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 151819 by mnjuly1970 last updated on 23/Aug/21

1. prove that : ∫_0 ^( ∞) (( e^( t) .ln(t ))/((1 + e^( t) )^( 2) )) dt=(1/2)(ln((π/2) )− γ )

$$ \\ $$$$\:\:\:\:\:\mathrm{1}.\:{prove}\:{that}\:: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \frac{\:{e}^{\:{t}} .{ln}\left({t}\:\right)}{\left(\mathrm{1}\:+\:{e}^{\:{t}} \right)^{\:\mathrm{2}} }\:{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\frac{\pi}{\mathrm{2}}\:\right)−\:\gamma\:\right)\: \\ $$$$\:\:\: \\ $$

Answered by Lordose last updated on 23/Aug/21

Ω = ∫_0 ^( ∞) ((e^t log(t))/((1+e^t )^2 ))dt = ∫_0 ^( ∞) ((e^(−t) log(t))/((1+e^(−t) )^2 ))dt Ω(a) = ∫_0 ^( ∞) ((t^a e^(−t) )/((1+e^(−t) )^2 ))dt = Σ_(n=1) ^∞ (−1)^(n+1) n∫_0 ^( ∞) t^a e^(−nt) dt Ω(a) =^(t=(x/n)) Σ_(n=1) ^∞ (((−1)^(n+1) )/n^a )∫_0 ^( ∞) x^(a+1−1) e^(−x) dx = 𝚪(a+1)𝛈(a) Ω′(a) = 𝚪(a+1)(𝛈′(a) + 𝛈(a)𝛙^((0)) (a+1)) Ω = Ω′(0) 𝛙^((0)) (1) = −𝛄, 𝛈(0) = (1/2), 𝛈′(0) = (1/2)log((𝛑/2)) 𝛀 = (1/2)(log((𝛑/2)) − 𝛄)

$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{e}^{\mathrm{t}} \mathrm{log}\left(\mathrm{t}\right)}{\left(\mathrm{1}+\mathrm{e}^{\mathrm{t}} \right)^{\mathrm{2}} }\mathrm{dt}\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{e}^{−\mathrm{t}} \mathrm{log}\left(\mathrm{t}\right)}{\left(\mathrm{1}+\mathrm{e}^{−\mathrm{t}} \right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$\Omega\left(\mathrm{a}\right)\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{t}^{\mathrm{a}} \mathrm{e}^{−\mathrm{t}} }{\left(\mathrm{1}+\mathrm{e}^{−\mathrm{t}} \right)^{\mathrm{2}} }\mathrm{dt}\:=\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} \mathrm{n}\int_{\mathrm{0}} ^{\:\infty} \mathrm{t}^{\mathrm{a}} \mathrm{e}^{−\mathrm{nt}} \mathrm{dt} \\ $$$$\Omega\left(\mathrm{a}\right)\:\overset{\mathrm{t}=\frac{\mathrm{x}}{\mathrm{n}}} {=}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}^{\mathrm{a}} }\int_{\mathrm{0}} ^{\:\infty} \mathrm{x}^{\mathrm{a}+\mathrm{1}−\mathrm{1}} \mathrm{e}^{−\mathrm{x}} \mathrm{dx}\:=\:\boldsymbol{\Gamma}\left(\mathrm{a}+\mathrm{1}\right)\boldsymbol{\eta}\left(\mathrm{a}\right) \\ $$$$\Omega'\left(\mathrm{a}\right)\:=\:\boldsymbol{\Gamma}\left(\mathrm{a}+\mathrm{1}\right)\left(\boldsymbol{\eta}'\left(\mathrm{a}\right)\:+\:\boldsymbol{\eta}\left(\mathrm{a}\right)\boldsymbol{\psi}^{\left(\mathrm{0}\right)} \left(\mathrm{a}+\mathrm{1}\right)\right) \\ $$$$\Omega\:=\:\Omega'\left(\mathrm{0}\right) \\ $$$$\boldsymbol{\psi}^{\left(\mathrm{0}\right)} \left(\mathrm{1}\right)\:=\:−\boldsymbol{\gamma},\:\boldsymbol{\eta}\left(\mathrm{0}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:\boldsymbol{\eta}'\left(\mathrm{0}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{log}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}\right) \\ $$$$\boldsymbol{\Omega}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{\mathrm{log}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)\:−\:\boldsymbol{\gamma}\right) \\ $$

Commented by mnjuly1970 last updated on 23/Aug/21

bravo mr lordose excellent... grateful..

$${bravo}\:{mr}\:{lordose}\:{excellent}... \\ $$$${grateful}.. \\ $$

Commented by Tawa11 last updated on 23/Aug/21

Weldone sir

$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

Commented by Tawa11 last updated on 23/Aug/21

Sir, please prescribe a book I can learn some stuffs like this. Thanks sir. I just want to be learning this sir.

$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{prescribe}\:\mathrm{a}\:\mathrm{book}\:\mathrm{I}\:\mathrm{can}\:\mathrm{learn}\:\mathrm{some}\:\mathrm{stuffs}\:\mathrm{like}\:\mathrm{this}. \\ $$$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{just}\:\mathrm{want}\:\mathrm{to}\:\mathrm{be}\:\mathrm{learning}\:\mathrm{this}\:\mathrm{sir}. \\ $$

Commented by Lordose last updated on 23/Aug/21

Advanced calculus explored Advanced Engineering Mathematics

$$\mathrm{Advanced}\:\mathrm{calculus}\:\mathrm{explored} \\ $$$$\mathrm{Advanced}\:\mathrm{Engineering}\:\mathrm{Mathematics} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com