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Question Number 38469 by maxmathsup by imad last updated on 25/Jun/18
calculate f(a) = ∫_(−∞) ^(+∞) ((sin(ax))/(x^2 +x+1))dx 2) find the value of ∫_(−∞) ^(+∞) ((sin(3x))/(x^2 +x+1))dx
$${calculate}\:\:{f}\left({a}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({ax}\right)}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}{dx} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left(\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}{dx} \\ $$
Commented by math khazana by abdo last updated on 28/Jun/18
1) f(a)= Im( ∫_(−∞) ^(+∞) (e^(iax) /(x^2 +x+1))dx) let consider the complex function ϕ(z)= (e^(iaz) /(z^2 +z +1)) the poles of ϕ are j=e^(i((2π)/3)) and j^− =e^(−((i2π)/3)) j=−(1/2) +i((√3)/2) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,j) but ϕ(z)=(e^(iaz) /((z−j)(z−j^− ))) ⇒Res(ϕ,j)= (e^(iaj) /(j−j^− )) = (e^(ia(−(1/2) +((i(√3))/2))) /(2i((√3)/2))) = ((e^(−((a(√3))/2)) {cos((a/2))−isin((a/2))})/(i(√3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((e^(−((a(√3))/2)) { cos((a/2))−i sin((a/2))})/(i(√3))) =((2π)/( (√3))) e^(−((a(√3))/2)) { cos((a/2))−i sin((a/2))}⇒ f(a)=Im( ∫_(−∞) ^(+∞) ϕ(z)dz)=−((2π)/3) e^(−a((√3)/2)) sin((a/2)) 2) ∫_(−∞) ^(+∞) ((sin(3x))/(x^2 +x+1))dx=f(3)=−((2π)/3)e^(−((3(√3))/2)) sin((3/2)).
$$\left.\mathrm{1}\right)\:{f}\left({a}\right)=\:{Im}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{iax}} }{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}{dx}\right)\:{let}\:{consider} \\ $$$${the}\:{complex}\:{function}\:\varphi\left({z}\right)=\:\frac{{e}^{{iaz}} }{{z}^{\mathrm{2}} \:+{z}\:+\mathrm{1}} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:{j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:{and}\:\overset{−} {{j}}={e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${j}=−\frac{\mathrm{1}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{j}\right)\:{but}\: \\ $$$$\varphi\left({z}\right)=\frac{{e}^{{iaz}} }{\left({z}−{j}\right)\left({z}−\overset{−} {{j}}\right)}\:\Rightarrow{Res}\left(\varphi,{j}\right)=\:\frac{{e}^{{iaj}} }{{j}−\overset{−} {{j}}} \\ $$$$=\:\frac{{e}^{{ia}\left(−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} }{\mathrm{2}{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\:\frac{{e}^{−\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}} \left\{{cos}\left(\frac{{a}}{\mathrm{2}}\right)−{isin}\left(\frac{{a}}{\mathrm{2}}\right)\right\}}{{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}} \left\{\:{cos}\left(\frac{{a}}{\mathrm{2}}\right)−{i}\:{sin}\left(\frac{{a}}{\mathrm{2}}\right)\right\}}{{i}\sqrt{\mathrm{3}}} \\ $$$$=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\:{e}^{−\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}} \left\{\:{cos}\left(\frac{{a}}{\mathrm{2}}\right)−{i}\:{sin}\left(\frac{{a}}{\mathrm{2}}\right)\right\}\Rightarrow \\ $$$${f}\left({a}\right)={Im}\left(\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\right)=−\frac{\mathrm{2}\pi}{\mathrm{3}}\:{e}^{−{a}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:{sin}\left(\frac{{a}}{\mathrm{2}}\right) \\ $$$$\left.\mathrm{2}\right)\:\int_{−\infty} ^{+\infty} \:\:\frac{{sin}\left(\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}{dx}=\mathrm{f}\left(\mathrm{3}\right)=−\frac{\mathrm{2}\pi}{\mathrm{3}}{e}^{−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}} \:{sin}\left(\frac{\mathrm{3}}{\mathrm{2}}\right). \\ $$

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