Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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..

$$\:\:{zendeh}\:{bashid}\left({be}\:{alive}\right)\:{mr}\:{payan}\:{grateful}.. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com