Question Number 115111 by mnjuly1970 last updated on 23/Sep/20

             ...mathematical  analysis...                        prove  that:                        Ω=∫_0 ^( 1) (((x^8 +1)ln(x))/(x^(10) −1)) dx=((π^2 ϕ^2 )/(25))  ✓            m.n.july 1970

Answered by Olaf last updated on 23/Sep/20

∫((1+x^8 )/(1−x^(10) ))dx = ∫(1+x^8 )Σ_(n=0) ^∞ x^(10n) dx  = Σ_(n=0) ^∞ ((x^(10n+1) /(10n+1))+(x^(10n+9) /(10n+9)))  Ω = [−Σ_(n=0) ^∞ ((x^(10n+1) /(10n+1))+(x^(10n+9) /(10n+9)))lnx]_0 ^1   +∫_0 ^1 Σ_(n=0) ^∞ ((x^(10n+1) /(10n+1))+(x^(10n+9) /(10n+9)))(dx/x)  Ω = ∫_0 ^1 Σ_(n=0) ^∞ ((x^(10n) /(10n+1))+(x^(10n+8) /(10n+9)))dx  Ω = Σ_(n=0) ^∞ [(x^(10n+1) /((10n+1)^2 ))+(x^(10n+9) /((10n+9)^2 ))]_0 ^1   Ω = Σ_(n=0) ^∞ [(1/((10n+1)^2 ))+(1/((10n+9)^2 ))]  work in progress...

Commented bymnjuly1970 last updated on 23/Sep/20

very nice  .thanks...

Commented bymaths mind last updated on 24/Sep/20

Ω=(1/(100))(Σ(1/((n+(1/(10)))^2 ))+Σ(1/((n+(9/(10)))^2 )))...sir olaf done all worck  =((Ψ_1 ((1/(10)))+Ψ_1 ((9/(10))))/(100))  Ψ_1 (z)+Ψ_1 (1−z)=(π^2 /(sin^2 (πz)))  ,z=(1/(10))  give usanswer

Answered by mathdave last updated on 24/Sep/20

solution  let  I=∫_0 ^1 (((x^8 +1)lnx)/(x^(10) −1))dx=−∫_0 ^1 ((x^8 lnx)/(1−x^(10) ))dx−∫_0 ^1 ((lnx)/(1−x^(10) ))dx  I=−Σ_(n=0) ^∞ ∫_0 ^1 x^8 .x^(10n) .lnxdx−Σ_(n=0) ^∞ ∫^1 _0 x^(10n) lnxdx  I=−Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ∫_0 ^1 x^(10n+8) .x^(a−1) dx−Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ∫_0 ^1 x^(10n) .x^(a−1) dx  I=−Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ((1/(10n+8+a)))−Σ_(n=0) ^∞ (∂/∂a)∣_(a=1) ((1/(10n+a)))  I=Σ_(n=0) ^∞ (1/((10n+9)^2 ))+Σ_(n=0) ^∞ (1/((10n+1)^2 ))=Σ_(n=−∞) ^∞ (1/((10n+1)^2 ))  I=Σ_(n=−∞) ^∞ (1/((10n+1)^2 ))=−(π/(100))lim_(z→−(1/(10))) (1/(1!))(d/dz)(cot(πz)  I=−(π/(100))lim_(z→−(1/(10)) ) −πcosec^2 (−(π/(10)))=(π^2 /(100))•(1/(sin^2 ((π/(10)))))  let ((90)/5)=18  let x=18   ∵90=5x=2x+3x  90−3x=2x    ,sin(90−3x)=sin(2x)  cos(3x)=sin(2x)  4cos^3 x−3cosx=2sinxcosx   ,4cos^2 x−3=2sinx  4(1−sin^2 x)−3=2sinx   ,4sin^2 x+2sinx−1=0  sinx=((−2±(√(4+16)))/8)=((−1+(√5))/4)  ∵sin18=sin((π/(10)))=((−1+(√5))/4)  sin^2 ((π/(10)))=[((−1+(√5))/4)]^2 =((3−(√5))/8)  I=(π^2 /(100))•(1/(sin^2 ((π/(10)))))=(π^2 /(100))•(8/((3−(√(5)))))=(π^2 /(25))•(1/(((3/2)−((√5)/2))))   by linear approximation method of  (a+b)^n =(a+nb)  I=(π^2 /(25))•((3/2)−((√5)/2))^(−1) =(π^2 /(25))((3/2)+((√5)/2))  but ((1/2)+((√5)/2))^2 =((3/2)+((√5)/2))     note φ^2 =((1/2)+((√5)/2))^2   ∵I=∫_0 ^1 (((x^8 +1)lnx)/(x^(10) −1))dx=(π^2 /(25))φ^2      Q.E.D  by mathdave(24/09/2020)

Commented bymnjuly1970 last updated on 24/Sep/20

very nice and perfect  mr   dave..thx a lot..

Commented byTawa11 last updated on 06/Sep/21

great