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Question Number 140055 by mnjuly1970 last updated on 03/May/21
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
$$\:\:\:\:\:\: \\ $$$$\:\:{prove}\:\:{that}\:: \\ $$$$\:\:\:\:\:\:\:\:\Omega:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−{e}^{−{x}} }{\mathrm{1}+{e}^{\mathrm{2}{x}} }\:.\frac{{dx}}{{x}}\:={ln}\left(\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}\sqrt{\mathrm{2}\pi}}\:\right) \\ $$$$\:\Theta:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}}\right)^{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} } \overset{??} {=}\:{e}^{\Omega} \\ $$$$\:\:\:\:\:\:\: \\ $$
Answered by Kamel last updated on 03/May/21
Commented by mnjuly1970 last updated on 04/May/21
thank you so much mr Kamel
$${thank}\:{you}\:{so}\:{much}\:{mr}\:{Kamel} \\ $$
Answered by Dwaipayan Shikari last updated on 03/May/21
ϑ(α)=∫_0 ^∞ ((1−e^(−αx) )/(1+e^(2x) )).(dx/x) ϑ′(α)=∫_0 ^∞ (e^(−αx) /(1+e^(2x) ))dx=Σ_(n=1) ^∞ ∫_0 ^∞ (−1)^(n+1) e^(−2nx−αx) =Σ_(n=1) ^∞ (((−1)^(n+1) )/(2n+α))=((1/(2+α))−(1/(4+α))+(1/(6+α))−(1/(8+α))+(1/(10+α))−(1/(12+α))+..) =(1/4)(Σ_(n=0) ^∞ (1/(n+(1/2)+(α/4)))−(1/(n+1+(α/4))))=(1/4)(ψ((α/4)+1)−ψ((α/4)+(1/2))) ϑ(α)=log(Γ((α/4)+1))−log(Γ((α/4)+(1/2)))+C ϑ(0)=log(Γ(1))−log(Γ((1/2)))+C=0⇒C=log((√π)) ϑ(1)=log(((Γ((5/4))(√π))/(Γ((3/4)))))=log(((Γ^2 ((1/4)))/( (√π))).(((√2)π)/4))=log(((Γ^2 ((1/4)))/(4(√(2π)))))
$$\vartheta\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{e}^{−\alpha{x}} }{\mathrm{1}+{e}^{\mathrm{2}{x}} }.\frac{{dx}}{{x}} \\ $$$$\vartheta'\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\alpha{x}} }{\mathrm{1}+{e}^{\mathrm{2}{x}} }{dx}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {e}^{−\mathrm{2}{nx}−\alpha{x}} \: \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}{n}+\alpha}=\left(\frac{\mathrm{1}}{\mathrm{2}+\alpha}−\frac{\mathrm{1}}{\mathrm{4}+\alpha}+\frac{\mathrm{1}}{\mathrm{6}+\alpha}−\frac{\mathrm{1}}{\mathrm{8}+\alpha}+\frac{\mathrm{1}}{\mathrm{10}+\alpha}−\frac{\mathrm{1}}{\mathrm{12}+\alpha}+..\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\alpha}{\mathrm{4}}}−\frac{\mathrm{1}}{{n}+\mathrm{1}+\frac{\alpha}{\mathrm{4}}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\psi\left(\frac{\alpha}{\mathrm{4}}+\mathrm{1}\right)−\psi\left(\frac{\alpha}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\vartheta\left(\alpha\right)={log}\left(\Gamma\left(\frac{\alpha}{\mathrm{4}}+\mathrm{1}\right)\right)−{log}\left(\Gamma\left(\frac{\alpha}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)+{C} \\ $$$$\vartheta\left(\mathrm{0}\right)={log}\left(\Gamma\left(\mathrm{1}\right)\right)−{log}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)+{C}=\mathrm{0}\Rightarrow{C}={log}\left(\sqrt{\pi}\right) \\ $$$$\vartheta\left(\mathrm{1}\right)={log}\left(\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}\right)\sqrt{\pi}}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\right)={log}\left(\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\:\sqrt{\pi}}.\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{4}}\right)={log}\left(\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}\sqrt{\mathrm{2}\pi}}\right) \\ $$
Commented by mnjuly1970 last updated on 03/May/21
zendeh bashid(be alive) mr payan grateful..
$$\:\:{zendeh}\:{bashid}\left({be}\:{alive}\right)\:{mr}\:{payan}\:{grateful}.. \\ $$

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