if-x-x-1-x-x-3-1-x-2-x-5-1-x-5-x-7-1-x-7-show-that-0-1-x-x-1-3-x-1-5-x-1-7-ln-1-x-x-25832pi-2-1575-soltion-let-the-generating-se

Question Number 111813 by mathdave last updated on 05/Sep/20

if ψ(x)=(x/(1−x))+(x^3 /(1−x^2 ))+(x^5 /(1−x^5 ))+(x^7 /(1−x^7 )) show that ∫_0 ^1 [ψ(x)+ψ((x)^(1/3) )+ψ((x)^(1/5) )+ψ((x)^(1/7) )]((ln((1/x)))/x)=((25832π^2 )/(1575)) soltion let the generating series form of ψ(x)=(x/(1−x))+(x^3 /(1−x^2 ))+(x^5 /(1−x^5 ))+(x^7 /(1−x^7 )) be ψ(x)=Σ_(i=0) ^∞ (x^(2i+1) /(1−x^(2i+1) )).........(1) from Ω=∫_0 ^1 ψ(x)+ψ((x)^(1/3) )+ψ((x)^(1/5) )+ψ((x)^(1/7) )((ln((1/x)))/x)dx Ω=−∫_0 ^1 [ψ(x)+ψ(x^(1/3) )+ψ(x^(1/5) )+ψ(x^(1/7) )]((lnx)/x)dx there the series form be Ω=∫_0 ^1 [Σ_(n=0) ^∞ ψ(x^(1/(2n+1)) )((lnx)/x)]dx.........(2) putting equation (1) into (2) Ω=−∫_0 ^1 Σ_(n=0) ^∞ Σ_(i=0) ^∞ [(((x^(2i+1) .x^(1/(2n+1)) )/(1−x^(2i+1) .x^(1/(2n+1)) )))((lnx)/x)]dx Ω=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 [((x^((2i+1)/(2n+1)) /(1−x^((2i+1)/(2n+1)) )))((lnx)/x)]dx let y=((2i+1)/(2n+1)) Ω=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 [((x^y /(1−x^y )))((lnx)/x)]dx=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 ((x^(y−1) /(1−x^y )))lnxdx let the series form of (1/(1−x^y ))=Σ_(m=0) ^∞ x^(my) Ω=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 (Σ_(m=0) ^∞ x^(my) .x^(y−1) lnx)dx (∂/∂a)∣_(a=0) Ω(a)=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ ∫_0 ^1 (x^(my) .x^(y−1) .x^a )dx (∂/∂a)∣_(a=0) Ω(a)=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ ∫_0 ^1 x^(my+y+a−1) dx (∂/∂a)∣_(a=0) Ω(a)=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ ((1/(my+y+a))) Ω^′ (0)=Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ [(1/((my+y)^2 ))]=(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 )Σ_(m=0) ^∞ (1/((1+m)^2 )) let z=1+m at m=0 ,z=1 Ω(0)=(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 )Σ_(z=1) ^∞ (1/z^2 )=(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 )ζ(2)=(π^2 /6)(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 ) but y=((2i+1)/(2n+1)) Ω(0)=(π^2 /6)(Σ_(n=0) ^∞ Σ_(i=0) ^∞ (1/((((2i+1)/(2n+1)))^2 )))=(π^2 /6)Σ_(n=0) ^∞ (2n+1)^2 Σ_(i=0) ^∞ (1/((2i+1)^2 )) Ω=(π^2 /6)[1^2 +3^2 +5^2 +7^2 ]×[(1/1^2 )+(1/3^2 )+(1/5^2 )+(1/7^2 )] Ω=(π^2 /6)(1+9+25+49)×[(1/1)+(1/9)+(1/(25))+(1/(49))] Ω=(π^2 /6)•(84)•((12916)/(11025))=((25832π^2 )/(1575)) ∵∫_0 ^1 [ψ(x)+ψ((x)^(1/3) )+ψ((x)^(1/5) )+ψ((x)^(1/7) )]((ln((1/x)))/x)=((25832π^2 )/(1575)) by mathew monday(05/09/2020)

