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

find the value of ∫_0 ^∞ ((xsin(3x))/((1+x^2 )^2 ))dx

$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{xsin}\left(\mathrm{3}{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$

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

let I = ∫_0 ^∞ ((x sin(3x))/((1+x^2 )^2 ))dx 2I = ∫_(−∞) ^(+∞) ((x sin(3x))/((1+x^2 )^2 ))dx =Im( ∫_(−∞) ^(+∞) ((x e^(3ix) )/((x^2 +1)^2 ))dx) let ϕ(z)= ((z e^(3iz) )/((z^2 +1)^2 )) ϕ(z)= ((z e^(3iz) )/((z−i)^2 (z+i)^2 )) the polrs of ϕ are i and −i (doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i)=lim_(z→i) (1/((2−1)!)) {(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { ((z e^(3iz) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((e^(3iz) +3iz e^(3iz) )(z+i)^2 −2(z+i)z e^(3iz) )/((z+i)^4 )) =lim_(z→i) (((1+3iz)e^(3iz) (z+i) −2z e^(3iz) )/((z+i)^3 )) =(((1−3)e^(−3) (2i) −2ie^(−3) )/((2i)^3 )) =(((−4i−2i)e^(−3) )/(−8i)) = ((6 e^(−3) )/8) =(3/4)e^(−3) 2I =Im( ∫_(−∞) ^(+∞) ϕ(z)dz) ⇒ I = (3/8)e^(−3) .

$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}\:{sin}\left(\mathrm{3}{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$\mathrm{2}{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{x}\:{sin}\left(\mathrm{3}{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:={Im}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{x}\:{e}^{\mathrm{3}{ix}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\right) \\ $$$${let}\:\varphi\left({z}\right)=\:\frac{{z}\:{e}^{\mathrm{3}{iz}} }{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)=\:\frac{{z}\:{e}^{\mathrm{3}{iz}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:{the}\:{polrs}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i} \\ $$$$\left({doubles}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:{but} \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \left\{\:\frac{{z}\:{e}^{\mathrm{3}{iz}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\left({e}^{\mathrm{3}{iz}} \:\:+\mathrm{3}{iz}\:{e}^{\mathrm{3}{iz}} \right)\left({z}+{i}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{i}\right){z}\:{e}^{\mathrm{3}{iz}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\left(\mathrm{1}+\mathrm{3}{iz}\right){e}^{\mathrm{3}{iz}} \left({z}+{i}\right)\:−\mathrm{2}{z}\:{e}^{\mathrm{3}{iz}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{1}−\mathrm{3}\right){e}^{−\mathrm{3}} \left(\mathrm{2}{i}\right)\:−\mathrm{2}{ie}^{−\mathrm{3}} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:=\frac{\left(−\mathrm{4}{i}−\mathrm{2}{i}\right){e}^{−\mathrm{3}} }{−\mathrm{8}{i}} \\ $$$$=\:\frac{\mathrm{6}\:{e}^{−\mathrm{3}} }{\mathrm{8}}\:=\frac{\mathrm{3}}{\mathrm{4}}{e}^{−\mathrm{3}} \\ $$$$\mathrm{2}{I}\:={Im}\left(\:\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\right)\:\Rightarrow\:{I}\:=\:\frac{\mathrm{3}}{\mathrm{8}}{e}^{−\mathrm{3}} . \\ $$

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