show that ∫_0 ^∞ x arctanh(e^(−αx) )dx=((7ζ(3))/(8α^2 ))


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 80764 by M±th+et£s last updated on 06/Feb/20

$s h o w t h a t$ $\int_{0}^{\infty} x a r c t a n h (e^{- α x}) d x = \frac{7 ζ (3)}{8 α^{2}}$
	Answered by ~blr237~ last updated on 06/Feb/20

	$l e t b e f (α) = \int_{0}^{\infty} x a r g t h (e^{- α x}) d x$ $f (α) e x i s t \Leftrightarrow \forall x > 0 e^{- α x} < 1 \Leftrightarrow α > 0 c a u s e a r g t h x = \frac{1}{2} l n (\frac{1 + x}{1 - x})$ $l e t s t a t e u = e^{- α x} \Leftrightarrow x = - \frac{l n u}{α}$ $f (α) = \frac{1}{2 α^{2}} \int_{0}^{1} - \frac{l n u}{u} l n (\frac{1 + u}{1 - u}) d u$ $2 α^{2} f (α) = \int_{0}^{1} l n u \frac{l n (1 - u)}{u} d u - \int_{0}^{1} l n u \frac{l n (1 + u)}{u} d u$ $2 α^{2} f (α) = - \underset{n = 1}{\sum^{\infty}} \int_{0}^{1} \frac{u^{n - 1}}{n} l n u d u - \underset{n = 1}{\sum^{\infty}} \int_{0}^{1} \frac{{(- u)}^{n - 1}}{n} l n u d u$ $2 α^{2} f (α) = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n^{3}}$ $2 α^{2} f (α) = \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n)}^{3}} + \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{3}} + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{2 n - 1}}{{(2 n)}^{3}} + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{2 n + 1 - 1}}{{(2 n + 1)}^{3}}$ $2 α^{2} f (α) = 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{3}} = 2 (\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} - \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n)}^{3}}) = 2 \times (1 - \frac{1}{8}) \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}}$ $s o f (α) = \frac{7 ζ (3)}{8 α^{2}}$
	Commented by M±th+et£s last updated on 06/Feb/20

	$t h a n k y o u s i r . n i c e s o l u t i o n$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum