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 137973 by mnjuly1970 last updated on 08/Apr/21

$$$$$$.......Advanced............Calculus.......$$$$provethat:::$$$$\mathit{\varphi }={\int }_0^1\frac{ln(1-x)}{1+x^2}dx=\frac{\pi }{8}ln\left(2\right)-G...✓$$$$whereGiscatalannumber...$$$$$$

Answered by EnterUsername last updated on 08/Apr/21

$${\int }_0^1\frac{ln(1-x)}{1+x^2}dx={\int }_0^{\frac{\pi }{4}}ln(1-tan\theta )d\theta$$$$={\int }_0^{\frac{\pi }{4}}ln(cosx-sinx)dx-{\int }_0^{\frac{\pi }{4}}lncosxdx$$$$={\int }_0^{\frac{\pi }{4}}ln\left[\sqrt{2}cos(x+\frac{\pi }{4})\right]dx-(\frac{G}{2}-\frac{\pi ln2}{4})$$$$=\frac{\pi }{8}ln2+{\int }_0^{\frac{\pi }{4}}ln\left(sinx\right)dx-(\frac{G}{2}-\frac{\pi ln2}{4})$$$$=\frac{\pi }{8}ln2-\frac{G}{2}-\frac{\pi ln2}{4}-(\frac{G}{2}-\frac{\pi ln2}{4})=\frac{\pi }{8}ln2-G$$

Commented by mnjuly1970 last updated on 08/Apr/21

$$grateful...$$

Answered by mathmax by abdo last updated on 09/Apr/21

$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^1\frac{\ln (1+\mathrm{ax})}{1+{\mathrm{x}}^2}\mathrm{dx}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^1\frac{\mathrm{x}}{(\mathrm{ax}+1)({\mathrm{x}}^2+1)}\mathrm{dx}$$$$=\frac{1}{\mathrm{a}}{\int }_0^1\frac{\mathrm{ax}+1-1}{(\mathrm{ax}+1)({\mathrm{x}}^2+1)}\mathrm{dx}=\frac{1}{\mathrm{a}}{\left[\mathrm{arctanx}\right]}_0^1-\frac{1}{\mathrm{a}}{\int }_0^1\frac{\mathrm{dx}}{(\mathrm{ax}+1)({\mathrm{x}}^2+1)}$$$$=\frac{\pi }{4\mathrm{a}}-\frac{1}{\mathrm{a}}{\int }_0^1\frac{\mathrm{dx}}{(\mathrm{ax}+1)({\mathrm{x}}^2+1)}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{(\mathrm{ax}+1)({\mathrm{x}}^2+1)}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\alpha }{\mathrm{ax}+1}+\frac{\mathrm{mx}+\mathrm{n}}{{\mathrm{x}}^2+1}$$$$\alpha =\frac{1}{\frac{1}{{\mathrm{a}}^2}+1}=\frac{{\mathrm{a}}^2}{1+{\mathrm{a}}^2}$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{xF}\left(\mathrm{x}\right)=0=\frac{\alpha }{\mathrm{a}}+\mathrm{m}\Rightarrow \mathrm{m}=-\frac{\mathrm{a}}{1+{\mathrm{a}}^2}$$$$\mathrm{F}\left(\mathrm{o}\right)=1=\alpha +\mathrm{n}\Rightarrow \mathrm{n}=1-\frac{{\mathrm{a}}^2}{1+{\mathrm{a}}^2}=\frac{1}{1+{\mathrm{a}}^2}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{{\mathrm{a}}^2}{({\mathrm{a}}^2+1)(\mathrm{ax}+1)}$$$$+\frac{-\frac{\mathrm{a}}{{\mathrm{a}}^2+1}\mathrm{x}+\frac{1}{1+{\mathrm{a}}^2}}{{\mathrm{x}}^2+1}\Rightarrow {\int }_0^1\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}=\frac{{\mathrm{a}}^2}{{\mathrm{a}}^2+1}{\int }_0^1\frac{\mathrm{dx}}{\mathrm{ax}+1}$$$$-\frac{1}{{\mathrm{a}}^2+1}{\int }_0^1\frac{\mathrm{ax}-1}{{\mathrm{x}}^2+1}\mathrm{dx}=\frac{\mathrm{a}}{{\mathrm{a}}^2+1}{\left[\ln (\mathrm{ax}+1)\right]}_0^1-\frac{\mathrm{a}}{2({\mathrm{a}}^2+1)}{\int }_0^1\frac{2\mathrm{x}}{{\mathrm{x}}^2+1}\mathrm{dx}$$$$+\frac{\pi }{4({\mathrm{a}}^2+1)}=\frac{\mathrm{aln}(\mathrm{a}+1)}{{\mathrm{a}}^2+1}-\frac{\mathrm{a}}{2({\mathrm{a}}^2+1)}\ln \left(2\right)+\frac{\pi }{4({\mathrm{a}}^2+1)}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\frac{\pi }{4\mathrm{a}}-\frac{\ln (\mathrm{a}+1)}{{\mathrm{a}}^2+1}+\frac{\ln 2}{2({\mathrm{a}}^2+1)}-\frac{\pi }{4\mathrm{a}({\mathrm{a}}^2+1)}$$$$=\frac{\pi }{4\mathrm{a}}(1-\frac{1}{{\mathrm{a}}^2+1})-\frac{\ln (\mathrm{a}+1)}{{\mathrm{a}}^2+1}+\frac{\ln 2}{2({\mathrm{a}}^2+1)}=\frac{\pi \mathrm{a}}{4}-\frac{\ln (\mathrm{a}+1)}{{\mathrm{a}}^2+1}+\frac{\ln 2}{2({\mathrm{a}}^2+1)}$$$$\Rightarrow {\int }_0^1{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)\mathrm{da}=\frac{\pi }{4}{\int }_0^1\mathrm{ada}-{\int }_0^1\frac{\ln (\mathrm{a}+1)}{{\mathrm{a}}^2+1}\mathrm{da}+\frac{\ln 2}{2}{\int }_0^1\frac{\mathrm{da}}{{\mathrm{a}}^2+1}$$$$=\frac{\pi }{4}\times \frac{1}{2}-{\int }_0^1\frac{\ln (1+\mathrm{x})}{{\mathrm{x}}^2+1}\mathrm{dx}+\frac{\ln 2}{2}.\frac{\pi }{4}$$$$=\frac{\pi }{8}-{\int }_0^1\frac{\ln (1+\mathrm{x})}{{\mathrm{x}}^2+1}\mathrm{dx}+\frac{\pi }{8}\ln 2=\mathrm{f}\left(1\right)={\int }_0^1\frac{\ln (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}\Rightarrow$$$$2{\int }_0^1\frac{\ln (1+\mathrm{x})}{{\mathrm{x}}^2+1}\mathrm{dx}=\frac{\pi }{8}+\frac{\pi }{8}\ln 2\Rightarrow {\int }_0^1\frac{\ln (1+\mathrm{x})}{{\mathrm{x}}^2+1}\mathrm{dx}=\frac{\pi }{16}+\frac{\pi }{16}\ln 2?$$$$\mathrm{if}\mathrm{we}\mathrm{want}{\int }_0^1\frac{\ln (1-\mathbf{x})}{{\mathbf{x}}^2+1}\mathbf{dx}\mathbf{we}\mathbf{use}\mathbf{the}\mathbf{parametric}$$$$\mathbf{f}\left(\mathbf{a}\right)={\int }_0^1\frac{\ln (1-\mathrm{ax})}{{\mathrm{x}}^2+1}\mathrm{dx}....\mathrm{be}\mathrm{continued}....$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com