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

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin \left(x\right)}{x^2+x+1}dx=?$$

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

$$\mathrm{\Psi }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{sinx}}{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}=\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{ix}}}{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}\right)\mathrm{let}$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}}{{\mathrm{z}}^2+\mathrm{z}+1}\mathrm{poles}\mathrm{of}\phi ?$$$${\mathrm{z}}^2+\mathrm{z}+1=0\to \mathrm{\Delta }=1-4=-3\Rightarrow \mathrm{the}\mathrm{roots}\mathrm{are}{\mathrm{z}}_1=\frac{-1+\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}}$$$${\mathrm{z}}_2=\frac{-1-\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{-2\frac{\mathrm{i}\pi }{3}}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}}{(\mathrm{z}-{\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}})(\mathrm{z}-{\mathrm{e}}^{-\frac{2\mathrm{i}\pi }{3}})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}})=2\mathrm{i}\pi .\frac{{\mathrm{e}}^{\mathrm{i}{\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}}}}{{\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}}-{\mathrm{e}}^{-\frac{2\mathrm{i}\pi }{3}}}$$$$=\frac{2\mathrm{i}\pi }{2\mathrm{i}\sin \left(\frac{2\pi }{3}\right)}\times {\mathrm{e}}^{\mathrm{i}(-\frac{1}{2}+\mathrm{i}\frac{\sqrt{3}}{2})}=\frac{\pi }{\frac{\sqrt{3}}{2}}\times {\mathrm{e}}^{-\frac{\sqrt{3}}{2}}{\mathrm{e}}^{-\frac{\mathrm{i}}{2}}$$$$=\frac{2\pi }{\sqrt{3}}{\mathrm{e}}^{-\frac{\sqrt{3}}{2}}\{\cos \left(\frac{1}{2}\right)-\mathrm{i}\sin \left(\frac{1}{2}\right)\}\Rightarrow$$$$\mathrm{\Psi }=\mathrm{Im}\left({\int }_{\mathrm{R}}\phi \left(\mathrm{z}\right)\mathrm{dz}\right)=-\frac{2\pi }{\sqrt{3}}\sin \left(\frac{1}{2}\right){\mathrm{e}}^{-\frac{\sqrt{3}}{2}}$$

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