ln-x-4-1-x-4-1-dx-

Question Number 152468 by talminator2856791 last updated on 28/Aug/21

$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{−\infty} ^{\:\infty} \:\frac{\mathrm{ln}\left(\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}\right)}{\:\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:\:{dx} \\ $$$$\: \\ $$

Answered by mindispower last updated on 28/Aug/21

∫_0 ^∞ ((ln(1+x^4 ))/( (√(1+x^4 ))))dx x^4 =u ∫_0 ^∞ ((ln(1+u))/( (√(1+u)))).u^(−(3/4)) (du/4)...A β(x,y)=∫_0 ^∞ (t^(y−1) /((1+t)^(x+y) ))dt ∂_x β(x,y)=∫_0 ^∞ ((t^(y−1) .−ln(1+t))/((1+t)^(x+y) ))dt A=−(1/4)β_x ((1/4),(1/4))=−(1/4)β((1/4),(1/4))(Ψ((1/4))−Ψ((1/2))) =−(1/4).((Γ^2 ((1/4)))/(Γ((1/2))))(Ψ((1/4))−Ψ((1/2))) =((Γ((1/4))Γ((5/4)))/(Γ((1/2))))((π/2)−ln(2))=((Γ((1/4))Γ((5/4)))/( (√π)))((π/2)−ln(2)) ∫_(−∞) ^∞ ((ln((√(1+x^4 ))))/( (√(1+x^4 ))))dx=((Γ((1/4))Γ((5/4)))/( (√π)))((π/2)−ln(2))

$$\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}{dx} \\ $$$${x}^{\mathrm{4}} ={u} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{u}\right)}{\:\sqrt{\mathrm{1}+{u}}}.{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \frac{{du}}{\mathrm{4}}…{A} \\ $$$$\beta\left({x},{y}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{y}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{{x}+{y}} }{dt} \\ $$$$\partial_{{x}} \beta\left({x},{y}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{y}−\mathrm{1}} .−{ln}\left(\mathrm{1}+{t}\right)}{\left(\mathrm{1}+{t}\right)^{{x}+{y}} }{dt} \\ $$$${A}=−\frac{\mathrm{1}}{\mathrm{4}}\beta_{{x}} \left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{4}}\right)=−\frac{\mathrm{1}}{\mathrm{4}}\beta\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}.\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\left(\frac{\pi}{\mathrm{2}}−{ln}\left(\mathrm{2}\right)\right)=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}{\:\sqrt{\pi}}\left(\frac{\pi}{\mathrm{2}}−{ln}\left(\mathrm{2}\right)\right) \\ $$$$\int_{−\infty} ^{\infty} \frac{{ln}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }\right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}{dx}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}{\:\sqrt{\pi}}\left(\frac{\pi}{\mathrm{2}}−{ln}\left(\mathrm{2}\right)\right) \\ $$$$ \\ $$$$ \\ $$

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