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mathematical-analysis-prove-that-0-1-x-8-1-ln-x-x-10-1-dx-pi-2-2-25-m-n-july-1970-




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
$$\:\:\:\:\:\:\:\:\:\:\:\:\:…{mathematical}\:\:{analysis}…\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:\:{that}: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left({x}^{\mathrm{8}} +\mathrm{1}\right){ln}\left({x}\right)}{{x}^{\mathrm{10}} −\mathrm{1}}\:{dx}=\frac{\pi^{\mathrm{2}} \varphi^{\mathrm{2}} }{\mathrm{25}}\:\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:{m}.{n}.{july}\:\mathrm{1970} \\ $$$$ \\ $$
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...
$$\int\frac{\mathrm{1}+{x}^{\mathrm{8}} }{\mathrm{1}−{x}^{\mathrm{10}} }{dx}\:=\:\int\left(\mathrm{1}+{x}^{\mathrm{8}} \right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{10}{n}} {dx} \\ $$$$=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{x}^{\mathrm{10}{n}+\mathrm{1}} }{\mathrm{10}{n}+\mathrm{1}}+\frac{{x}^{\mathrm{10}{n}+\mathrm{9}} }{\mathrm{10}{n}+\mathrm{9}}\right) \\ $$$$\Omega\:=\:\left[−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{x}^{\mathrm{10}{n}+\mathrm{1}} }{\mathrm{10}{n}+\mathrm{1}}+\frac{{x}^{\mathrm{10}{n}+\mathrm{9}} }{\mathrm{10}{n}+\mathrm{9}}\right)\mathrm{ln}{x}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$+\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{x}^{\mathrm{10}{n}+\mathrm{1}} }{\mathrm{10}{n}+\mathrm{1}}+\frac{{x}^{\mathrm{10}{n}+\mathrm{9}} }{\mathrm{10}{n}+\mathrm{9}}\right)\frac{{dx}}{{x}} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{x}^{\mathrm{10}{n}} }{\mathrm{10}{n}+\mathrm{1}}+\frac{{x}^{\mathrm{10}{n}+\mathrm{8}} }{\mathrm{10}{n}+\mathrm{9}}\right){dx} \\ $$$$\Omega\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{{x}^{\mathrm{10}{n}+\mathrm{1}} }{\left(\mathrm{10}{n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{{x}^{\mathrm{10}{n}+\mathrm{9}} }{\left(\mathrm{10}{n}+\mathrm{9}\right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\Omega\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\left(\mathrm{10}{n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{10}{n}+\mathrm{9}\right)^{\mathrm{2}} }\right] \\ $$$${work}\:{in}\:{progress}… \\ $$
Commented by mnjuly1970 last updated on 23/Sep/20
very nice  .thanks...
$${very}\:{nice}\:\:.{thanks}… \\ $$
Commented by maths 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
$$\Omega=\frac{\mathrm{1}}{\mathrm{100}}\left(\Sigma\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{10}}\right)^{\mathrm{2}} }+\Sigma\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{9}}{\mathrm{10}}\right)^{\mathrm{2}} }\right)…{sir}\:{olaf}\:{done}\:{all}\:{worck} \\ $$$$=\frac{\Psi_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{10}}\right)+\Psi_{\mathrm{1}} \left(\frac{\mathrm{9}}{\mathrm{10}}\right)}{\mathrm{100}} \\ $$$$\Psi_{\mathrm{1}} \left({z}\right)+\Psi_{\mathrm{1}} \left(\mathrm{1}−{z}\right)=\frac{\pi^{\mathrm{2}} }{{sin}^{\mathrm{2}} \left(\pi{z}\right)}\:\:,{z}=\frac{\mathrm{1}}{\mathrm{10}} \\ $$$${give}\:{usanswer}\: \\ $$$$ \\ $$
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)
$${solution} \\ $$$${let} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({x}^{\mathrm{8}} +\mathrm{1}\right)\mathrm{ln}{x}}{{x}^{\mathrm{10}} −\mathrm{1}}{dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{8}} \mathrm{ln}{x}}{\mathrm{1}−{x}^{\mathrm{10}} }{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\mathrm{1}−{x}^{\mathrm{10}} }{dx} \\ $$$${I}=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{8}} .{x}^{\mathrm{10}{n}} .\mathrm{ln}{xdx}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{\mathrm{0}} {\int}^{\mathrm{1}} {x}^{\mathrm{10}{n}} \mathrm{ln}{xdx} \\ $$$${I}=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{10}{n}+\mathrm{8}} .