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0-1-log-1-x-x-2-x-3-x-4-1-1-x-1-x-2-1-x-3-1-x-4-dx-

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Question Number 77570 by TawaTawa last updated on 08/Jan/20
∫_( 0) ^( 1) log(((1 + x + x^2 + x^3 + x^4 )/( (√(1 + (1/x) + (1/x^2 ) + (1/x^3 ) + (1/x^4 )))))) dx
$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\:\mathrm{log}\left(\frac{\mathrm{1}\:+\:\mathrm{x}\:+\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{x}^{\mathrm{4}} }{\:\sqrt{\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }}}\right)\:\mathrm{dx} \\ $$
Answered by MJS last updated on 08/Jan/20
=∫_0 ^1 ln (x^2 (√(x^4 +x^3 +x^2 +x+1))) dx= =2∫_0 ^1 ln x dx+(1/2)∫_0 ^1 ln (x^4 +x^3 +x^2 +x+1) dx 2∫_0 ^1 ln x dx=2[x(ln x −1)]_0 ^1 = [lim_(x→0) x(ln x −1) =0] =−2 (1/2)∫_0 ^1 ln (x^4 +x^3 +x^2 +x+1) dx= [α_j =e^(i((2πj)/5)) ; j=1, 2, 3, 4] =(1/2)∫_0 ^1 ln (Π_(j=1) ^4 (x−α_j )) dx= =(1/2)Σ_(j=1) ^4 (∫_0 ^1 ln (x−α_j ) dx)= =(1/2)[Σ_(j=1) ^4 ((x−α_j )ln (x−α_j ))−2x]_0 ^1 =... ...=((√5)/4)ln ((1+(√5))/2) +(5/8)ln 5 +((√(25+10(√5)))/(20))π−2 ⇒ answer is ((√5)/4)ln ((1+(√5))/2) +(5/8)ln 5 +((√(25+10(√5)))/(20))π−4
$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{ln}\:\left({x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right)\:{dx}= \\ $$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{ln}\:{x}\:{dx}+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{ln}\:\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:{dx} \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{ln}\:{x}\:{dx}=\mathrm{2}\left[{x}\left(\mathrm{ln}\:{x}\:−\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} = \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}\left(\mathrm{ln}\:{x}\:−\mathrm{1}\right)\:=\mathrm{0}\right] \\ $$$$\:\:\:\:\:=−\mathrm{2} \\ $$$$ \\ $$$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{ln}\:\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\alpha_{{j}} =\mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi{j}}{\mathrm{5}}} ;\:{j}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\right] \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{ln}\:\left(\underset{{j}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}\left({x}−\alpha_{{j}} \right)\right)\:{dx}= \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\underset{{j}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\left(\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{ln}\:\left({x}−\alpha_{{j}} \right)\:{dx}\right)= \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\underset{{j}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\left(\left({x}−\alpha_{{j}} \right)\mathrm{ln}\:\left({x}−\alpha_{{j}} \right)\right)−\mathrm{2}{x}\right]_{\mathrm{0}} ^{\mathrm{1}} =… \\ $$$$\:\:\:\:\:…=\frac{\sqrt{\mathrm{5}}}{\mathrm{4}}\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:+\frac{\mathrm{5}}{\mathrm{8}}\mathrm{ln}\:\mathrm{5}\:+\frac{\sqrt{\mathrm{25}+\mathrm{10}\sqrt{\mathrm{5}}}}{\mathrm{20}}\pi−\mathrm{2} \\ $$$$ \\ $$$$\Rightarrow \\ $$$$\mathrm{answer}\:\mathrm{is}\:\frac{\sqrt{\mathrm{5}}}{\mathrm{4}}\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:+\frac{\mathrm{5}}{\mathrm{8}}\mathrm{ln}\:\mathrm{5}\:+\frac{\sqrt{\mathrm{25}+\mathrm{10}\sqrt{\mathrm{5}}}}{\mathrm{20}}\pi−\mathrm{4} \\ $$
Commented by TawaTawa last updated on 08/Jan/20
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by TawaTawa last updated on 12/Jan/20
Please sir. How did you get: e^(i((2πj)/5) ) ; j = 1, 2, 3, 4
$$\mathrm{Please}\:\mathrm{sir}.\:\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}:\:\:\:\:\mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi\mathrm{j}}{\mathrm{5}}\:} ;\:\:\:\:\:\:\:\mathrm{j}\:\:=\:\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4} \\ $$
Commented by MJS last updated on 12/Jan/20
(x^4 +x^3 +x^2 +x+1)(x−1)=x^5 −1 ⇒ the roots are the 5^(th) roots of unity =e^(i((2πj)/5)) with j=0, 1, 2, 3, 4 but j=0 ⇔ x−1=0
$$\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)={x}^{\mathrm{5}} −\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{the}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity} \\ $$$$=\mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi{j}}{\mathrm{5}}} \:\mathrm{with}\:{j}=\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\:\mathrm{but}\:{j}=\mathrm{0}\:\Leftrightarrow\:{x}−\mathrm{1}=\mathrm{0} \\ $$

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