prove that : Ω:= ∫_0 ^( ∞) ((1−e^(−x) )/(1+e^(2x) )) .(dx/x) =ln(((Γ^2 ((1/4)))/(4(√(2π)))) ) Θ:= Π_(n=1) ^∞ (((2n+1)/(2n)))^((−1)^(n+1) ) =^(??) e^Ω


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Question Number 140055 by mnjuly1970 last updated on 03/May/21

$p r o v e t h a t :$ $Ω := \int_{0}^{\infty} \frac{1 - e^{- x}}{1 + e^{2 x}} . \frac{d x}{x} = l n (\frac{Γ^{2} (\frac{1}{4})}{4 \sqrt{2 π}})$ $Θ := \underset{n = 1}{\prod^{\infty}} {(\frac{2 n + 1}{2 n})}^{{(- 1)}^{n + 1}} \overset{? ?}{=} e^{Ω}$
	Answered by Kamel last updated on 03/May/21

	Commented by mnjuly1970 last updated on 04/May/21

	$t h a n k y o u s o m u c h m r K a m e l$
	Answered by Dwaipayan Shikari last updated on 03/May/21

	$ϑ (α) = \int_{0}^{\infty} \frac{1 - e^{- α x}}{1 + e^{2 x}} . \frac{d x}{x}$ $ϑ^{'} (α) = \int_{0}^{\infty} \frac{e^{- α x}}{1 + e^{2 x}} d x = \underset{n = 1}{\sum^{\infty}} \int_{0}^{\infty} {(- 1)}^{n + 1} e^{- 2 n x - α x}$ $= \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{2 n + α} = (\frac{1}{2 + α} - \frac{1}{4 + α} + \frac{1}{6 + α} - \frac{1}{8 + α} + \frac{1}{10 + α} - \frac{1}{12 + α} + . .)$ $= \frac{1}{4} (\underset{n = 0}{\sum^{\infty}} \frac{1}{n + \frac{1}{2} + \frac{α}{4}} - \frac{1}{n + 1 + \frac{α}{4}}) = \frac{1}{4} (ψ (\frac{α}{4} + 1) - ψ (\frac{α}{4} + \frac{1}{2}))$ $ϑ (α) = l o g (Γ (\frac{α}{4} + 1)) - l o g (Γ (\frac{α}{4} + \frac{1}{2})) + C$ $ϑ (0) = l o g (Γ (1)) - l o g (Γ (\frac{1}{2})) + C = 0 \Rightarrow C = l o g (\sqrt{π})$ $ϑ (1) = l o g (\frac{Γ (\frac{5}{4}) \sqrt{π}}{Γ (\frac{3}{4})}) = l o g (\frac{Γ^{2} (\frac{1}{4})}{\sqrt{π}} . \frac{\sqrt{2} π}{4}) = l o g (\frac{Γ^{2} (\frac{1}{4})}{4 \sqrt{2 π}})$
	Commented by mnjuly1970 last updated on 03/May/21

	$z e n d e h b a s h i d (b e a l i v e) m r p a y a n g r a t e f u l . .$
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