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 143230 by qaz last updated on 11/Jun/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{3\mathrm{n}+1}}{3\mathrm{n}+1}=?$$

Answered by mathmax by abdo last updated on 11/Jun/21

$$\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=\mathrm{o}}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{3\mathrm{n}+1}}{3\mathrm{n}+1}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{3\mathrm{x}}=\frac{1}{1-{\mathrm{x}}^3}\mathrm{if}\mid \mathrm{x}\mid <1\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\frac{\mathrm{dt}}{1-{\mathrm{t}}^3}+\mathrm{c}\mathrm{but}\mathrm{c}=\mathrm{f}\left(0\right)=0\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\frac{\mathrm{dt}}{1-{\mathrm{t}}^3}$$$$\mathrm{let}\mathrm{H}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{t}}^3-1}\Rightarrow \mathrm{H}\left(\mathrm{t}\right)=\frac{1}{(\mathrm{t}-1)({\mathrm{t}}^2+\mathrm{t}+1)}=\frac{\mathrm{a}}{\mathrm{t}-1}+\frac{\mathrm{bt}+\mathrm{c}}{{\mathrm{t}}^2+\mathrm{t}+1}$$$$\mathrm{a}=\frac{1}{3},{\lim }_{\mathrm{t}\to +\mathrm{\infty }}\mathrm{tH}\left(\mathrm{t}\right)=0=\mathrm{a}+\mathrm{b}\Rightarrow \mathrm{b}=-\frac{1}{3}$$$$\mathrm{H}\left(0\right)=-1=-\mathrm{a}+\mathrm{c}\Rightarrow \mathrm{c}=\mathrm{a}-1=-\frac{2}{3}\Rightarrow \mathrm{H}\left(\mathrm{x}\right)=\frac{1}{3(\mathrm{t}-1)}+\frac{-\frac{1}{3}\mathrm{t}-\frac{2}{3}}{{\mathrm{t}}^2+\mathrm{t}+1}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}(-\frac{1}{3(\mathrm{t}-1)}+\frac{1}{3}\frac{\mathrm{t}+2}{{\mathrm{t}}^2+\mathrm{t}+1})\mathrm{dt}$$$$=-\frac{1}{3}{\int }_0^{\mathrm{x}}\frac{\mathrm{dt}}{\mathrm{t}-1}+\frac{1}{6}{\int }_0^{\mathrm{x}}\frac{2\mathrm{t}+1+3}{{\mathrm{t}}^2+\mathrm{t}+1}\mathrm{dt}$$$$=-\frac{1}{3}{[\log \mid \mathrm{t}-1\mid ]}_0^{\mathrm{x}}+\frac{1}{6}{\left[\log ({\mathrm{t}}^2+\mathrm{t}+1)\right]}_0^{\mathrm{x}}+\frac{1}{2}{\int }_0^{\mathrm{x}}\frac{\mathrm{dt}}{{\mathrm{t}}^2+\mathrm{t}+1}$$$$=-\frac{1}{3}\log \mid \mathrm{x}-1\mid +\frac{1}{6}\log ({\mathrm{x}}^2+\mathrm{x}+1)+\frac{1}{2}{\int }_0^{\mathrm{x}}\frac{\mathrm{dt}}{{\mathrm{t}}^2+\mathrm{t}+1}$$$$\mathrm{we}\mathrm{have}{\int }_0^{\mathrm{x}}\frac{\mathrm{dt}}{{\mathrm{t}}^2+\mathrm{t}+1}={\int }_0^{\mathrm{x}}\frac{\mathrm{dt}}{{(\mathrm{t}+\frac{1}{2})}^2+\frac{3}{4}}$$$${=}_{\mathrm{t}+\frac{1}{2}=\frac{\sqrt{3}}{2}\mathrm{y}}\frac{4}{3}{\int }_{\frac{1}{\sqrt{3}}}^{\frac{2\mathrm{x}+1}{\sqrt{3}}}\frac{1}{{\mathrm{y}}^2+1}.\frac{\sqrt{3}}{2}\mathrm{dy}=\frac{2}{\sqrt{3}}{\left[\mathrm{arctany}\right]}_{\frac{1}{\sqrt{3}}}^{\frac{2\mathrm{x}+1}{\sqrt{3}}}$$$$=\frac{2}{\sqrt{3}}(\arctan \left(\frac{2\mathrm{x}+1}{\sqrt{3}}\right)-\arctan \left(\frac{1}{\sqrt{3}}\right))\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{3\mathrm{n}+1}}{3\mathrm{n}+1}=-\frac{1}{3}\log \mid \mathrm{x}-1\mid +\frac{1}{6}\log ({\mathrm{x}}^2+\mathrm{x}+1)+\frac{1}{\sqrt{3}}(\arctan \left(\frac{2\mathrm{x}+1}{\sqrt{3}}\right)-\frac{\pi }{6})-\mathrm{x}$$

Answered by mr W last updated on 11/Jun/21

$$for\mid x\mid <1:$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^{3n}=\frac{x^3}{1-x^3}=\frac{1}{1-x^3}-1$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^x{x}^{3n}dx={\int }_0^x(\frac{1}{1-x^3}-1)dx$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{3n+1}}{3n+1}={\int }_0^x\frac{dx}{1-x^3}-x$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{3n+1}}{3n+1}=\frac{1}{6}\ln \frac{x^2+x+1}{x^2-2x+1}+\frac{1}{\sqrt{3}}({\tan }^{-1}\frac{2x+1}{\sqrt{3}}-\frac{\pi }{6})-x$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com