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Question Number 156177 by amin96 last updated on 08/Oct/21

∫_(−∞) ^∞ ((sin (x))/(x^2 +x+1))dx=?

$$\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left({x}\right)}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}=? \\ $$

Answered by mathmax by abdo last updated on 10/Oct/21

Ψ=∫_(−∞) ^(+∞) ((sinx)/(x^2 +x+1))dx =Im(∫_(−∞) ^(+∞) (e^(ix) /(x^2 +x+1))dx) let ϕ(z)=(e^(iz) /(z^2 +z+1)) poles of ϕ? z^2 +z +1=0 →Δ=1−4=−3 ⇒the roots are z_1 =((−1+i(√3))/2)=e^((2iπ)/3) z_2 =((−1−i(√3))/2) =e^(−2((iπ)/3)) ⇒ϕ(z)=(e^(iz) /((z−e^((2iπ)/3) )(z−e^(−((2iπ)/3)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^((2iπ)/3) ) =2iπ.(e^(i e^((2iπ)/3) ) /(e^((2iπ)/3) −e^(−((2iπ)/3)) )) =((2iπ)/(2i sin(((2π)/3))))×e^(i(−(1/2)+i((√3)/2))) =(π/((√3)/2))×e^(−((√3)/2)) e^(−(i/2)) =((2π)/( (√3))) e^(−((√3)/2)) {cos((1/2))−i sin((1/2))} ⇒ Ψ=Im(∫_R ϕ(z)dz) =−((2π)/( (√3)))sin((1/2))e^(−((√3)/2))

$$\Psi=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{sinx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\mathrm{dx}\:=\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{ix}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}+\mathrm{1}}\mathrm{dx}\right)\:\mathrm{let} \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{z}+\mathrm{1}}\:\:\mathrm{poles}\:\mathrm{of}\:\varphi? \\ $$$$\mathrm{z}^{\mathrm{2}} \:+\mathrm{z}\:+\mathrm{1}=\mathrm{0}\:\rightarrow\Delta=\mathrm{1}−\mathrm{4}=−\mathrm{3}\:\Rightarrow\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{z}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \\ $$$$\mathrm{z}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{−\mathrm{2}\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} }{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)\:=\mathrm{2i}\pi.\frac{\mathrm{e}^{\mathrm{i}\:\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} } }{\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} −\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} } \\ $$$$=\frac{\mathrm{2i}\pi}{\mathrm{2i}\:\mathrm{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)}×\mathrm{e}^{\mathrm{i}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \:=\frac{\pi}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}×\mathrm{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\mathrm{e}^{−\frac{\mathrm{i}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\:\mathrm{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\left\{\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{i}\:\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right\}\:\Rightarrow \\ $$$$\Psi=\mathrm{Im}\left(\int_{\mathrm{R}} \varphi\left(\mathrm{z}\right)\mathrm{dz}\right)\:=−\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$

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