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 167213 by mnjuly1970 last updated on 09/Mar/22

solve I= ∫_0 ^( (1/2)) ((ln^( 2) (x))/(1−x)) dx =?

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{solve} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\mathcal{I}=\:\int_{\mathrm{0}} ^{\:\frac{\mathrm{1}}{\mathrm{2}}} \frac{{ln}^{\:\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}}\:{dx}\:=? \\ $$$$ \\ $$

Answered by mindispower last updated on 10/Mar/22

∫_0 ^(1/2) x^k ln^2 (x)dx=((ln^2 (2))/(k+1)).(1/2^(k+1) )−(2/(k+1))∫_0 ^(1/2) x^k l(x)dx =((ln^2 (2))/(2^(k+1) (k+1)))(−(2/(k+1))[−((ln(2())/(2^(k+1) (k+1)))−(1/((k+1)^2 2^(k+1) ))) =((ln^2 (2))/(2^(k+1) (k+1)))−((2ln(2))/(2^(k+1) (k+1)^2 ))+(2/((k+1)^3 2^(k+1) )) I=∫_0 ^(1/2) ((ln^2 (x))/(1−x))dx=Σ_(k≥0) ∫_0 ^(1/2) x^k ln^2 (x)dx =ln^2 (2)Σ_(k≥0) ((1/2))^(k+1) .(1/(k+1))+2ln(2).Σ((((1/2))^(k+1) )/((k+1)^2 ))+2Σ((((1/2))^(k+1) )/((k+1)^3 )) =ln^2 (2).ln(2)+2ln(2)Li_2 ((1/2))+2Li_3 ((1/2)) ln^3 (2)+2ln(2)((π^2 /(12))−((ln^2 (2))/2))+(1/(12))(21ζ(3)+4ln^3 (2)−2π^2 ln(2)) =(7/4)ζ(3)+((ln^3 (2))/3)

$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {x}^{{k}} {ln}^{\mathrm{2}} \left({x}\right){dx}=\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{{k}+\mathrm{1}}.\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }−\frac{\mathrm{2}}{{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {x}^{{k}} {l}\left({x}\right){dx} \\ $$$$=\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}^{{k}+\mathrm{1}} \left({k}+\mathrm{1}\right)}\left(−\frac{\mathrm{2}}{{k}+\mathrm{1}}\left[−\frac{{ln}\left(\mathrm{2}\left(\right)\right.}{\mathrm{2}^{{k}+\mathrm{1}} \left({k}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{2}^{{k}+\mathrm{1}} }\right)\right. \\ $$$$=\frac{{l}\overset{\mathrm{2}} {{n}}\left(\mathrm{2}\right)}{\mathrm{2}^{{k}+\mathrm{1}} \left({k}+\mathrm{1}\right)}−\frac{\mathrm{2}{ln}\left(\mathrm{2}\right)}{\mathrm{2}^{{k}+\mathrm{1}} \left({k}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{2}}{\left({k}+\mathrm{1}\right)^{\mathrm{3}} \mathrm{2}^{{k}+\mathrm{1}} } \\ $$$$\mathcal{I}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}}{dx}=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {x}^{{k}} {ln}^{\mathrm{2}} \left({x}\right){dx} \\ $$$$={ln}^{\mathrm{2}} \left(\mathrm{2}\right)\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}+\mathrm{1}} .\frac{\mathrm{1}}{{k}+\mathrm{1}}+\mathrm{2}{ln}\left(\mathrm{2}\right).\Sigma\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}+\mathrm{1}} }{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{2}\Sigma\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}+\mathrm{1}} }{\left({k}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$={ln}^{\mathrm{2}} \left(\mathrm{2}\right).{ln}\left(\mathrm{2}\right)+\mathrm{2}{ln}\left(\mathrm{2}\right){Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{2}{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${ln}^{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{2}{ln}\left(\mathrm{2}\right)\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{21}\zeta\left(\mathrm{3}\right)+\mathrm{4}{ln}^{\mathrm{3}} \left(\mathrm{2}\right)−\mathrm{2}\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)\right) \\ $$$$=\frac{\mathrm{7}}{\mathrm{4}}\zeta\left(\mathrm{3}\right)+\frac{{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 10/Mar/22

thank you so much sir power...

$$\:\:\:\:{thank}\:{you}\:{so}\:{much}\:{sir} \\ $$$$\:\:\:{power}... \\ $$

Commented by mindispower last updated on 10/Mar/22

Withe Pleasur sir Have a nice Day

$${Withe}\:{Pleasur}\:{sir} \\ $$$${Have}\:{a}\:{nice}\:{Day} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com