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 33129 by prof Abdo imad last updated on 10/Apr/18

$$1)findthevalueof{u}_n={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(nx\right)}{4+x^2}dx$$$$2)findthenatureof\mathrm{\Sigma }{u}_n.$$

Commented by prof Abdo imad last updated on 12/Apr/18

$$letcalculate{u}_{n}byredidustheorem.$$$$u_n=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{inx}}{4+x^2}dx\right)letconsider$$$$\phi \left(z\right)=\frac{e^{inz}}{z^2+4}.polesof\phi ?$$$$\phi \left(z\right)=\frac{e^{inz}}{(z-2i)(z+2i)}sothepolesare$$$$2iand-2i$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,2i)$$$$=\frac{e^{in\left(2i\right)}}{4i}=\frac{e^{-2n}}{4i}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{-2n}}{4i}$$$$=\frac{\pi }{2}{e}^{-2n}\Rightarrow u_n=\frac{\pi }{2}{e}^{-2n}$$$$2)\sum _{n=0}^{+\mathrm{\infty }}{u}_n=\frac{\pi }{2}\sum _{n=0}^{\mathrm{\infty }}{\left(e^{-2}\right)}^n=\frac{\pi }{2}\frac{1}{1-e^{-2}}$$$$=\frac{\pi }{2(1-\frac{1}{e^2})}=\frac{\pi e^2}{2(e^2-1)}.$$

Commented by prof Abdo imad last updated on 12/Apr/18

$$remarkIm\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{inx}}{x^2+4}dx\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin\left(nx\right)}{x^2+4}dx=0$$$$brcausethefunctionisodd.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com