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Question Number 124024 by hatakekakashi1729gmailcom last updated on 30/Nov/20

Answered by Dwaipayan Shikari last updated on 30/Nov/20

(x^x )^((x^x )^((x^(x)) )...) ) =t x^x =y y^y^(y..) =t⇒y^t =t⇒tlog(y)=log(t) e^(log(t)) log(y)=log(t) −log(t)e^(−log(t)) =−log(y) log(t)=−W_0 (−log(y)) t=e^(−W_0 (−logy)) =e^(−W_0 (−xlogx)) =−((W_0 (−xlogx))/(xlogx)) W_0 (Φ)=Σ_(n=1) ^∞ (((−1)^(n−1) n^(n−1) )/(n!))Φ^n ∫_0 ^1 ((W_0 (−xlogx))/(−xlogx))dx =∫_0 ^1 Σ_(n=1) ^∞ (−1)^n n^(n−1) .(((−xlogx)^(n−1) )/(n!))dx =Σ_(n=1) ^∞ ∫_0 ^1 (−1)^2 n^(n−1) (((xlogx)^n )/(n!))dx =Σ_(n=1) ^∞ (n^(n−1) /(n!))∫_0 ^1 x^n log^n x dx logx=p⇒(1/x)=(dt/dx) =Σ_(n=1) ^∞ (n^(n−1) /(n!))∫_(−∞) ^0 e^((n+1)p) p^n dx p(n+1)=−Λ⇒(n+1)=−(dΛ/dp) =Σ_(n=1) ^∞ (((−1)^n n^(n−1) )/(n!n^(n+1) ))∫_0 ^∞ Λ^n e^(−Λ) dΛ =Σ_(n=1) ^∞ (((−1)^(n+1) )/(n^2 .n!)).Γ(n+1)=Σ_(n=1) ^∞ (((−1)^(n+1) )/n^2 )=(π^2 /(12))

(xx)(xx)(xx))...=txx=yyyy..=t⇒yt=t⇒tlog(y)=log(t)elog(t)log(y)=log(t)−log(t)e−log(t)=−log(y)log(t)=−W0(−log(y))t=e−W0(−logy)=e−W0(−xlogx)=−W0(−xlogx)xlogxW0(Φ)=∑∞n=1(−1)n−1nn−1n!Φn∫01W0(−xlogx)−xlogxdx=∫01∑∞n=1(−1)nnn−1.(−xlogx)n−1n!dx=∑∞n=1∫01(−1)2nn−1(xlogx)nn!dx=∑∞n=1nn−1n!∫01xnlognxdxlogx=p⇒1x=dtdx=∑∞n=1nn−1n!∫−∞0e(n+1)ppndxp(n+1)=−Λ⇒(n+1)=−dΛdp=∑∞n=1(−1)nnn−1n!nn+1∫0∞Λne−ΛdΛ=∑∞n=1(−1)n+1n2.n!.Γ(n+1)=∑∞n=1(−1)n+1n2=π212

Commented by hatakekakashi1729gmailcom last updated on 02/Dec/20

Thanks

Thanks

Commented by hatakekakashi1729gmailcom last updated on 30/Nov/20

don′t mind do you have another method

don′tminddoyouhaveanothermethod

Commented by Dwaipayan Shikari last updated on 30/Nov/20

W_0 (x) is Lambert W function W_0 (x)=Σ_(n≥1) ^∞ (((−n)^(n−1) )/(n!))x^n

W0(x)isLambertWfunctionW0(x)=∑∞n⩾1(−n)n−1n!xn

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