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Question Number 119570 by mnjuly1970 last updated on 25/Oct/20

$. . . \underset{♣}{\overset{♣}{♣}} n i c e c a l c u l u s \underset{♣}{\overset{♣}{♣}} . . .$ $p r o v e t h a t ::$ $Ω = \int_{0}^{\infty} e^{- 2 x} l n (\frac{1 + e^{- x}}{1 - e^{- x}}) = 1$ $. . . ★ M . N . 1970 ★ . . .$
	Answered by mindispower last updated on 25/Oct/20

	$e^{- 2 x} = e^{- x} * e^{- x}$ $l e t t = e^{- x} \Rightarrow d t = - e^{- x} d x$ $\Leftrightarrow= \int_{0}^{1} t l n (\frac{1 + t}{1 - t}) d t = - \int_{0}^{1} t l n (\frac{1 - t}{1 + t}) d t$ $x = \frac{1 - t}{1 + t} \Rightarrow t = \frac{1 - x}{1 + x}, d t = \frac{- 2}{{(1 + x)}^{2}} d x$ $\Leftrightarrow - 2 \int_{0}^{1} \frac{1 - x}{{(1 + x)}^{3}} l n (x) d x$ $= - 2 {[(\frac{1}{(1 + x)} - \frac{1}{{(1 + x)}^{2}}) l n (x)]}_{0}^{1} + 2 \int (\frac{1}{(1 + x) x} - \frac{1}{x {(1 + x)}^{2}}) d x$ $= 2 \int_{0}^{1} \frac{1}{{(1 + x)}^{2}} = 2 {[- \frac{1}{(1 + x)}]}_{0}^{1} = 2 . \frac{1}{2} = 1$
	Commented by mnjuly1970 last updated on 25/Oct/20

	$t h a n k y o u m r p o w e r . .$
	Answered by mathmax by abdo last updated on 25/Oct/20
	$I =∫_0 ^∞ e^(−2x) ln(((1+e^(−x) )/(1−e^(−x) )))dx ⇒I =∫_0 ^∞ e^(−2x) ln(1+e^(−x) )dx−∫_0 ^∞ e^(−2x) ln(1−e^(−x) )dx =H−K H =∫_0 ^∞ e^(−2x) ln(1+e^(−x) )dx =_(e^(−x) =t) −∫_0 ^1 t^2 ln(1+t)(((−dt)/t))dt =∫_0 ^1 tln(1+t)dt =[(t^2 /2)ln(1+t)]_0 ^1 −∫_0 ^1 (t^2 /(2(1+t)))dt =((ln2)/2)−(1/2)∫_0 ^1 ((t^2 −1+1)/(t+1))dt =((ln2)/2)−(1/2){∫_0 ^1 (t−1)dt+∫_0 ^1 (dt/(t+1))} =((ln2)/2)−(1/2){[(t^2 /2)−t]_0 ^1 +ln(2)} =−(1/2)(−(1/2))=(1/4) K=∫_0 ^∞ e^(−2x) ln(1−e^(−x) )dx =_(e^(−x) =t) −∫_0 ^1 t^2 ln(1−t)(((−dt)/t)) =∫_0 ^1 tln(1−t)dt =[(1/2)(t^2 −1)ln(1−t)]_0 ^1 −∫_0 ^1 ((t^2 −1)/2)×((−1)/(1−t))dt =−(1/2)∫_0 ^1 (t+1)dt =−(1/2)[(t^2 /2)+t]_0 ^1 =−(1/2)((3/2))=−(3/4) ⇒ I =H−K =(1/4)+(3/4) ⇒ ★ I =1★$
	$I = \int_{0}^{\infty} e^{- 2 x} \ln (\frac{1 + e^{- x}}{1 - e^{- x}}) dx \Rightarrow I = \int_{0}^{\infty} e^{- 2 x} \ln (1 + e^{- x}) dx - \int_{0}^{\infty} e^{- 2 x} \ln (1 - e^{- x}) dx$ $= H - K$ $H = \int_{0}^{\infty} e^{- 2 x} \ln (1 + e^{- x}) dx =_{e^{- x} = t} - \int_{0}^{1} t^{2} \ln (1 + t) (\frac{- dt}{t}) dt$ $= \int_{0}^{1} tln (1 + t) dt = {[\frac{t^{2}}{2} \ln (1 + t)]}_{0}^{1} - \int_{0}^{1} \frac{t^{2}}{2 (1 + t)} dt$ $= \frac{\ln 2}{2} - \frac{1}{2} \int_{0}^{1} \frac{t^{2} - 1 + 1}{t + 1} dt = \frac{\ln 2}{2} - \frac{1}{2} {\int_{0}^{1} (t - 1) dt + \int_{0}^{1} \frac{dt}{t + 1}}$ $= \frac{\ln 2}{2} - \frac{1}{2} {{[\frac{t^{2}}{2} - t]}_{0}^{1} + \ln (2)} = - \frac{1}{2} (- \frac{1}{2}) = \frac{1}{4}$ $K = \int_{0}^{\infty} e^{- 2 x} \ln (1 - e^{- x}) dx =_{e^{- x} = t} - \int_{0}^{1} t^{2} \ln (1 - t) (\frac{- dt}{t})$ $= \int_{0}^{1} tln (1 - t) dt = {[\frac{1}{2} (t^{2} - 1) \ln (1 - t)]}_{0}^{1} - \int_{0}^{1} \frac{t^{2} - 1}{2} \times \frac{- 1}{1 - t} dt$ $= - \frac{1}{2} \int_{0}^{1} (t + 1) dt = - \frac{1}{2} {[\frac{t^{2}}{2} + t]}_{0}^{1} = - \frac{1}{2} (\frac{3}{2}) = - \frac{3}{4} \Rightarrow$ $I = H - K = \frac{1}{4} + \frac{3}{4} \Rightarrow ★ I = 1 ★$
	Commented by mnjuly1970 last updated on 25/Oct/20

	$t h a n k y o u m r m a x . .$
	Commented by Bird last updated on 25/Oct/20

	$y o u a r e w e l c o m e$
	Answered by mnjuly1970 last updated on 25/Oct/20

	Commented by mnjuly1970 last updated on 25/Oct/20

	$e d i t t i o n :$ $1 : r e c a l l :: l n (\frac{1 + z}{1 - z}) = 2 \underset{n = 0}{\sum^{\infty}} \frac{z^{2 n + 1}}{2 n + 1}$ $2 : Ω \overset{t e l e s c o p i c s e r i e s}{=} \frac{1}{2 (0) + 1} - {l i m}_{n \to \infty} (\frac{1}{2 n + 3}) = 1$ $. . . m . n . j u l y . 1970 . . .$
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