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 68409 by mathmax by abdo last updated on 10/Sep/19

$$calculate{\int }_0^{+\mathrm{\infty }}\frac{arctan\left(x^2\right)}{1+x^2}dx$$

Commented by mathmax by abdo last updated on 11/Sep/19

$$letI={\int }_0^{\mathrm{\infty }}\frac{arctan\left(x^2\right)}{1+x^2}dxchangementx=\frac{1}{t}give$$$$I=-{\int }_0^{\mathrm{\infty }}\frac{arctan\left(\frac{1}{t^2}\right)}{1+\frac{1}{t^2}}(-\frac{dt}{t^2})={\int }_0^{\mathrm{\infty }}\frac{\frac{\pi }{2}-arctan\left(t^2\right)}{t^2+1}dt$$$$=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2}-{\int }_0^{\mathrm{\infty }}\frac{arctan\left(t^2\right)}{1+t^2}=\frac{{\pi }^2}{4}-I\Rightarrow 2I=\frac{{\pi }^2}{4}\Rightarrow I=\frac{{\pi }^2}{8}$$

Commented by mathmax by abdo last updated on 11/Sep/19

$$anotherway2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan\left(x^2\right)}{x^2+1}dxletW\left(z\right)=\frac{arctan\left(z^2\right)}{z^2+1}$$$$residustheoremgive{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi Res(W,i)$$$$=2i\pi \times \frac{\mid arctan\left(i^2\right)\mid }{2i}=\pi \times \mid -\frac{\pi }{4}\mid =\frac{{\pi }^2}{4}\Rightarrow I=\frac{{\pi }^2}{8}(x\to \frac{arctan\left(x^2\right)}{1+x^2}$$$$ispositiveon[0,+\mathrm{\infty }\left[\right)$$

Commented by mathmax by abdo last updated on 11/Sep/19

$$wemustuseaoppositcontourinthiscase$$

Answered by mind is power last updated on 10/Sep/19

$$={\int }_0^{+\mathrm{\infty }}\frac{arctg\left(x^2\right)}{1+x^2}dx={\int }_0^{+\mathrm{\infty }}\frac{arctg\left(\frac{1}{x^2}\right)}{x^2+1}={\int }_0^{+\mathrm{\infty }}\frac{\frac{\pi }{2}-arctg\left(x^2\right)}{x^2+1}$$$$\Rightarrow 2{\int }_0^{+\mathrm{\infty }}\frac{arctg\left(x^2\right)}{x^2+1}dx={\int }_0^{\mathrm{\infty }}\frac{\pi }{2(x^2+1)}dx=\frac{{\pi }^2}{4}$$$$\Rightarrow {\int }_0^{+\mathrm{\infty }}\frac{arctg\left(x^2\right)}{1+x^2}dx=\frac{{\pi }^2}{8}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com