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

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}} \dots \dots \dots (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 \dots \dots \dots (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 \dots 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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