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Question Number 142275 by mnjuly1970 last updated on 29/May/21

.......Advanced ...∗∗∗∗∗ ...Integral...... Prove that :: Φ :=∫_0 ^( 1) ((1−x)/((1−x+x^2 )log(x)))dx= proof:: Φ:=∫_0 ^( 1) ((1−x^2 )/((1−x^3 )log(x)))dx f (a):= ∫_0 ^( 1) ((1−x^a )/((1−x^3 )log(x))) Φ := f (2) ........✓ f ′(a):=∫_0 ^( 1) ((−x^a log(x))/((1−x^3 )log(x)))=∫_0 ^( 1) ((−x^a )/(1−x^3 ))dx (★) (★):: x^3 =y ⇒ f ′(a):=(1/3)∫_0 ^( 1) ((y^((−2)/3) −y^((a/3)−(2/3)) )/(1−y))dy :=(1/3)∫_0 ^( 1) ((y^((−2)/3) −1+1−y^((a/3)−(1/3)) )/(1−y))dy :=(1/3)(ψ((a/3)+(2/3))−ψ((2/3))) f (a):=log(Γ((a/3)+(2/3)))−(a/3)ψ((2/3))+C f (0):=0=log(Γ((2/3)))+C C :=−log(Γ((2/3))) Φ:= f (2)=log(Γ((4/3)))−(2/3) ψ((2/3))−log(Γ((2/3))) :=log(((Γ((4/3)))/(Γ((2/3)))))−(2/3)ψ((2/3)) ....✓

$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:.......{Advanced}\:\:...\ast\ast\ast\ast\ast\:...{Integral}...... \\ $$$$\:\:\:\:\:\:{Prove}\:\:{that}\:::\:\:\:\Phi\::=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{x}}{\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right){log}\left({x}\right)}{dx}= \\ $$$$\:\:\:{proof}:: \\ $$$$\:\:\:\:\:\:\Phi:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{3}} \right){log}\left({x}\right)}{dx} \\ $$$$\:\:\:\:\:\:{f}\:\left({a}\right):=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{x}^{{a}} }{\left(\mathrm{1}−{x}^{\mathrm{3}} \right){log}\left({x}\right)} \\ $$$$\:\:\:\:\:\:\:\Phi\::=\:{f}\:\left(\mathrm{2}\right)\:........\checkmark \\ $$$$\:\:\:\:\:\:\:{f}\:'\left({a}\right):=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{−{x}^{{a}} {log}\left({x}\right)}{\left(\mathrm{1}−{x}^{\mathrm{3}} \right){log}\left({x}\right)}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{−{x}^{{a}} }{\mathrm{1}−{x}^{\mathrm{3}} }{dx}\:\:\left(\bigstar\right) \\ $$$$\:\:\:\:\left(\bigstar\right)::\:\:{x}^{\mathrm{3}} ={y}\:\Rightarrow\:{f}\:'\left({a}\right):=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{y}^{\frac{−\mathrm{2}}{\mathrm{3}}} −{y}^{\frac{{a}}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{1}−{y}}{dy} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{y}^{\frac{−\mathrm{2}}{\mathrm{3}}} −\mathrm{1}+\mathrm{1}−{y}^{\frac{{a}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{1}−{y}}{dy} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{3}}\left(\psi\left(\frac{{a}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\right)−\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right) \\ $$$$\:\:\:\:\:\:\:{f}\:\left({a}\right):={log}\left(\Gamma\left(\frac{{a}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\right)\right)−\frac{{a}}{\mathrm{3}}\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+{C} \\ $$$$\:\:\:\:\:\:\:\:{f}\:\left(\mathrm{0}\right):=\mathrm{0}={log}\left(\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right)+{C} \\ $$$$\:\:\:\:\:\:\:\:\:{C}\::=−{log}\left(\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\Phi:=\:{f}\:\left(\mathrm{2}\right)={log}\left(\Gamma\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\right)−\frac{\mathrm{2}}{\mathrm{3}}\:\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−{log}\left(\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\::={log}\left(\frac{\Gamma\left(\frac{\mathrm{4}}{\mathrm{3}}\right)}{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}\right)−\frac{\mathrm{2}}{\mathrm{3}}\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:....\checkmark \\ $$$$\:\:\:\:\:\:\:\: \\ $$

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