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Question Number 122431 by danielasebhofoh last updated on 17/Nov/20

Answered by mindispower last updated on 18/Nov/20

$$let{x}^x=t$$$$t^{t^{t^{t....}}}=f\left(t\right)$$$$\Rightarrow t^{f\left(t\right)}=f\left(t\right)$$$$ln\left(f\left(t\right)\right)=f\left(t\right)ln\left(t\right)$$$$\Rightarrow \frac{1}{f\left(t\right)}ln\left(\frac{1}{f\left(t\right)}\right)=-ln\left(t\right)$$$$\Rightarrow ln\left(\frac{1}{f\left(t\right)}\right)e^{ln\left(\frac{1}{f\left(t\right)}\right)}=-ln\left(t\right)$$$$\Rightarrow ln\left(\frac{1}{f\left(t\right)}\right)=W(-ln\left(t\right))$$$$f\left(t\right)=e^{-w(-ln\left(t\right))}=-\frac{W(-ln\left(t\right))}{ln\left(t\right)}$$$${\int }_0^1-\frac{W(-xln\left(x\right))}{xln\left(x\right)}dx=A$$$$W\left(z\right)=\sum _{n⩾1}\frac{{(-n)}^{n-1}}{n!}{z}^n$$$$A=-{\int }_0^1\sum _{n⩾1}\frac{{(-n)}^{n-1}}{n!}\frac{{(-xln\left(x\right))}^n}{xln\left(x\right)}dx$$$$-\sum _{n⩾1}\frac{{(-n)}^{n-1}}{n!}{\int }_0^1{(-1)}^n{x}^{n-1}{ln}^{n-1}\left(x\right)dx$$$$let-ln\left(x\right)=t\Rightarrow x=e^{-t}$$$$\iff -\sum _{n⩾1}\frac{{(-n)}^{n-1}}{n!}{\int }_0^{\mathrm{\infty }}{e}^{-(n-1)t}{\left(t\right)}^{n-1}{e}^{-t}dt$$$$u=nt$$$$=\sum _{n⩾1}\frac{{(-n)}^{n-1}}{n!}{\int }_0^{\mathrm{\infty }}{e}^{-u}{\left(\frac{1}{n}u\right)}^{n-1}.\frac{du}{n}$$$$=\sum _{n⩾1}\frac{{(-1)}^{n-1}{n}^{n-1}.}{n!.n^n}{\int }_0^{\mathrm{\infty }}{u}^{n-1}{e}^{-u}du$$$$=\sum _{n⩾1}\frac{{(-1)}^{n-1}}{n.n!}.\mathrm{\Gamma }\left(n\right)=\sum _{n⩾1}\frac{{(-1)}^{n-1}}{n.n!}(n-1)!$$$$=\sum _{n⩾1}\frac{{(-1)}^{n-1}}{n^2}=\frac{\zeta \left(2\right)}{2}=\frac{{\pi }^2}{12}$$

Commented by hatakekakashi1729gmailcom last updated on 30/Nov/20

$$let{x}^x=t$$$$t^{t^{t^{t....}}}=f\left(t\right)$$$$\Rightarrow t^{f\left(t\right)}=f\left(t\right)$$$$ln\left(f\left(t\right)\right)=f\left(t\right)ln\left(t\right)$$$$\Rightarrow \frac{1}{f\left(t\right)}ln\left(\frac{1}{f\left(t\right)}\right)=-ln\left(t\right)$$$$\Rightarrow ln\left(\frac{1}{f\left(t\right)}\right)e^{ln\left(\frac{1}{f\left(t\right)}\right)}=-ln\left(t\right)$$$$\Rightarrow ln\left(\frac{1}{f\left(t\right)}\right)=W(-ln\left(t\right))???????$$$$f\left(t\right)=e^{-w(-ln\left(t\right))}=-\frac{W(-ln\left(t\right))}{ln\left(t\right)}$$$${\int }_0^1-\frac{W(-xln\left(x\right))}{xln\left(x\right)}dx=A$$$$W\left(z\right)=\sum _{n⩾1}\frac{{(-n)}^{n-1}}{n!}{z}^n????????????$$$$A=-{\int }_0^1\sum _{n⩾1}\frac{{(-n)}^{n-1}}{n!}\frac{{(-xln\left(x\right))}^n}{xln\left(x\right)}dx$$$$-\sum _{n⩾1}\frac{{(-n)}^{n-1}}{n!}{\int }_0^1{(-1)}^n{x}^{n-1}{ln}^{n-1}\left(x\right)dx$$$$let-ln\left(x\right)=t\Rightarrow x=e^{-t}$$$$\iff -\sum _{n⩾1}\frac{{(-n)}^{n-1}}{n!}{\int }_0^{\mathrm{\infty }}{e}^{-(n-1)t}{\left(t\right)}^{n-1}{e}^{-t}dt$$$$u=nt$$$$=\sum _{n⩾1}\frac{{(-n)}^{n-1}}{n!}{\int }_0^{\mathrm{\infty }}{e}^{-u}{\left(\frac{1}{n}u\right)}^{n-1}.\frac{du}{n}$$$$=\sum _{n⩾1}\frac{{(-1)}^{n-1}{n}^{n-1}.}{n!.n^n}{\int }_0^{\mathrm{\infty }}{u}^{n-1}{e}^{-u}du$$$$=\sum _{n⩾1}\frac{{(-1)}^{n-1}}{n.n!}.\mathrm{\Gamma }\left(n\right)=\sum _{n⩾1}\frac{{(-1)}^{n-1}}{n.n!}(n-1)!$$$$=\sum _{n⩾1}\frac{{(-1)}^{n-1}}{n^2}=\frac{\zeta \left(2\right)}{2}=\frac{{\pi }^2}{12}$$

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