Prove that if Li_2 (x)=Σ_(n=1) (x^n /n^2 ) then ∀ x Li_2 (x)+Li_2 (1−x) = (π^2 /6) −ln(x)ln(1−x) ∀ x∉[0:1] Li_2 (x)+Li_2 ((1/x)) = −(π^2 /6) −[ln(−x)]^2 Find A=Σ_(n=1) ^∞ (ϕ^n /n^2 ) and B=Σ


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Question Number 68270 by ~ À ® @ 237 ~ last updated on 08/Sep/19

$P r o v e t h a t i f {L i}_{2} (x) = \sum_{n = 1} \frac{x^{n}}{n^{2}} t h e n$ $\forall x {L i}_{2} (x) + {L i}_{2} (1 - x) = \frac{π^{2}}{6} - l n (x) l n (1 - x)$ $\forall x \notin [0 : 1] {L i}_{2} (x) + {L i}_{2} (\frac{1}{x}) = - \frac{π^{2}}{6} - {[l n (- x)]}^{2}$ $F i n d A = \underset{n = 1}{\sum^{\infty}} \frac{φ^{n}}{n^{2}} a n d B = \underset{n = 1}{\sum^{\infty}} \frac{2^{n}}{n^{2}}$
Commented by mathmax by abdo last updated on 10/Sep/19

$l e t f (x) = l_{i^{2}} (x) + l_{i_{2}} (1 - x) a n d g (x) = \frac{π^{2}}{6} - l n (x) l n (1 - x)$ $w i t h ∣ x ∣< 1 w e {h a v e f}^{^{'}} (x) = l_{i_{2}}^{^{'}} (x) - l_{i_{2}}^{'} (1 - x) b u t l^{^{'}} i_{2} (x)$ $= \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{n} = \frac{1}{x} \sum_{n = 1}^{\infty} \frac{x^{n}}{n} = - \frac{1}{x} l n (1 - x) a n d$ $l_{i_{2}}^{^{'}} (1 - x) = \frac{1}{1 - x} l n (x) \Rightarrow f^{^{'}} (x) = - \frac{1}{x} l n (1 - x) + \frac{1}{1 - x} l n x$ $g^{^{'}} (x) = - \frac{1}{x} l n (1 - x) - (l n (x) \frac{- 1}{1 - x}) = - \frac{1}{x} l n (1 - x) + \frac{1}{1 - x} l n (x)$ $\Rightarrow f (x) = g (x) + c$ ${l i m}_{x \to 1} f (x) = L_{i_{2}} (1) + L_{i_{2}} (0) = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6} \Rightarrow c = 0$ ${l i m}_{x \to 1} g (x) = \frac{π^{2}}{6} - {l i m}_{x \to 1} l n (x) l n (1 - x)$ $1 - x = t \Rightarrow {l i m}_{x \to 1} l n (x) l n (1 - x) = {l i m}_{t \to 0} l n (1 - t) l n t$ $= {l i m}_{t \to 0} t l n (t) \frac{l n (1 - t)}{t} = 0 \Rightarrow c = 0 \Rightarrow$ $L_{i_{2}} (x) + L_{i_{2}} (1 - x) = - l n (x) l n (1 - x)$ $p e r h a p s t b e r e i s a e r r o r i n t h e q u e s t i o n . . .!$
Commented by ~ À ® @ 237 ~ last updated on 10/Sep/19

$T h a n k s f o r t h i s s i r b u t t h e r e i s n o n e e r r o r$
Commented by ~ À ® @ 237 ~ last updated on 10/Sep/19

$Y o u h a v e a l r e a d y p r o v e d t h e e q u a l i t y w h e n s h o w i n g t h a t c = 0$ $I n y o u r c o n c l u s i o n y o u f o r g o t \frac{π^{2}}{6} i n t h e e x p r e s s i o n o f g (x)$
	Answered by mind is power last updated on 09/Sep/19

	$a r e y o u s u r f o r {l i}_{2} (x) + {l i}_{2} (\frac{1}{x}) = - \frac{π^{2}}{6_{}} - {[l n (- x)]}^{2} . . .$
	Commented by ~ À ® @ 237 ~ last updated on 10/Sep/19

	$I f o r g e t s o m e t h i n g h e r e s i r : S o r r y$ $I t i s {L i}_{2} (x) + {L i}_{2} (\frac{1}{x}) = -_{} \frac{π^{2}}{6} - \frac{1}{2} {[l n (- x)]}^{2}$
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