{e}^{−{W}\left(−{y}\right)} +{c} \\ $$$$=−{e}^{−{W}\left(−{ln}\left({x}\right)\right)} +{c}=\int\frac{{x}^{{x}^{…^{\mathrm{2}} } } }{{x}\left(\mathrm{1}−{x}^{{x}^{{x}^{.} } } {ln}\left({x}\right)\right.}{dx} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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