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Question Number 36689 by prof Abdo imad last updated on 04/Jun/18

$$1)findthevalueof{\int }_0^{1}ln(1-x^3)dxthen$$$$findthevalueof{\int }_0^{1}ln(1+x+x^2)dx$$$$2)findthevalueof{\int }_0^{1}ln(1+x^3)dxthen$$$$calculate{\int }_0^{1}ln(1-x+x^2)dx$$

Commented by maxmathsup by imad last updated on 04/Aug/18

$$1)letA={\int }_0^{1}ln(1-x^3)dxwehave{ln}^{{}^{\prime }}{(1-u)}^{}=-\frac{1}{1-u}=-\sum _{n=0}^{\mathrm{\infty }}{u}^n$$$$with\mid u\mid <1\Rightarrow ln(1-u)=-\sum _{n=0}^{\mathrm{\infty }}\frac{u^{n+1}}{n+1}=-\sum _{n=1}^{\mathrm{\infty }}\frac{u^n}{n}\Rightarrow$$$$A=-{\int }_0^1\left(\sum _{n=1}^{\mathrm{\infty }}\frac{x^{3n}}{n}\right)dx=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}{\int }_0^1{x}^{3n}dx=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(3n+1)}\Rightarrow$$$$-\frac{A}{3}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{3n(3n+1)}=\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{3n}-\frac{1}{3n+1})let$$$$S_n=\sum _{k=1}^n(\frac{1}{3k}-\frac{1}{3k+1})\Rightarrow S_n=\frac{1}{3}{H}_n-\sum _{k=1}^n\frac{1}{3k+1}but$$$$\sum _{k=1}^n\frac{1}{3k+1}?wehave\sum _{k=1}^n\frac{1}{k}=\sum _{k=3p}\frac{1}{k}+\sum _{k=3p+1}\frac{1}{k}+\sum _{k=3p+2}\frac{1}{k}$$$$=\sum _{p=1}^{\left[\frac{n}{3}\right]}\frac{1}{3p}+\sum _{p=0}^{\left[\frac{n-1}{3}\right]}\frac{1}{3p+1}+\sum _{p=0}^{\left[\frac{n-2}{3}\right]}\frac{1}{3p+2}but$$$$\sum _{p=0}^{\left[\frac{n-2}{3}\right]}\frac{1}{3p+2}{=}_{p=j-1}\sum _{j=1}^{\left[\frac{n-2}{3}\right]-1}\frac{1}{3j-1}\Rightarrow$$$$\sum _{p=0}^{\left[\frac{n-1}{3}\right]}\frac{1}{3p+1}+\sum _{p=1}^{\left[\frac{n-5}{3}\right]}\frac{1}{3p-1}=H_n-\frac{1}{3}{H}_{\left[\frac{n}{3}\right]}...becontinued...$$$$$$

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