$${if}\: \\ $$$$\psi\left({x}\right)=\frac{{x}}{\mathrm{1}−{x}}+\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{5}} }{\mathrm{1}−{x}^{\mathrm{5}} }+\frac{{x}^{\mathrm{7}} }{\mathrm{1}−{x}^{\mathrm{7}} } \\ $$$${show}\:{that}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left[\psi\left({x}\right)+\psi\left(\sqrt[{\mathrm{3}}]{{x}}\right)+\psi\left(\sqrt[{\mathrm{5}}]{{x}}\right)+\psi\left(\sqrt[{\mathrm{7}}]{{x}}\right)\right]\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}}=\frac{\mathrm{25832}\pi^{\mathrm{2}} }{\mathrm{1575}} \\ $$$${soltion}\: \\ $$$${let}\:{the}\:{generating}\:{series}\:{form}\:{of} \\ $$$$\psi\left({x}\right)=\frac{{x}}{\mathrm{1}−{x}}+\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{5}} }{\mathrm{1}−{x}^{\mathrm{5}} }+\frac{{x}^{\mathrm{7}} }{\mathrm{1}−{x}^{\mathrm{7}} } \\ $$$${be}\:\psi\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{i}+\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{2}{i}+\mathrm{1}} }………\left(\mathrm{1}\right) \\ $$$${from}\:\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \psi\left({x}\right)+\psi\left(\sqrt[{\mathrm{3}}]{{x}}\right)+\psi\left(\sqrt[{\mathrm{5}}]{{x}}\right)+\psi\left(\sqrt[{\mathrm{7}}]{{x}}\right)\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}}{dx} \\ $$$$\Omega=−\int_{\mathrm{0}} ^{\mathrm{1}} \left[\psi\left({x}\right)+\psi\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)+\psi\left({x}^{\frac{\mathrm{1}}{\mathrm{5}}} \right)+\psi\left({x}^{\frac{\mathrm{1}}{\mathrm{7}}} \right)\right]\frac{\mathrm{ln}{x}}{{x}}{dx} \\ $$$${there}\:{the}\:{series}\:{form}\:{be} \\ $$$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\psi\left({x}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} \right)\frac{\mathrm{ln}{x}}{{x}}\right]{dx}………\left(\mathrm{2}\right) \\ $$$${putting}\:{equation}\:\left(\mathrm{1}\right)\:{into}\:\left(\mathrm{2}\right) \\ $$$$\Omega=−\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left(\frac{{x}^{\mathrm{2}{i}+\mathrm{1}} .{x}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} }{\mathrm{1}−{x}^{\mathrm{2}{i}+\mathrm{1}} .{x}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} }\right)\frac{\mathrm{ln}{x}}{{x}}\right]{dx} \\ $$$$\Omega=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left[\left(\frac{{x}^{\frac{\mathrm{2}{i}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} }{\mathrm{1}−{x}^{\frac{\mathrm{2}{i}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} }\right)\frac{\mathrm{ln}{x}}{{x}}\right]{dx} \\ $$$${let}\:{y}=\frac{\mathrm{2}{i}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Omega=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left[\left(\frac{{x}^{{y}} }{\mathrm{1}−{x}^{{y}} }\right)\frac{\mathrm{ln}{x}}{{x}}\right]{dx}=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{x}^{{y}−\mathrm{1}} }{\mathrm{1}−{x}^{{y}} }\right)\mathrm{ln}{xdx} \\ $$$${let}\:{the}\:{series}\:{form}\:{of}\:\:\:\frac{\mathrm{1}}{\mathrm{1}−{x}^{{y}} }=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{my}} \\ $$$$\Omega=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{my}} .{x}^{{y}−\mathrm{1}} \mathrm{ln}{x}\right){dx} \\ $$$$\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{0}} \Omega\left({a}\right)=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{{my}} .{x}^{{y}−\mathrm{1}} .{x}^{{a}} \right){dx} \\ $$$$\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{0}} \Omega\left({a}\right)=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{my}+{y}+{a}−\mathrm{1}} {dx} \\ $$$$\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{0}} \Omega\left({a}\right)=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{my}+{y}+{a}}\right) \\ $$$$\Omega^{'} \left(\mathrm{0}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\left({my}+{y}\right)^{\mathrm{2}} }\right]=\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\right)\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{1}+{m}\right)^{\mathrm{2}} } \\ $$$${let}\:{z}=\mathrm{1}+{m}\:{at}\:{m}=\mathrm{0}\:\:,{z}=\mathrm{1} \\ $$$$\Omega\left(\mathrm{0}\right)=\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\right)\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\underset{{z}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{z}^{\mathrm{2}} }=\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\right)\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\right)\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$${but}\:{y}=\frac{\mathrm{2}{i}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Omega\left(\mathrm{0}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\frac{\mathrm{2}{i}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)^{\mathrm{2}} }\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{i}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\left[\mathrm{1}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \right]×\left[\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{2}} }\right] \\ $$$$\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\left(\mathrm{1}+\mathrm{9}+\mathrm{25}+\mathrm{49}\right)×\left[\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{25}}+\frac{\mathrm{1}}{\mathrm{49}}\right] \\ $$$$\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\bullet\left(\mathrm{84}\right)\bullet\frac{\mathrm{12916}}{\mathrm{11025}}=\frac{\mathrm{25832}\pi^{\mathrm{2}} }{\mathrm{1575}} \\ $$$$\because\int_{\mathrm{0}} ^{\mathrm{1}} \left[\psi\left({x}\right)+\psi\left(\sqrt[{\mathrm{3}}]{{x}}\right)+\psi\left(\sqrt[{\mathrm{5}}]{{x}}\right)+\psi\left(\sqrt[{\mathrm{7}}]{{x}}\right)\right]\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}}=\frac{\mathrm{25832}\pi^{\mathrm{2}} }{\mathrm{1575}} \\ $$$${by}\:{mathew}\:{monday}\left(\mathrm{05}/\mathrm{09}/\mathrm{2020}\right) \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 05/Sep/20