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 39119 by math khazana by abdo last updated on 02/Jul/18

$$calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^{2}cos\left(4x\right)}{{(x^2+1)}^2}dx$$

Commented by math khazana by abdo last updated on 03/Jul/18

$$letI={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^{2}cos\left(4x\right)}{{(x^2+1)}^2}dx$$$$I=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2{e}^{i4x}}{{(x^2+1)}^2}\right)let\phi \left(z\right)=\frac{z^2{e}^{i4z}}{{(z^2+1)}^2}$$$$wehave\phi \left(z\right)=\frac{z^2{e}^{i4z}}{{(z-i)}^2{(z+i)}^2}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)but$$$$Res(\phi ,i)={lim}_{z\to i}{\left\{{(z-i)}^2\phi \left(z\right)\right\}}^{\left(1\right)}$$$$={lim}_{z\to i}{\left\{\frac{z^2{e}^{i4z}}{{(z+i)}^2}\right\}}^{\left(1\right)}$$$$={lim}_{z\to i}\frac{(2z{e}^{i4z}+4{iz}^2{e}^{i4z}){(z+i)}^2-2(z+i)z^2{e}^{i4z}}{{(z+i)}^4}$$$$={lim}_{z\to i}\frac{(2z+4{iz}^2)e^{i4z}(z+i)-2z^2{e}^{i4z}}{{(z+i)}^3}$$$$=\frac{(2i-4i)e^{-4}\left(2i\right)+2e^{-4}}{{\left(2i\right)}^3}=\frac{4e^{-4}+2e^{-4}}{-8i}$$$$=\frac{6e^{-4}}{-8i}=i\frac{3}{4}{e}^{-4}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{3i}{4}{e}^{-4}$$$$=-\frac{3\pi }{2}{e}^{-4}\Rightarrow I=-\frac{3\pi }{2}{e}^{-4}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com