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 61453 by maxmathsup by imad last updated on 02/Jun/19

$$find{\int }_0^1\frac{ln\left(x\right)ln(1+x)}{x}dx$$

Answered by Smail last updated on 02/Jun/19

$$A={\int }_0^1\frac{ln\left(x\right)ln(1+x)}{x}dx$$$$={\int }_0^1\frac{lnx}{x}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}{x}^n}{n}dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{n}{\int }_0^1{x}^{n-1}ln\left(x\right)dx$$$$Byparts$$$$u=lnx\Rightarrow u^{\prime }=\frac{1}{x}$$$$v^{\prime }=x^{n-1}\Rightarrow v=\frac{x^n}{n}$$$$A=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{n}({\left[\frac{x^{n}ln\left(x\right)}{n}\right]}_0^1-{\int }_0^1\frac{x^{n-1}}{n}dx)$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{n}\left(\frac{1}{n}{\left[\frac{x^n}{n}\right]}_0^1\right)$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{n^3}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2n+1)}^3}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\left(2n\right)}^3}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2n+1)}^3}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\left(2n\right)}^3}-2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\left(2n\right)}^3}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}-\frac{2}{8}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}$$$$=\frac{3}{4}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}$$$$A=\frac{3}{4}\zeta \left(3\right)\approx 0.9013$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com