Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

$$\frac{{d}}{{dx}}{Li}_{{s}+\mathrm{1}} \left({x}\right)=\frac{{li}_{{s}} \left({x}\right)}{{x}} \\ $$$${f}\left({x}\right)={li}_{\mathrm{3}} \left({x}\right)+{li}_{\mathrm{3}} \left(\mathrm{1}−{x}\right)+{li}_{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right) \\ $$$${f}'\left({x}\right)=\frac{{li}_{\mathrm{2}} \left({x}\right)}{{x}}−\frac{{li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{Li}_{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right).\frac{{x}}{{x}−\mathrm{1}} \\ $$$$=\frac{\left(\mathrm{1}−{x}\right){li}_{\mathrm{2}} \left({x}\right)−{xli}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)−{li}_{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)}{{x}\left(\mathrm{1}−{x}\right)} \\ $$$${Li}_{\mathrm{2}} \left({x}\right)+{li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)..\mathrm{1} \\ $$$${li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)+{li}_{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=−\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}..\mathrm{2} \\ $$$${f}'\left({x}\right)=\frac{\left(\mathrm{1}−{x}\right)\left({li}_{\mathrm{2}} \left({x}\right)+{li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)\right)−\mathrm{li}_{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)−\mathrm{li}_{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}{x}}+\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}{x}\left(\mathrm{1}−{x}\right)} \\ $$$${f}\left({x}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{ln}\left({x}\right)+\int{ln}^{\mathrm{2}} \left({x}\right)\left(\frac{{x}+\left(\mathrm{1}−{x}\right)}{\mathrm{2}{x}\left(\mathrm{1}−{x}\right)}\right){dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{ln}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{ln}^{\mathrm{2}} \left({x}\right)}{{x}}{dx}+\int\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}\left(\mathrm{1}−{x}\right)}{d}\underset{={A}} {{x}} \\ $$$${A}\:{by}\:{part} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{ln}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{6}}{ln}^{\mathrm{3}} \left({x}\right)+−\frac{{ln}\left(\mathrm{1}−{x}\right){ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}+/\int{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right).\frac{\mathrm{1}}{{x}}\underset{{missing}\:{one}\:} {/} \\ $$$${f}\left({x}\right)={Li}_{\mathrm{3}} \left({x}\right)+{Li}_{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)+{li}_{\mathrm{3}} \left(\mathrm{1}−{x}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{ln}\left({x}\right)+\frac{{ln}^{\mathrm{3}} \left({x}\right)}{\mathrm{6}}−\frac{{ln}^{\mathrm{2}} \left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{\mathrm{2}}+{c} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)={Li}_{\mathrm{3}} \left(\mathrm{1}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\zeta\left(\mathrm{3}\right)={c} \\ $$$$\Rightarrow{Li}_{\mathrm{3}} \left({x}\right)+{li}_{\mathrm{3}} \left(\mathrm{1}−{x}\right)+{li}_{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\zeta\left(\mathrm{3}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{ln}\left({x}\right)+\frac{{ln}^{\mathrm{3}} \left({x}\right)}{\mathrm{6}}−\frac{{ln}^{\mathrm{2}} \left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{\mathrm{2}}\:{what}\:{we}\:{want}\: \\ $$$${too}\:{bee}\:{contibued}\:{the}\:{A}\:{integral}\:{missing} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com