...advanced calculus... evaluate ::: Ω=∫_(0 ) ^( (1/2)) ((Arctanh(x))/x)dx=?


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Question Number 128922 by mnjuly1970 last updated on 11/Jan/21

$. . . a d v a n c e d c a l c u l u s . . .$ $e v a l u a t e :::$ $Ω = \int_{0}^{\frac{1}{2}} \frac{Arctanh (x)}{x} d x = ?$
	Answered by Dwaipayan Shikari last updated on 11/Jan/21

	${t a n h}^{- 1} (x) = y$ $\Rightarrow x = \frac{e^{y} - e^{- y}}{e^{y} + e^{- y}} \Rightarrow {x e}^{2 y} + x = e^{2 y} - 1 \Rightarrow e^{2 y} = \frac{1 + x}{1 - x}$ $y = \frac{1}{2} l o g (\frac{1 + x}{1 - x})$ $\frac{1}{2} \int_{0}^{\frac{1}{2}} \frac{l o g (1 + x)}{x} - \frac{l o g (1 - x)}{x} d x$ $= \frac{1}{2} \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \frac{1}{n} \int_{0}^{\frac{1}{2}} x^{n - 1} d x + \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{\frac{1}{2}} x^{n - 1} d x$ $= \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{n^{2}} {(\frac{1}{2})}^{n} + \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} {(\frac{1}{2})}^{n}$ $= \underset{n = 0}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{2}} {(\frac{1}{2})}^{2 n + 1} = \frac{1}{8} \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + \frac{1}{2})}^{2}} {(\frac{1}{4})}^{n}$ $= \frac{1}{8} \underset{n = 0}{\sum^{\infty}} \frac{{(1)}_{n}}{{(n + \frac{1}{2})}^{2} n!} {(\frac{1}{4})}^{n}$
	Commented by Dwaipayan Shikari last updated on 11/Jan/21

	$I t h i n k i t c a n b e f r a n s f o r m e d i n t o H y p e r g e o m e t r i c f u n c t i o n s$ $T r y i n g t o f i n d t h a t . .$
	Commented by mnjuly1970 last updated on 11/Jan/21

	$t h a n k y o u s o m u c h . .$ ${l i}_{2} (x) = - \int_{0}^{x} \frac{l n (1 - t)}{t} d t$ ${l i}_{2} (x) + {l i}_{2} (1 - x) = \frac{π^{2}}{6} - l n (x) l n (1 - x)$ $o f c o u r s e y o u k n o w$ $t h e s e b e t t e r t h a n t h a t m e .$ $i j u s t m e n t i o n e d i t . . . (d i l o g a r i t h m f u n c t i o n p r o p e r t i e s) . .$
	Commented by Dwaipayan Shikari last updated on 11/Jan/21

	$O r$ $- \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} {(- \frac{1}{2})}^{n} + \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} {(\frac{1}{2})}^{n}$ $= - \frac{1}{2} {L i}_{2} (- \frac{1}{2}) + \frac{1}{2} {L i}_{2} (\frac{1}{2})$
	Commented by mnjuly1970 last updated on 11/Jan/21

	$v e r y n i c e$ $t h a n k y o u m r d w a i p a y a n$
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