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 184925 by cortano1 last updated on 14/Jan/23

Commented by Frix last updated on 14/Jan/23

$$\mathrm{I}\mathrm{think}\mathrm{that}$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{\tan }^{-1}ax-{\tan }^{-1}bx}{x}dx=\frac{\pi }{2}\ln \frac{a}{b}$$

Answered by ARUNG_Brandon_MBU last updated on 14/Jan/23

$$I={\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(ax\right)-{\tan }^{-1}\left(bx\right)}{x}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(ax\right)}{x}dx-{\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(bx\right)}{x}dx=I\left(a\right)-I\left(b\right)$$$$\Rightarrow I{}^{\prime }\left(a\right)-I{}^{\prime }\left(b\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{1+{\left(ax\right)}^2}-{\int }_0^{\mathrm{\infty }}\frac{dx}{1+{\left(bx\right)}^2}$$$$=\frac{1}{a}{\left[{\tan }^{-1}\left(ax\right)\right]}_0^{\mathrm{\infty }}-\frac{1}{b}{\left[{\tan }^{-1}\left(bx\right)\right]}_0^{\mathrm{\infty }}$$$$=\frac{\pi }{2a}-\frac{\pi }{2b}\Rightarrow I\left(a\right)-I\left(b\right)=\frac{\pi }{2}\ln a-\frac{\pi }{2}\ln b+C$$$$I\left(1\right)-I\left(1\right)=0=C\Rightarrow I=I\left(a\right)-I\left(b\right)=\frac{\pi }{2}\ln \left(\frac{a}{b}\right)$$$${\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(\pi x\right)-{\tan }^{-1}\left(2x\right)}{x}dx=I\left(\pi \right)-I\left(2\right)=\frac{\pi }{2}\ln \left(\frac{\pi }{2}\right)★$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com