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 118710 by eric last updated on 19/Oct/20

Answered by mathmax by abdo last updated on 19/Oct/20

$$\mathrm{M}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{1+{\mathrm{x}}^3}\mathrm{dx}={\int }_0^1\frac{\mathrm{lnx}}{1+{\mathrm{x}}^3}\mathrm{dx}+{\int }_1^{\mathrm{\infty }}\frac{\mathrm{lnx}}{1+{\mathrm{x}}^3}\mathrm{dx}(\to \mathrm{x}=\frac{1}{\mathrm{t}})$$$$={\int }_0^1\frac{\mathrm{lnx}}{1+{\mathrm{x}}^3}\mathrm{dx}-{\int }_0^1\frac{-\mathrm{lnt}}{1+\frac{1}{{\mathrm{t}}^3}}\times \left(\frac{-\mathrm{dt}}{{\mathrm{t}}^2}\right)$$$$={\int }_0^1\frac{\mathrm{lnx}}{1+{\mathrm{x}}^3}\mathrm{dx}-{\int }_0^1\frac{\mathrm{tlnt}}{1+{\mathrm{t}}^3}\mathrm{dt}\mathrm{we}\mathrm{have}$$$${\int }_0^1\frac{\mathrm{lnx}}{1+{\mathrm{x}}^3}\mathrm{dx}={\int }_0^1\mathrm{lnx}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{3\mathrm{n}}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{3\mathrm{n}}\mathrm{lnx}\mathrm{dx}$$$${\mathrm{U}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{3\mathrm{n}}\mathrm{lnxdx}{=}_{\mathrm{byparts}}{\left[\frac{{\mathrm{x}}^{3\mathrm{n}+1}}{3\mathrm{n}+1}\mathrm{lnx}\right]}_0^1-{\int }_0^1\frac{{\mathrm{x}}^{3\mathrm{n}}}{3\mathrm{n}+1}\mathrm{dx}$$$$=-\frac{1}{{(3\mathrm{n}+1)}^2}\Rightarrow {\int }_0^1\frac{\mathrm{lnx}}{1+{\mathrm{x}}^3}\mathrm{dx}=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(3\mathrm{n}+1)}^2}$$$${\int }_0^1\frac{\mathrm{xlnx}}{1+{\mathrm{x}}^3}\mathrm{dx}={\int }_0^1\mathrm{xlnx}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{3\mathrm{n}}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{3\mathrm{n}+1}\mathrm{lnx}\mathrm{dx}$$$${\mathrm{V}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{3\mathrm{n}+1}\mathrm{lnx}\mathrm{dx}={\left[\frac{{\mathrm{x}}^{3\mathrm{n}+2}}{3\mathrm{n}+2}\mathrm{lnx}\right]}_0^1-{\int }_0^1\frac{{\mathrm{x}}^{3\mathrm{n}+1}}{3\mathrm{n}+2}\mathrm{dx}$$$$=-\frac{1}{{(3\mathrm{n}+2)}^2}\Rightarrow {\int }_0^1\frac{\mathrm{xlnx}}{1+{\mathrm{x}}^3}\mathrm{dx}=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(3\mathrm{n}+2)}^2}\Rightarrow$$$$\mathrm{M}=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(3\mathrm{n}+1)}^2}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(3\mathrm{n}+2)}^2}\mathrm{rest}\mathrm{calculus}\mathrm{of}\mathrm{those}$$$$\mathrm{series}...\mathrm{be}\mathrm{continued}...$$

Commented by mathdave last updated on 19/Oct/20

$$letmecomplettheremainingpart$$$$M=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{(3n+1)}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{(3n+2)}=-\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{(3n+1)}$$$$\because M=-\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{(3n+1)}=-(-\frac{\pi }{3^2}\underset{z\to -\frac{1}{3}}{\lim }\frac{1}{(2-1)!}\cdot \frac{\partial }{\partial z}\left(\mathrm{cosec}\left(\pi z\right)\right))$$$$M=\frac{\pi }{9}\underset{z\to -\frac{1}{3}}{\lim }(-\pi \cot \left(\pi z\right)\mathrm{cosec}\left(\pi z\right))$$$$M=-\frac{{\pi }^2}{9}\cot (-\frac{\pi }{3})\mathrm{cosec}(-\frac{\pi }{3})=-\frac{{\pi }^2}{9}\left(\frac{2}{3}\right)=-\frac{2{\pi }^2}{27}$$$$\because M={\int }_0^{\mathrm{\infty }}\frac{\ln x}{1+x^3}dx=-\frac{2{\pi }^2}{27}$$$$notethis$$$$\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(ax+1)}^n}=-\frac{\pi }{a^n}\underset{z\to -\frac{1}{a}}{\lim }\frac{1}{(n-1)!}\cdot \frac{{\partial }^{(n-1)}}{\partial z^{(n-1)}}\mathrm{cosec}\left(\pi z\right)$$

Answered by mnjuly1970 last updated on 19/Oct/20

$$assume:f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{x^a}{1+x^3}dx$$$$goal:M=f{}^{\prime }\left(0\right)=?$$$$eulerreflection$$$$formula...$$$$f\left(a\right)\stackrel{x^3=t}{=}\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{t^{(\frac{a-2}{3}+1)-1}}{1+t}dt=\frac{1}{3}\beta (\frac{a+1}{3},1-\frac{a+1}{3})$$$$=\frac{1}{3}\mathrm{\Gamma }\left(\frac{a+1}{3}\right)\mathrm{\Gamma }(1-\frac{a+1}{3})$$$$=\frac{1}{3}\ast \frac{\pi }{sin(\frac{\pi }{3}+\frac{\pi a}{3}})$$$$\therefore f{}^{\prime }\left(a\right)=\frac{\pi }{3}\ast \frac{-\frac{\pi }{3}cos(\frac{\pi }{3}+\frac{\pi a}{3})}{{sin}^2(\frac{\pi }{3}+\frac{\pi a}{3})}$$$$\therefore M=f{}^{\prime }\left(0\right)=\frac{-{\pi }^2}{9}\ast \frac{\frac{1}{2}}{\frac{3}{4}}=-\frac{2{\pi }^2}{27}✓$$$$...m.n.1970...$$$$$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com