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Question Number 33986 by abdo imad last updated on 28/Apr/18

$$find{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(tx\right)}{{(1+x^2)}^2}dxwitht⩾0$$

Commented by math khazana by abdo last updated on 01/May/18

$$letintroducethecomplexfunction$$$$\phi \left(z\right)=\frac{e^{itz}}{{(1+z^2)}^2}wehave$$$$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(tx\right)}{{(1+x^2)}^2}dx=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{itx}}{{(1+x^2)}^2}dx\right)$$$$\phi \left(z\right)=\frac{e^{itz}}{{(z-i)}^2{(z+i)}^2}sothepolesof\phi are$$$$iand-i\left(doubles\right)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)$$$$Res(\phi ,i)={lim}_{z\to i}\frac{1}{(2-1)!}{\left({(z-i)}^{2}f\left(z\right)\right)}^{{}^{\prime }}$$$$={lim}_{z\to i}{\left(\frac{e^{itz}}{{(z+i)}^2}\right)}^{{}^{\prime }}$$$$={lim}_{z\to i}\frac{{ite}^{itz}{(z+i)}^2-2(z+i)e^{itz}}{{(z+i)}^4}$$$$={lim}_{z\to i}\frac{(z+i)it{e}^{itz}-2e^{itz}}{{(z+i)}^3}$$$$=\frac{\left(2i\right)it{e}^{-t}-2e^{-t}}{{\left(2i\right)}^3}=\frac{-4e^{-t}}{-8i}=\frac{1}{2i}{e}^{-t}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{-t}}{2i}=\pi e^{-t}\Rightarrow I=\pi e^{-t}$$$$$$

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