Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 141752 by 0731619 last updated on 23/May/21

Answered by qaz last updated on 23/May/21

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}x}{(x+1)(x^2+1)}dx$$$$={\int }_0^{\pi /2}\frac{u}{\tan u+1}du$$$$={\int }_0^{\pi /2}\frac{u\cos u}{\sin u+\cos u}du.......................\left(1\right)$$$$={\int }_0^{\pi /2}\frac{(\frac{\pi }{2}-u)\sin u}{\sin u+\cos u}du$$$$=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\sin u}{\sin u+\cos u}du-{\int }_0^{\pi /2}\frac{u\sin u}{\sin u+\cos u}du$$$$=\frac{{\pi }^2}{8}-{\int }_0^{\pi /2}\frac{u\sin u}{\sin u+\cos u}du.................\left(2\right)$$$$\left(1\right)+\left(2\right)$$$$\Rightarrow 2\mathrm{\Omega }=\frac{{\pi }^2}{8}+{\int }_0^{\pi /2}\frac{u(\cos u-\sin u)}{\sin u+\cos u}du$$$$=\frac{{\pi }^2}{8}-{\int }_0^{\pi /2}ln(\sin u+\cos u)du$$$$=\frac{{\pi }^2}{8}-{\int }_0^{\pi /2}ln\left(\cos u(\tan u+1)\right)du$$$$=\frac{{\pi }^2}{8}+\frac{\pi }{2}ln2-{\int }_0^{\pi /2}ln(1+\tan u)du$$$$=\frac{{\pi }^2}{8}+\frac{\pi }{2}ln2-{\int }_0^{\mathrm{\infty }}\frac{ln(1+u)}{1+u^2}du$$$$=\frac{{\pi }^2}{8}+\frac{\pi }{2}ln2-2{\int }_0^1\frac{ln(1+u)}{1+u^2}du+{\int }_0^1\frac{lnu}{1+u^2}du$$$$=\frac{{\pi }^2}{8}+\frac{\pi }{2}ln2-2\cdot \frac{\pi }{8}ln2-G$$$$=\frac{{\pi }^2}{8}+\frac{\pi }{4}ln2-G$$$$\Rightarrow \mathrm{\Omega }=\frac{{\pi }^2}{16}+\frac{\pi }{8}ln2-\frac{G}{2}$$

Commented by 0731619 last updated on 23/May/21

$$tanks$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com