0-ln-x-e-x-e-x-dx-

Question Number 155207 by mathdanisur last updated on 26/Sep/21

𝛀 =∫_( 0) ^( ∞) ((ln(x))/(e^x + e^(−x) )) dx = ?

$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{e}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{e}^{−\boldsymbol{\mathrm{x}}} }\:\mathrm{dx}\:=\:? \\ $$

Answered by mindispower last updated on 27/Sep/21

Ω=f′(0),f′(a)=∫_0 ^∞ (x^a /(e^x +e^(−x) ))dx =∫_0 ^∞ x^a Σ_k (−1)^k e^(−(1+2k)x) dx =Σ_(k≥0) ∫_0 ^∞ ((x^a e^(−x) )/((1+2k)^(a+1) ))dx=Σ_(k≥0) (−1)^k ((Γ(1+a))/((1+2k)^(a+1) )) =((Γ(1+a))/2^(1+a) )Σ_(k≥0) (1/((k+(1/4))^(a+1) ))−(1/((k+(3/4))^(a+1) )) =((Γ(1+a))/2^(1+a) )(ζ((1/4),a+1)−ζ((3/4),a+1)) know using Steiljis constante

$$\Omega={f}'\left(\mathrm{0}\right),{f}'\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{a}} }{{e}^{{x}} +{e}^{−{x}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {x}^{{a}} \underset{{k}} {\sum}\left(−\mathrm{1}\right)^{{k}} {e}^{−\left(\mathrm{1}+\mathrm{2}{k}\right){x}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{a}} {e}^{−{x}} }{\left(\mathrm{1}+\mathrm{2}{k}\right)^{{a}+\mathrm{1}} }{dx}=\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} \frac{\Gamma\left(\mathrm{1}+{a}\right)}{\left(\mathrm{1}+\mathrm{2}{k}\right)^{{a}+\mathrm{1}} } \\ $$$$=\frac{\Gamma\left(\mathrm{1}+{a}\right)}{\mathrm{2}^{\mathrm{1}+{a}} }\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{{a}+\mathrm{1}} }−\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{3}}{\mathrm{4}}\right)^{{a}+\mathrm{1}} } \\ $$$$=\frac{\Gamma\left(\mathrm{1}+{a}\right)}{\mathrm{2}^{\mathrm{1}+{a}} }\left(\zeta\left(\frac{\mathrm{1}}{\mathrm{4}},{a}+\mathrm{1}\right)−\zeta\left(\frac{\mathrm{3}}{\mathrm{4}},{a}+\mathrm{1}\right)\right) \\ $$$${know}\:{using}\:{Steiljis}\:{constante} \\ $$$$ \\ $$

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