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Question Number 161706 by mnjuly1970 last updated on 21/Dec/21

J =∫_0 ^( 1) (( 1−x)/(( 1+x +x^( 2) + x^( 3) )ln(x))) dx=?

$$\: \\ $$$$\mathrm{J}\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{1}−{x}}{\left(\:\mathrm{1}+{x}\:+{x}^{\:\mathrm{2}} +\:{x}^{\:\mathrm{3}} \:\right){ln}\left({x}\right)}\:{dx}=? \\ $$$$ \\ $$

Answered by Ar Brandon last updated on 21/Dec/21

f(α)=∫_0 ^1 ((1−x^α )/((1+x+x^2 +x^3 )lnx))dx⇒f(1)=∫_0 ^1 ((1−x)/((1+x+x^2 +x^3 )lnx))dx f ′(α)=−∫_0 ^1 (x^α /(1+x+x^2 +x^3 ))dx=−∫_0 ^1 ((x^α (1−x))/(1−x^4 ))dx =∫_0 ^1 ((x^(α+1) −x^α )/(1−x^4 ))dx=(1/4)∫_0 ^1 ((x^((α/4)−(1/2)) −x^((α/4)−(3/4)) )/(1−x))dx =(1/4)(ψ((α/4)+(1/4))−ψ((α/4)+(1/2))) ⇒f(α)=ln(Γ((α/4)+(1/4)))−ln(Γ((α/4)+(1/2)))+C f(0)=0=ln(Γ((1/4)))−ln(Γ((1/2)))+C ⇒f(α)=ln(Γ((α/4)+(1/4)))−ln(Γ((α/4)+(1/2)))+ln(Γ((1/2)))−ln(Γ((1/4))) ⇒f(1)=ln(Γ((1/2)))−ln(Γ((3/4)))+ln(Γ((1/2)))−ln(Γ((1/4))) =ln(Γ^2 ((1/2)))−ln(Γ((1/4))Γ((3/4)))=lnπ−ln((π/(sin((π/4))))) =lnπ−ln(π(√2))=ln((π/(π(√2))))=−((ln2)/2)

$${f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{\alpha} }{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \right)\mathrm{ln}{x}}{dx}\Rightarrow{f}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \right)\mathrm{ln}{x}}{dx} \\ $$$${f}\:'\left(\alpha\right)=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\alpha} }{\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\alpha} \left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}^{\mathrm{4}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\alpha+\mathrm{1}} −{x}^{\alpha} }{\mathrm{1}−{x}^{\mathrm{4}} }{dx}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\frac{\alpha}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}} −{x}^{\frac{\alpha}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}} }{\mathrm{1}−{x}}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\psi\left(\frac{\alpha}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\right)−\psi\left(\frac{\alpha}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\Rightarrow{f}\left(\alpha\right)=\mathrm{ln}\left(\Gamma\left(\frac{\alpha}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\right)−\mathrm{ln}\left(\Gamma\left(\frac{\alpha}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)+{C} \\ $$$$\:\:\:\:\:{f}\left(\mathrm{0}\right)=\mathrm{0}=\mathrm{ln}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right)−\mathrm{ln}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)+{C} \\ $$$$\Rightarrow{f}\left(\alpha\right)=\mathrm{ln}\left(\Gamma\left(\frac{\alpha}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\right)−\mathrm{ln}\left(\Gamma\left(\frac{\alpha}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)+\mathrm{ln}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)−\mathrm{ln}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right) \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{ln}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)−\mathrm{ln}\left(\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right)+\mathrm{ln}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)−\mathrm{ln}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{ln}\left(\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)−\mathrm{ln}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right)=\mathrm{ln}\pi−\mathrm{ln}\left(\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{ln}\pi−\mathrm{ln}\left(\pi\sqrt{\mathrm{2}}\right)=\mathrm{ln}\left(\frac{\pi}{\pi\sqrt{\mathrm{2}}}\right)=−\frac{\mathrm{ln2}}{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 21/Dec/21

very nice solution mr brandon thank you so much

$$\:\:\:\:\:{very}\:{nice}\:{solution}\:{mr}\:{brandon} \\ $$$${thank}\:{you}\:{so}\:{much} \\ $$

Commented by Ar Brandon last updated on 21/Dec/21

You're welcome Sir.

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