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Question Number 130167 by Lordose last updated on 23/Jan/21

∫_0 ^( ∞) ((ln(u)e^(−u) )/((1+e^(−u) )^2 ))du

$$\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{ln}\left(\mathrm{u}\right)\mathrm{e}^{−\mathrm{u}} }{\left(\mathrm{1}+\mathrm{e}^{−\mathrm{u}} \right)^{\mathrm{2}} }\mathrm{du} \\ $$$$ \\ $$$$ \\ $$

Answered by mindispower last updated on 26/Jan/21

∫_0 ^∞ (x^s /((1+e^(−x) )))e^(−x) dx=f(s) (1/(1+x))=Σ(−1)^k x^k ⇒(x/((1+x)^2 ))=Σ_(k≥1) k(−1)^(k+1) x^k ∫_0 ^∞ x^s Σ_(k≥1) k(−1)^(k+1) e^(−kx) dx kx=t ∫_0 ^∞ Σ_(k≥1) ((t^s (−1)^(k+1) e^(−x) dx)/k^s ) =Σ_(k≥1) (((−1)^(k+1) )/k^s )∫_0 ^∞ t^s e^(−x) dx =Σ_(k≥1) (((−1)^(k+1) )/k^s )Γ(s+1)=Γ(s+1)(1−(1/2^(s−1) ))ζ(s) f(s)=Γ(s+1)(1−(1/2^(s−1) ))ζ(s) f′(s)=(Ψ(s+1)(1−(1/2^(s−1) ))+Γ(s+1)(ln(2)2^(1−s) ))ζ(s) +ζ′(s)(Γ(s+1)(1−(1/2^(s−1) ))) Γ(1)=1,Ψ(1)=−γ ζ(0)=−(1/2),ζ′(0)=−((ln(2π))/2) we get (−γ(−1)+2ln(2)).−(1/2)+−((ln(2π))/2)(−2) =(γ/2)+ln(2π)−ln(2)=(γ/2)+ln(((2π)/2))=(γ/2)+ln(π)

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{s}} }{\left(\mathrm{1}+{e}^{−{x}} \right)}{e}^{−{x}} {dx}={f}\left({s}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}}=\Sigma\left(−\mathrm{1}\right)^{{k}} {x}^{{k}} \\ $$$$\Rightarrow\frac{{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }=\underset{{k}\geqslant\mathrm{1}} {\sum}{k}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} {x}^{{k}} \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{{s}} \underset{{k}\geqslant\mathrm{1}} {\sum}{k}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} {e}^{−{kx}} {dx} \\ $$$${kx}={t} \\ $$$$\int_{\mathrm{0}} ^{\infty} \underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{t}^{{s}} \left(−\mathrm{1}\right)^{{k}+\mathrm{1}} {e}^{−{x}} {dx}}{{k}^{{s}} } \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{k}^{{s}} }\int_{\mathrm{0}} ^{\infty} {t}^{{s}} {e}^{−{x}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{k}^{{s}} }\Gamma\left({s}+\mathrm{1}\right)=\Gamma\left({s}+\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{s}−\mathrm{1}} }\right)\zeta\left({s}\right) \\ $$$${f}\left({s}\right)=\Gamma\left({s}+\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{s}−\mathrm{1}} }\right)\zeta\left({s}\right) \\ $$$${f}'\left({s}\right)=\left(\Psi\left({s}+\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{s}−\mathrm{1}} }\right)+\Gamma\left({s}+\mathrm{1}\right)\left({ln}\left(\mathrm{2}\right)\mathrm{2}^{\mathrm{1}−{s}} \right)\right)\zeta\left({s}\right) \\ $$$$+\zeta'\left({s}\right)\left(\Gamma\left({s}+\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{s}−\mathrm{1}} }\right)\right) \\ $$$$\Gamma\left(\mathrm{1}\right)=\mathrm{1},\Psi\left(\mathrm{1}\right)=−\gamma \\ $$$$\zeta\left(\mathrm{0}\right)=−\frac{\mathrm{1}}{\mathrm{2}},\zeta'\left(\mathrm{0}\right)=−\frac{{ln}\left(\mathrm{2}\pi\right)}{\mathrm{2}} \\ $$$${we}\:{get} \\ $$$$\left(−\gamma\left(−\mathrm{1}\right)+\mathrm{2}{ln}\left(\mathrm{2}\right)\right).−\frac{\mathrm{1}}{\mathrm{2}}+−\frac{{ln}\left(\mathrm{2}\pi\right)}{\mathrm{2}}\left(−\mathrm{2}\right) \\ $$$$=\frac{\gamma}{\mathrm{2}}+{ln}\left(\mathrm{2}\pi\right)−{ln}\left(\mathrm{2}\right)=\frac{\gamma}{\mathrm{2}}+{ln}\left(\frac{\mathrm{2}\pi}{\mathrm{2}}\right)=\frac{\gamma}{\mathrm{2}}+{ln}\left(\pi\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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