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

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