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

\dots m a t h e m a t i c a l a n a l y s i s \dots

p r o v e t h a t :

Ω = \int_{0}^{1} \frac{(x^{8} + 1) l n (x)}{x^{10} - 1} d x = \frac{π^{2} φ^{2}}{25} ✓

m . n . j u l y 1970

Answered by Olaf last updated on 23/Sep/20

\int \frac{1 + x^{8}}{1 - x^{10}} d x = \int (1 + x^{8}) \underset{n = 0}{\sum^{\infty}} x^{10 n} d x

= \underset{n = 0}{\sum^{\infty}} (\frac{x^{10 n + 1}}{10 n + 1} + \frac{x^{10 n + 9}}{10 n + 9})

Ω = {[- \underset{n = 0}{\sum^{\infty}} (\frac{x^{10 n + 1}}{10 n + 1} + \frac{x^{10 n + 9}}{10 n + 9}) \ln x]}_{0}^{1}

+ \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} (\frac{x^{10 n + 1}}{10 n + 1} + \frac{x^{10 n + 9}}{10 n + 9}) \frac{d x}{x}

Ω = \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} (\frac{x^{10 n}}{10 n + 1} + \frac{x^{10 n + 8}}{10 n + 9}) d x

Ω = \underset{n = 0}{\sum^{\infty}} {[\frac{x^{10 n + 1}}{{(10 n + 1)}^{2}} + \frac{x^{10 n + 9}}{{(10 n + 9)}^{2}}]}_{0}^{1}

Ω = \underset{n = 0}{\sum^{\infty}} [\frac{1}{{(10 n + 1)}^{2}} + \frac{1}{{(10 n + 9)}^{2}}]

w o r k i n p r o g r e s s \dots

Commented by mnjuly1970 last updated on 23/Sep/20

v e r y n i c e . t h a n k s \dots

Commented by maths mind last updated on 24/Sep/20

Ω = \frac{1}{100} (Σ \frac{1}{{(n + \frac{1}{10})}^{2}} + Σ \frac{1}{{(n + \frac{9}{10})}^{2}}) \dots s i r o l a f d o n e a l l w o r c k

= \frac{Ψ_{1} (\frac{1}{10}) + Ψ_{1} (\frac{9}{10})}{100}

Ψ_{1} (z) + Ψ_{1} (1 - z) = \frac{π^{2}}{{s i n}^{2} (π z)}, z = \frac{1}{10}

g i v e u s a n s w e r

Answered by mathdave last updated on 24/Sep/20

s o l u t i o n

l e t

I = \int_{0}^{1} \frac{(x^{8} + 1) \ln x}{x^{10} - 1} d x = - \int_{0}^{1} \frac{x^{8} \ln x}{1 - x^{10}} d x - \int_{0}^{1} \frac{\ln x}{1 - x^{10}} d x

I = - \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} x^{8} . x^{10 n} . \ln x d x - \underset{n = 0}{\sum^{\infty}} {\int_{0}}^{1} x^{10 n} \ln x d x

I = - \underset{n = 0}{\sum^{\infty}} \frac{\partial}{\partial a} ∣_{a = 1} \int_{0}^{1} x^{10 n + 8} . x^{a - 1} d x - \underset{n = 0}{\sum^{\infty}} \frac{\partial}{\partial a} ∣_{a = 1} \int_{0}^{1} x^{10 n} . x^{a - 1} d x

I = - \underset{n = 0}{\sum^{\infty}} \frac{\partial}{\partial a} ∣_{a = 1} (\frac{1}{10 n + 8 + a}) - \underset{n = 0}{\sum^{\infty}} \frac{\partial}{\partial a} ∣_{a = 1} (\frac{1}{10 n + a})

I = \underset{n = 0}{\sum^{\infty}} \frac{1}{{(10 n + 9)}^{2}} + \underset{n = 0}{\sum^{\infty}} \frac{1}{{(10 n + 1)}^{2}} = \underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(10 n + 1)}^{2}}

I = \underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(10 n + 1)}^{2}} = - \frac{π}{100} \lim_{z \to - \frac{1}{10}} \frac{1}{1!} \frac{d}{d z} (\cot (π z)

I = - \frac{π}{100} \lim_{z \to - \frac{1}{10}} - π {cosec}^{2} (- \frac{π}{10}) = \frac{π^{2}}{100} ∙ \frac{1}{\sin^{2} (\frac{π}{10})}

l e t \frac{90}{5} = 18 l e t x = 18 ∵ 90 = 5 x = 2 x + 3 x

90 - 3 x = 2 x, \sin (90 - 3 x) = \sin (2 x)

\cos (3 x) = \sin (2 x)

4 co s^{3} x - 3 \cos x = 2 \sin x \cos x, {4 \cos}^{2} x - 3 = 2 \sin x

4 (1 - \sin^{2} x) - 3 = 2 \sin x, {4 \sin}^{2} x + 2 \sin x - 1 = 0

\sin x = \frac{- 2 \pm \sqrt{4 + 16}}{8} = \frac{- 1 + \sqrt{5}}{4}

∵ \sin 18 = \sin (\frac{π}{10}) = \frac{- 1 + \sqrt{5}}{4}

\sin^{2} (\frac{π}{10}) = {[\frac{- 1 + \sqrt{5}}{4}]}^{2} = \frac{3 - \sqrt{5}}{8}

I = \frac{π^{2}}{100} ∙ \frac{1}{\sin^{2} (\frac{π}{10})} = \frac{π^{2}}{100} ∙ \frac{8}{(3 - \sqrt{5)}} = \frac{π^{2}}{25} ∙ \frac{1}{(\frac{3}{2} - \frac{\sqrt{5}}{2})}

b y l i n e a r a p p r o x i m a t i o n m e t h o d o f

{(a + b)}^{n} = (a + n b)

I = \frac{π^{2}}{25} ∙ {(\frac{3}{2} - \frac{\sqrt{5}}{2})}^{- 1} = \frac{π^{2}}{25} (\frac{3}{2} + \frac{\sqrt{5}}{2})

b u t {(\frac{1}{2} + \frac{\sqrt{5}}{2})}^{2} = (\frac{3}{2} + \frac{\sqrt{5}}{2}) n o t e ϕ^{2} = {(\frac{1}{2} + \frac{\sqrt{5}}{2})}^{2}

∵ I = \int_{0}^{1} \frac{(x^{8} + 1) \ln x}{x^{10} - 1} d x = \frac{π^{2}}{25} ϕ^{2} Q . E . D

b y m a t h d a v e (24 / 09 / 2020)

Commented by mnjuly1970 last updated on 24/Sep/20

v e r y n i c e a n d p e r f e c t m r

d a v e . . t h x a l o t . .

Commented by Tawa11 last updated on 06/Sep/21

great

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