{x}^{{a}−\mathrm{1}} {dx}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{10}{n}} .{x}^{{a}−\mathrm{1}} {dx} \\ $$$${I}=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{10}{n}+\mathrm{8}+{a}}\right)−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{10}{n}+{a}}\right) \\ $$$${I}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{10}{n}+\mathrm{9}\right)^{\mathrm{2}} }+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{10}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{10}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${I}=\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{10}{n}+\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\pi}{\mathrm{100}}\underset{{z}\rightarrow−\frac{\mathrm{1}}{\mathrm{10}}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{1}!}\frac{{d}}{{dz}}\left(\mathrm{cot}\left(\pi{z}\right)\right. \\ $$$${I}=−\frac{\pi}{\mathrm{100}}\underset{{z}\rightarrow−\frac{\mathrm{1}}{\mathrm{10}}\:} {\mathrm{lim}}−\pi\mathrm{cosec}^{\mathrm{2}} \left(−\frac{\pi}{\mathrm{10}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{100}}\bullet\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{10}}\right)} \\ $$$${let}\:\frac{\mathrm{90}}{\mathrm{5}}=\mathrm{18}\:\:{let}\:{x}=\mathrm{18}\:\:\:\because\mathrm{90}=\mathrm{5}{x}=\mathrm{2}{x}+\mathrm{3}{x} \\ $$$$\mathrm{90}−\mathrm{3}{x}=\mathrm{2}{x}\:\:\:\:,\mathrm{sin}\left(\mathrm{90}−\mathrm{3}{x}\right)=\mathrm{sin}\left(\mathrm{2}{x}\right) \\ $$$$\mathrm{cos}\left(\mathrm{3}{x}\right)=\mathrm{sin}\left(\mathrm{2}{x}\right) \\ $$$$\mathrm{4co}{s}^{\mathrm{3}} {x}−\mathrm{3cos}{x}=\mathrm{2sin}{x}\mathrm{cos}{x}\:\:\:,\mathrm{4cos}^{\mathrm{2}} {x}−\mathrm{3}=\mathrm{2sin}{x} \\ $$$$\mathrm{4}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\right)−\mathrm{3}=\mathrm{2sin}{x}\:\:\:,\mathrm{4sin}^{\mathrm{2}} {x}+\mathrm{2sin}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}{x}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{16}}}{\mathrm{8}}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\because\mathrm{sin18}=\mathrm{sin}\left(\frac{\pi}{\mathrm{10}}\right)=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{10}}\right)=\left[\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\right]^{\mathrm{2}} =\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{100}}\bullet\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{10}}\right)}=\frac{\pi^{\mathrm{2}} }{\mathrm{100}}\bullet\frac{\mathrm{8}}{\left(\mathrm{3}−\sqrt{\left.\mathrm{5}\right)}\right.}=\frac{\pi^{\mathrm{2}} }{\mathrm{25}}\bullet\frac{\mathrm{1}}{\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)}\: \\ $$$${by}\:{linear}\:{approximation}\:{method}\:{of} \\ $$$$\left({a}+{b}\right)^{{n}} =\left({a}+{nb}\right) \\ $$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{25}}\bullet\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{−\mathrm{1}} =\frac{\pi^{\mathrm{2}} }{\mathrm{25}}\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$${but}\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:\:\:\:\:{note}\:\phi^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\because{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({x}^{\mathrm{8}} +\mathrm{1}\right)\mathrm{ln}{x}}{{x}^{\mathrm{10}} −\mathrm{1}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{25}}\phi^{\mathrm{2}} \:\:\:\:\:{Q}.{E}.{D} \\ $$$${by}\:{mathdave}\left(\mathrm{24}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by mnjuly1970 last updated on 24/Sep/20
very nice and perfect  mr   dave..thx a lot..
$${very}\:{nice}\:{and}\:{perfect}\:\:{mr}\: \\ $$$${dave}..{thx}\:{a}\:{lot}.. \\ $$
Commented by Tawa11 last updated on 06/Sep/21
great
$$\mathrm{great} \\ $$

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