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Question Number 27619 by abdo imad last updated on 11/Jan/18

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

Commented by abdo imad last updated on 12/Jan/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{cos\left(2x\right)}{{(1+x^2)}^2}dx$$$$I=\frac{1}{2}{\int }_R\frac{cos\left(2x\right)}{{(1+x^2)}^2}dx=\frac{1}{2}Re\left({\int }_R\frac{e^{i2x}}{{(1+x^2)}^2}dx\right)$$$$letintroducethecomplexfunction$$$$f\left(z\right)=\frac{e^{i\left(2z\right)}}{{(1+z^2)}^2}wehavef\left(z\right)=\frac{e^{i\left(2z\right)}}{{(z-i)}^2{(z+i)}^2}sothepoles$$$$offareiand-i\left(doublepoles\right)$$$${\int }_{R}f\left(z\right)dz=2i\pi Res(f,i)(wedonttake-ibecauseIm(-i)<0)$$$$Re(f,i)={lim}_{z->i}\frac{1}{(2-1)!}\frac{d}{dz}\left({(z-i)}^{2}f\left(z\right)\right)$$$$={lim}_{z->i}\frac{d}{dz}\left(\frac{e^{i\left(2z\right)}}{{(z+i)}^2}\right)=$$$$={lim}_{z->i}\frac{2i{e}^{i\left(2z\right)}{(z+i)}^2-2(z+i)e^{i\left(2z\right)}}{{(z+i)}^4}$$$$={lim}_{z->i}\frac{2i{e}^{i\left(2z\right)}(z+i)-2e^{i\left(2z\right)}}{{(z+i)}^3}$$$$=\frac{2i{e}^{-2}\left(2i\right)-2e^{-2}}{{\left(2i\right)}^3}=\frac{-4e^{-2}-2e^{-2}}{-8i}=\frac{6e^{-2}}{8i}=\frac{3e^{-2}}{4i}$$$${\int }_{R}f\left(z\right)dz=2i\pi .\frac{3e^{-2}}{4i}=\frac{3\pi }{2}{e}^{-2}$$$$I=\frac{1}{2}{\int }_R^{}f\left(z\right)dz=\frac{3\pi }{4}{e}^{-2}=\frac{3\pi }{4e^2}.$$

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