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Question-116996

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Question Number 116996 by mnjuly1970 last updated on 08/Oct/20
Answered by mindispower last updated on 09/Oct/20
the last is li_3 ((1/x)−1)∫?
$${the}\:{last}\:{is}\:{li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)\int? \\ $$
Commented by mnjuly1970 last updated on 09/Oct/20
no li_3 (1−(1/x))
$${no}\: \\ $$$${li}_{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right) \\ $$
Commented by mnjuly1970 last updated on 09/Oct/20
it′s proof is very nice .
$${it}'{s}\:{proof}\:\:{is}\:\:{very}\:\:{nice}\:. \\ $$
Answered by mindispower last updated on 19/Oct/20
(d/dx)Li_(s+1) (x)=((li_s (x))/x) f(x)=li_3 (x)+li_3 (1−x)+li_3 (1−(1/x)) f′(x)=((li_2 (x))/x)−((li_2 (1−x))/(1−x))+(1/x^2 )Li_2 (1−(1/x)).(x/(x−1)) =(((1−x)li_2 (x)−xli_2 (1−x)−li_2 (1−(1/x)))/(x(1−x))) Li_2 (x)+li_2 (1−x)=(π^2 /6)−ln(x)ln(1−x)..1 li_2 (1−x)+li_2 (1−(1/x))=−((ln^2 (x))/2)..2 f′(x)=(((1−x)(li_2 (x)+li_2 (1−x))−li_2 (1−x)−li_2 (1−(1/x)))/(x(1−x))) =(π^2 /(6x))+((ln^2 (x))/(2x(1−x))) f(x)=(π^2 /6)ln(x)+∫ln^2 (x)(((x+(1−x))/(2x(1−x))))dx =(π^2 /6)ln(x)+(1/2)∫((ln^2 (x))/x)dx+∫((ln^2 (x))/(2(1−x)))dx_(=A) A by part =(π^2 /6)ln(x)+(1/6)ln^3 (x)+−((ln(1−x)ln^2 (x))/2)+/∫ln(x)ln(1−x).(1/x)/_(missing one ) f(x)=Li_3 (x)+Li_3 (1−(1/x))+li_3 (1−x) =(π^2 /6)ln(x)+((ln^3 (x))/6)−((ln^2 (x)ln(1−x))/2)+c lim_(x→1) f(x)=Li_3 (1)=Σ_(n≥1) (1/n^3 )=ζ(3)=c ⇒Li_3 (x)+li_3 (1−x)+li_3 (1−(1/x)) =ζ(3)+(π^2 /6)ln(x)+((ln^3 (x))/6)−((ln^2 (x)ln(1−x))/2) what we want too bee contibued the A integral missing
$$\frac{{d}}{{dx}}{Li}_{{s}+\mathrm{1}} \left({x}\right)=\frac{{li}_{{s}} \left({x}\right)}{{x}} \\ $$$${f}\left({x}\right)={li}_{\mathrm{3}} \left({x}\right)+{li}_{\mathrm{3}} \left(\mathrm{1}−{x}\right)+{li}_{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right) \\ $$$${f}'\left({x}\right)=\frac{{li}_{\mathrm{2}} \left({x}\right)}{{x}}−\frac{{li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{Li}_{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right).\frac{{x}}{{x}−\mathrm{1}} \\ $$$$=\frac{\left(\mathrm{1}−{x}\right){li}_{\mathrm{2}} \left({x}\right)−{xli}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)−{li}_{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)}{{x}\left(\mathrm{1}−{x}\right)} \\ $$$${Li}_{\mathrm{2}} \left({x}\right)+{li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)..\mathrm{1} \\ $$$${li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)+{li}_{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=−\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}..\mathrm{2} \\ $$$${f}'\left({x}\right)=\frac{\left(\mathrm{1}−{x}\right)\left({li}_{\mathrm{2}} \left({x}\right)+{li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)\right)−\mathrm{li}_{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)−\mathrm{li}_{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}{x}}+\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}{x}\left(\mathrm{1}−{x}\right)} \\ $$$${f}\left({x}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{ln}\left({x}\right)+\int{ln}^{\mathrm{2}} \left({x}\right)\left(\frac{{x}+\left(\mathrm{1}−{x}\right)}{\mathrm{2}{x}\left(\mathrm{1}−{x}\right)}\right){dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{ln}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{ln}^{\mathrm{2}} \left({x}\right)}{{x}}{dx}+\int\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}\left(\mathrm{1}−{x}\right)}{d}\underset{={A}} {{x}} \\ $$$${A}\:{by}\:{part} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{ln}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{6}}{ln}^{\mathrm{3}} \left({x}\right)+−\frac{{ln}\left(\mathrm{1}−{x}\right){ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}+/\int{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right).\frac{\mathrm{1}}{{x}}\underset{{missing}\:{one}\:} {/} \\ $$$${f}\left({x}\right)={Li}_{\mathrm{3}} \left({x}\right)+{Li}_{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)+{li}_{\mathrm{3}} \left(\mathrm{1}−{x}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{ln}\left({x}\right)+\frac{{ln}^{\mathrm{3}} \left({x}\right)}{\mathrm{6}}−\frac{{ln}^{\mathrm{2}} \left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{\mathrm{2}}+{c} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)={Li}_{\mathrm{3}} \left(\mathrm{1}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\zeta\left(\mathrm{3}\right)={c} \\ $$$$\Rightarrow{Li}_{\mathrm{3}} \left({x}\right)+{li}_{\mathrm{3}} \left(\mathrm{1}−{x}\right)+{li}_{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\zeta\left(\mathrm{3}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{ln}\left({x}\right)+\frac{{ln}^{\mathrm{3}} \left({x}\right)}{\mathrm{6}}−\frac{{ln}^{\mathrm{2}} \left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{\mathrm{2}}\:{what}\:{we}\:{want}\: \\ $$$${too}\:{bee}\:{contibued}\:{the}\:{A}\:{integral}\:{missing} \\ $$$$ \\ $$

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