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Question Number 38723 by maxmathsup by imad last updated on 28/Jun/18

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{xsin\left(3x\right)}{{(1+x^2)}^2}dx$$

Commented by math khazana by abdo last updated on 30/Jun/18

$$letI={\int }_0^{\mathrm{\infty }}\frac{xsin\left(3x\right)}{{(1+x^2)}^2}dx$$$$2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{xsin\left(3x\right)}{{(1+x^2)}^2}dx=Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x{e}^{3ix}}{{(x^2+1)}^2}dx\right)$$$$let\phi \left(z\right)=\frac{z{e}^{3iz}}{{(z^2+1)}^2}$$$$\phi \left(z\right)=\frac{z{e}^{3iz}}{{(z-i)}^2{(z+i)}^2}thepolrsof\phi areiand-i$$$$\left(doubles\right)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)but$$$$Res(\phi ,i)={lim}_{z\to i}\frac{1}{(2-1)!}{\left\{{(z-i)}^2\phi \left(z\right)\right\}}^{\left(1\right)}$$$$={lim}_{z\to i}{\left\{\frac{z{e}^{3iz}}{{(z+i)}^2}\right\}}^{\left(1\right)}$$$$={lim}_{z\to i}\frac{(e^{3iz}+3iz{e}^{3iz}){(z+i)}^2-2(z+i)z{e}^{3iz}}{{(z+i)}^4}$$$$={lim}_{z\to i}\frac{(1+3iz)e^{3iz}(z+i)-2z{e}^{3iz}}{{(z+i)}^3}$$$$=\frac{(1-3)e^{-3}\left(2i\right)-2{ie}^{-3}}{{\left(2i\right)}^3}=\frac{(-4i-2i)e^{-3}}{-8i}$$$$=\frac{6e^{-3}}{8}=\frac{3}{4}{e}^{-3}$$$$2I=Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz\right)\Rightarrow I=\frac{3}{8}{e}^{-3}.$$

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