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)


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

$i f$ $ψ (x) = \frac{x}{1 - x} + \frac{x^{3}}{1 - x^{2}} + \frac{x^{5}}{1 - x^{5}} + \frac{x^{7}}{1 - x^{7}}$ $s h o w t h a t$ $\int_{0}^{1} [ψ (x) + ψ (\sqrt[3]{x}) + ψ (\sqrt[5]{x}) + ψ (\sqrt[7]{x})] \frac{\ln (\frac{1}{x})}{x} = \frac{25832 π^{2}}{1575}$ $s o l t i o n$ $l e t t h e g e n e r a t i n g s e r i e s f o r m o f$ $ψ (x) = \frac{x}{1 - x} + \frac{x^{3}}{1 - x^{2}} + \frac{x^{5}}{1 - x^{5}} + \frac{x^{7}}{1 - x^{7}}$ $b e ψ (x) = \underset{i = 0}{\sum^{\infty}} \frac{x^{2 i + 1}}{1 - x^{2 i + 1}} . . . . . . . . . (1)$ $f r o m Ω = \int_{0}^{1} ψ (x) + ψ (\sqrt[3]{x}) + ψ (\sqrt[5]{x}) + ψ (\sqrt[7]{x}) \frac{\ln (\frac{1}{x})}{x} d x$ $Ω = - \int_{0}^{1} [ψ (x) + ψ (x^{\frac{1}{3}}) + ψ (x^{\frac{1}{5}}) + ψ (x^{\frac{1}{7}})] \frac{\ln x}{x} d x$ $t h e r e t h e s e r i e s f o r m b e$ $Ω = \int_{0}^{1} [\underset{n = 0}{\sum^{\infty}} ψ (x^{\frac{1}{2 n + 1}}) \frac{\ln x}{x}] d x . . . . . . . . . (2)$ $p u t t i n g e q u a t i o n (1) i n t o (2)$ $Ω = - \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}} [(\frac{x^{2 i + 1} . x^{\frac{1}{2 n + 1}}}{1 - x^{2 i + 1} . x^{\frac{1}{2 n + 1}}}) \frac{\ln x}{x}] d x$ $Ω = - \underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}} \int_{0}^{1} [(\frac{x^{\frac{2 i + 1}{2 n + 1}}}{1 - x^{\frac{2 i + 1}{2 n + 1}}}) \frac{\ln x}{x}] d x$ $l e t y = \frac{2 i + 1}{2 n + 1}$ $Ω = - \underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}} \int_{0}^{1} [(\frac{x^{y}}{1 - x^{y}}) \frac{\ln x}{x}] d x = - \underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}} \int_{0}^{1} (\frac{x^{y - 1}}{1 - x^{y}}) \ln x d x$ $l e t t h e s e r i e s f o r m o f \frac{1}{1 - x^{y}} = \underset{m = 0}{\sum^{\infty}} x^{m y}$ $Ω = - \underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}} \int_{0}^{1} (\underset{m = 0}{\sum^{\infty}} x^{m y} . x^{y - 1} \ln x) d x$ $\frac{\partial}{\partial a} ∣_{a = 0} Ω (a) = - \underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}} \underset{m = 0}{\sum^{\infty}} \int_{0}^{1} (x^{m y} . x^{y - 1} . x^{a}) d x$ $\frac{\partial}{\partial a} ∣_{a = 0} Ω (a) = - \underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}} \underset{m = 0}{\sum^{\infty}} \int_{0}^{1} x^{m y + y + a - 1} d x$ $\frac{\partial}{\partial a} ∣_{a = 0} Ω (a) = - \underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}} \underset{m = 0}{\sum^{\infty}} (\frac{1}{m y + y + a})$ $Ω^{^{'}} (0) = \underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}} \underset{m = 0}{\sum^{\infty}} [\frac{1}{{(m y + y)}^{2}}] = (\underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}}) \frac{1}{y^{2}} \underset{m = 0}{\sum^{\infty}} \frac{1}{{(1 + m)}^{2}}$ $l e t z = 1 + m a t m = 0, z = 1$ $Ω (0) = (\underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}}) \frac{1}{y^{2}} \underset{z = 1}{\sum^{\infty}} \frac{1}{z^{2}} = (\underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}}) \frac{1}{y^{2}} ζ (2) = \frac{π^{2}}{6} (\underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}}) \frac{1}{y^{2}}$ $b u t y = \frac{2 i + 1}{2 n + 1}$ $Ω (0) = \frac{π^{2}}{6} (\underset{n = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}} \frac{1}{{(\frac{2 i + 1}{2 n + 1})}^{2}}) = \frac{π^{2}}{6} \underset{n = 0}{\sum^{\infty}} {(2 n + 1)}^{2} \underset{i = 0}{\sum^{\infty}} \frac{1}{{(2 i + 1)}^{2}}$ $Ω = \frac{π^{2}}{6} [1^{2} + 3^{2} + 5^{2} + 7^{2}] \times [\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}}]$ $Ω = \frac{π^{2}}{6} (1 + 9 + 25 + 49) \times [\frac{1}{1} + \frac{1}{9} + \frac{1}{25} + \frac{1}{49}]$ $Ω = \frac{π^{2}}{6} ∙ (84) ∙ \frac{12916}{11025} = \frac{25832 π^{2}}{1575}$ $∵ \int_{0}^{1} [ψ (x) + ψ (\sqrt[3]{x}) + ψ (\sqrt[5]{x}) + ψ (\sqrt[7]{x})] \frac{\ln (\frac{1}{x})}{x} = \frac{25832 π^{2}}{1575}$ $b y m a t h e w m o n d a y (05 / 09 / 2020)$
Commented by mnjuly1970 last updated on 05/Sep/20

$v e r y e x c e l l e n t a n d v e r y$ $c o n t e n t f u l . . . p . b . u . y$
Commented by Don08q last updated on 05/Sep/20

$G o o d j o b$
Commented by Tawa11 last updated on 06/Sep/21

$great sir$
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