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Question Number 36203 by prof Abdo imad last updated on 30/May/18

let f(t) = ∫_0 ^∞ ((cos(tx))/((2+x^2 )^2 ))dx 1) find a simple form of f(t) 2) calculate ∫_0 ^∞ ((cos(3x))/((2+x^2 )^2 ))dx

$${let}\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left({tx}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$

Commented by prof Abdo imad last updated on 01/Jun/18

we have 2f(t) = ∫_(−∞) ^(+∞) ((cos(tx))/((2+x^2 )^2 ))dx =Re( ∫_(−∞) ^(+∞) (e^(itx) /((2+x^2 )^2 ))dx) let introduce the complex?function ϕ(z) = (e^(itz) /((2+z^2 )^2 )) ϕ(z) = (e^(itz) /((z−i(√2))^2 (z +i(√2))^2 )) so the poles of ϕ are i(√2) and −i(√2) (doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res( ϕ,i(√2)) Res(ϕ,i(√2)) =lim_(z→i(√2)) { (z−i(√2))^2 ϕ(z)}^′ =lim_(z→i(√2)) { (e^(itz) /((z+i(√2))^2 ))}^′ =lim_(z→i(√2)) { ((it e^(itz) (z+i(√2))^2 −2(z+i(√2))e^(itz) )/((z +i(√2))^4 ))} =lim_(z→i(√2)) e^(itz) ((it(z +i(√2)) −2)/((z +i(√2))^3 )) =e^(it(i(√2))) ((it(2i(√2)) −2)/((2i(√2))^3 )) = ((−2t(√2)−2)/(−8i(2(√2)))) e^(−t(√2)) = ((t(√2) +1)/(8i(√2))) e^(−t(√2)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((1+t(√2))/(8i(√2))) e^(−t(√2)) = (π/(4(√2)))(1+t(√2)) e^(−t(√2)) but we have 2f(t)= Re( ∫_(−∞) ^(+∞) ϕ(z)dz) ⇒ ★ f(t) = (π/(8(√2))) (1+t(√2))e^(−t(√2)) ★ 2) we have ∫_0 ^∞ ((cos(3x))/((2+x^2 )^2 ))dx =f(3) ⇒ ∫_0 ^∞ ((cos(3x))/((2+x^2 )^2 ))dx = (π/(8(√2)))( 1+3(√2))e^(−3(√2)) .

$${we}\:{have}\:\:\mathrm{2}{f}\left({t}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({tx}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{itx}} }{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\right)\:{let}\:{introduce}\:{the} \\ $$$${complex}?{function}\:\varphi\left({z}\right)\:=\:\frac{{e}^{{itz}} }{\left(\mathrm{2}+{z}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{itz}} }{\left({z}−{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \left({z}\:+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$${i}\sqrt{\mathrm{2}}\:\:{and}\:−{i}\sqrt{\mathrm{2}}\:\left({doubles}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\:\varphi,{i}\sqrt{\mathrm{2}}\right) \\ $$$${Res}\left(\varphi,{i}\sqrt{\mathrm{2}}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \left\{\:\left({z}−{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{'} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \:\:\left\{\:\:\frac{{e}^{{itz}} }{\left({z}+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\right\}^{'} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \:\:\left\{\:\frac{{it}\:{e}^{{itz}} \left({z}+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:\:−\mathrm{2}\left({z}+{i}\sqrt{\mathrm{2}}\right){e}^{{itz}} }{\left({z}\:+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{4}} }\right\} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \:\:\:{e}^{{itz}} \:\frac{{it}\left({z}\:+{i}\sqrt{\mathrm{2}}\right)\:−\mathrm{2}}{\left({z}\:+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$={e}^{{it}\left({i}\sqrt{\mathrm{2}}\right)} \:\frac{{it}\left(\mathrm{2}{i}\sqrt{\mathrm{2}}\right)\:−\mathrm{2}}{\left(\mathrm{2}{i}\sqrt{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$=\:\frac{−\mathrm{2}{t}\sqrt{\mathrm{2}}−\mathrm{2}}{−\mathrm{8}{i}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)}\:{e}^{−{t}\sqrt{\mathrm{2}}} \:\:=\:\:\frac{{t}\sqrt{\mathrm{2}}\:+\mathrm{1}}{\mathrm{8}{i}\sqrt{\mathrm{2}}}\:{e}^{−{t}\sqrt{\mathrm{2}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:\frac{\mathrm{1}+{t}\sqrt{\mathrm{2}}}{\mathrm{8}{i}\sqrt{\mathrm{2}}}\:{e}^{−{t}\sqrt{\mathrm{2}}} \\ $$$$=\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\left(\mathrm{1}+{t}\sqrt{\mathrm{2}}\right)\:{e}^{−{t}\sqrt{\mathrm{2}}} \:\:{but}\:{we}\:{have} \\ $$$$\mathrm{2}{f}\left({t}\right)=\:{Re}\left(\:\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\right)\:\Rightarrow \\ $$$$\bigstar\:{f}\left({t}\right)\:=\:\frac{\pi}{\mathrm{8}\sqrt{\mathrm{2}}}\:\left(\mathrm{1}+{t}\sqrt{\mathrm{2}}\right){e}^{−{t}\sqrt{\mathrm{2}}} \:\bigstar \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:={f}\left(\mathrm{3}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:=\:\frac{\pi}{\mathrm{8}\sqrt{\mathrm{2}}}\left(\:\mathrm{1}+\mathrm{3}\sqrt{\mathrm{2}}\right){e}^{−\mathrm{3}\sqrt{\mathrm{2}}} \:. \\ $$

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