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Question Number 33986 by abdo imad last updated on 28/Apr/18
find ∫_(−∞) ^(+∞) ((cos(tx))/((1+x^2 )^2 )) dx with t≥0
$${find}\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({tx}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:{with}\:{t}\geqslant\mathrm{0} \\ $$
Commented by math khazana by abdo last updated on 01/May/18
let introduce the complex function ϕ(z)= (e^(itz) /((1+z^2 )^2 )) we have I =∫_(−∞) ^(+∞) ((cos(tx))/((1+x^2 )^2 ))dx= Re( ∫_(−∞) ^(+∞) (e^(itx) /((1+x^2 )^2 ))dx) ϕ(z)= (e^(itz) /((z−i)^2 (z+i)^2 )) so the poles of ϕ are i and−i( doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((2−1)!)) ((z−i)^2 f(z))^′ =lim_(z→i) ( (e^(itz) /((z+i)^2 )))^′ =lim_(z→i) ((ite^(itz) (z+i)^2 −2(z+i)e^(itz) )/((z+i)^4 )) =lim_(z→i) (((z+i)it e^(itz) −2 e^(itz) )/((z+i)^3 )) =(((2i)it e^(−t) −2 e^(−t) )/((2i)^3 )) = ((−4 e^(−t) )/(−8i)) = (1/(2i)) e^(−t) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(−t) /(2i)) =π e^(−t) ⇒ I= π e^(−t)
$${let}\:{introduce}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)=\:\frac{{e}^{{itz}} }{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{we}\:{have} \\ $$$${I}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({tx}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{itx}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\right) \\ $$$$\varphi\left({z}\right)=\:\:\frac{{e}^{{itz}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$${i}\:{and}−{i}\left(\:{doubles}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\left(\left({z}−{i}\right)^{\mathrm{2}} {f}\left({z}\right)\right)^{'} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\left(\:\:\frac{{e}^{{itz}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right)^{'} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{{ite}^{{itz}} \:\left({z}+{i}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{i}\right){e}^{{itz}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\left({z}+{i}\right){it}\:{e}^{{itz}} \:−\mathrm{2}\:{e}^{{itz}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{2}{i}\right){it}\:{e}^{−{t}} \:\:\:−\mathrm{2}\:{e}^{−{t}} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:=\:\frac{−\mathrm{4}\:{e}^{−{t}} }{−\mathrm{8}{i}}\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\:{e}^{−{t}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−{t}} }{\mathrm{2}{i}}\:=\pi\:{e}^{−{t}} \:\:\Rightarrow\:{I}=\:\pi\:{e}^{−{t}} \: \\ $$$$ \\ $$

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