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 164395 by amin96 last updated on 16/Jan/22

Answered by mnjuly1970 last updated on 17/Jan/22

Commented by mnjuly1970 last updated on 17/Jan/22

$$thankyousomuchmydearfriend$$$$siramin....yashasinazerbaijan$$

Commented by amin96 last updated on 17/Jan/22

$$\mathbf{Bravo}\mathbf{my}\mathbf{dear}\mathbf{sir}\mathbf{thank}\mathbf{you}$$

Answered by Lordose last updated on 17/Jan/22

$$$$$$\mathrm{\Omega }={\int }_0^1\frac{\ln (1+\mathrm{x})\ln (1-\mathrm{x})}{1+\mathrm{x}}\mathrm{dx}$$$$\mathrm{\Omega }=-\frac{1}{2}({\int }_0^1{\ln }^2\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)\frac{\mathrm{dx}}{1+\mathrm{x}}-{\int }_0^1\frac{{\ln }^2(1-\mathrm{x})}{1+\mathrm{x}}\mathrm{dx}-{\int }_0^1\frac{{\ln }^2(1+\mathrm{x})}{1+\mathrm{x}}\mathrm{dx})$$$$\mathrm{\Omega }=-\frac{1}{2}(\mathrm{A}-\mathrm{B}-\mathrm{C})$$$$\mathrm{A}\stackrel{\mathrm{x}=\frac{1-\mathrm{x}}{1+\mathrm{x}}}{=}2{\int }_0^1\frac{{\ln }^2\left(\mathrm{x}\right)}{1+\mathrm{x}}=2\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}-1}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}{\ln }^2\left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{A}\stackrel{\mathbf{IBP}\times 2}{=}2\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}-1}}{{\mathrm{n}}^3}=2\mathit{\eta }\left(3\right)$$$$\mathrm{B}\stackrel{\mathrm{x}=1-\mathrm{x}}{=}{\int }_0^1\frac{{\ln }^2\left(\mathrm{x}\right)}{2(1-\frac{\mathrm{x}}{2})}\mathrm{dx}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{1}{2}\right)}^{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}{\ln }^2\left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{B}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{1}{2}\right)}^{\mathrm{n}}\cdot \frac{1}{{\mathrm{n}}^3}=2{\mathbf{Li}}_3\left(\frac{1}{2}\right)$$$$\mathrm{C}\stackrel{\mathrm{x}=1+\mathrm{x}}{=}{\int }_1^2\frac{{\ln }^2\left(\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\stackrel{\mathrm{x}=\ln \left(\mathrm{x}\right)}{=}{\int }_0^{\ln \left(2\right)}{\mathrm{x}}^2\mathrm{dx}=\frac{{\ln }^3\left(2\right)}{3}$$$$\mathrm{\Omega }=-\frac{1}{2}(2\mathit{\eta }\left(3\right)-2{\mathbf{Li}}_3\left(\frac{1}{2}\right)-\frac{{\ln }^3\left(2\right)}{3})$$$$\mathrm{\Omega }=-\frac{3}{8}\mathit{\zeta }\left(3\right)+(\frac{{\ln }^3\left(2\right)}{6}-\frac{{\mathit{\pi }}^2\ln \left(2\right)}{12}+\frac{7}{8}\mathit{\zeta }\left(3\right))+\frac{{\ln }^3\left(2\right)}{6}$$$$\mathbf{\Omega }=\frac{{\mathbf{ln}}^3\left(2\right)}{3}-\frac{{\mathit{\pi }}^2\mathbf{ln}\left(2\right)}{12}+\frac{\mathit{\zeta }\left(3\right)}{8}$$$$\mathit{\varnothing }\mathbf{sE}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com