...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


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Question Number 115111 by mnjuly1970 last updated on 23/Sep/20

$. . . m a t h e m a t i c a l a n a l y s i s . . .$ $p r o v e t h a t :$ $Ω = \int_{0}^{1} \frac{(x^{8} + 1) l n (x)}{x^{10} - 1} d x = \frac{π^{2} φ^{2}}{25} ✓$ $m . n . j u l y 1970$
	Answered by Olaf last updated on 23/Sep/20

	$\int \frac{1 + x^{8}}{1 - x^{10}} d x = \int (1 + x^{8}) \underset{n = 0}{\sum^{\infty}} x^{10 n} d x$ $= \underset{n = 0}{\sum^{\infty}} (\frac{x^{10 n + 1}}{10 n + 1} + \frac{x^{10 n + 9}}{10 n + 9})$ $Ω = {[- \underset{n = 0}{\sum^{\infty}} (\frac{x^{10 n + 1}}{10 n + 1} + \frac{x^{10 n + 9}}{10 n + 9}) \ln x]}_{0}^{1}$ $+ \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} (\frac{x^{10 n + 1}}{10 n + 1} + \frac{x^{10 n + 9}}{10 n + 9}) \frac{d x}{x}$ $Ω = \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} (\frac{x^{10 n}}{10 n + 1} + \frac{x^{10 n + 8}}{10 n + 9}) d x$ $Ω = \underset{n = 0}{\sum^{\infty}} {[\frac{x^{10 n + 1}}{{(10 n + 1)}^{2}} + \frac{x^{10 n + 9}}{{(10 n + 9)}^{2}}]}_{0}^{1}$ $Ω = \underset{n = 0}{\sum^{\infty}} [\frac{1}{{(10 n + 1)}^{2}} + \frac{1}{{(10 n + 9)}^{2}}]$ $w o r k i n p r o g r e s s . . .$
	Commented by mnjuly1970 last updated on 23/Sep/20

	$v e r y n i c e . t h a n k s . . .$
	Commented by maths mind last updated on 24/Sep/20

	$Ω = \frac{1}{100} (Σ \frac{1}{{(n + \frac{1}{10})}^{2}} + Σ \frac{1}{{(n + \frac{9}{10})}^{2}}) . . . s i r o l a f d o n e a l l w o r c k$ $= \frac{Ψ_{1} (\frac{1}{10}) + Ψ_{1} (\frac{9}{10})}{100}$ $Ψ_{1} (z) + Ψ_{1} (1 - z) = \frac{π^{2}}{{s i n}^{2} (π z)}, z = \frac{1}{10}$ $g i v e u s a n s w e r$
	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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