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Question Number 39033 by maxmathsup by imad last updated on 01/Jul/18

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

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

$$letI={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{xsin\left(2x\right)}{{(1+x^2)}^2}dx$$$$I=Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x{e}^{ix}}{{(1+x^2)}^2}dx\right)let\phi \left(z\right)=\frac{z{e}^{iz}}{{(1+z^2)}^2}$$$$\phi \left(z\right)=\frac{z{e}^{iz}}{{(z-i)}^2{(z+i)}^2}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)$$$$Res(\phi ,i)={lim}_{z\to i}\frac{1}{(2-1)!}{\left\{{(z-i)}^2\phi \left(z\right)\right\}}^{\left(1\right)}$$$$={lim}_{z\to i}{\left\{\frac{z{e}^{iz}}{{(z+i)}^2}\right\}}^{\left(1\right)}$$$$={lim}_{z\to i}\frac{(e^{iz}+iz{e}^{iz}){(z+i)}^2-2(z+i)z{e}^{iz}}{{(z+i)}^4}$$$$={lim}_{z\to i}\frac{(1+iz)e^{ix}(z+i)-2{ze}^{iz}}{{(z+i)}^3}$$$$=\frac{-2i{e}^{-1}}{{\left(2i\right)}^3}=\frac{-2i{e}^{-1}}{-8i}=\frac{e^{-1}}{4}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{-1}}{4}=\frac{i\pi e^{-1}}{2}\Rightarrow$$$$I=Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz\right)=\frac{\pi }{2e}.$$

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

$$theQisfindI={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{xsin\left(x\right)}{{(1+x^2)}^2}dx$$

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