prove-that-0-1-e-x-1-e-2x-dx-x-ln-2-1-4-4-2pi-n-1-2n-1-2n-1-n-1-e-

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 . .