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Question Number 131957 by rs4089 last updated on 10/Feb/21

$$Evaluate{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{sinx}{x^2+2x+2}dx$$

Answered by Dwaipayan Shikari last updated on 10/Feb/21

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{sinx}{{(x+1)}^2+1}x+1=u$$$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{sinucos1-sin1cosu}{u^2+1}du$$$$=cos1{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{sinu}{u^2+1}-sin1{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{cosu}{u^2+1}du$$$$=0-\frac{\pi }{e}sin\left(1\right)=-\frac{\pi }{e}sin\left(1\right)$$

Commented by mnjuly1970 last updated on 10/Feb/21

$$verynicemrpayan...$$

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

$$\mathrm{let}\mathrm{f}\left(\lambda \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\sin \left(\lambda \mathrm{x}\right)}{{\mathrm{x}}^2+2\mathrm{x}+2}\mathrm{dx}\mathrm{with}\lambda >0\Rightarrow \mathrm{f}\left(\lambda \right)=\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{i}\lambda \mathrm{x}}}{{\mathrm{x}}^2+2\mathrm{x}+2}\mathrm{dx}\right)$$$$\mathrm{\Phi }\left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{i}\lambda \mathrm{z}}}{{\mathrm{z}}^2+2\mathrm{z}+2}\mathrm{we}\mathrm{have}{\mathrm{z}}^2+2\mathrm{z}+2=0\Rightarrow {(\mathrm{z}+1)}^2+1=0\Rightarrow$$$${(\mathrm{z}+1)}^2=-1\Rightarrow \mathrm{z}+1=\stackrel{-}{+}\mathrm{i}\Rightarrow \mathrm{z}=-1\stackrel{-}{+}\mathrm{i}\mathrm{si}\mathrm{the}\mathrm{poles}\mathrm{of}\mathrm{\Phi }\mathrm{are}$$$${\mathrm{z}}_1=-1+\mathrm{i}\mathrm{and}{\mathrm{z}}_2=-1-\mathrm{i}$$$${\int }_{\mathrm{R}}\mathrm{\Phi }\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\mathrm{\Phi },{\mathrm{z}}_1)=2\mathrm{i}\pi .\frac{{\mathrm{e}}^{\mathrm{i}\lambda {\mathrm{z}}_1}}{2\mathrm{i}}=\pi {\mathrm{e}}^{\mathrm{i}\lambda (-1+\mathrm{i})}$$$$=\pi {\mathrm{e}}^{-\mathrm{i}\lambda -\lambda }=\pi {\mathrm{e}}^{-\lambda }\{\cos \left(\lambda \right)-\mathrm{isin}\left(\lambda \right)\}\Rightarrow \mathrm{f}\left(\lambda \right)=-\pi {\mathrm{e}}^{-\lambda }\sin \left(\lambda \right)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{sinx}}{{\mathrm{x}}^2+2\mathrm{x}+2}\mathrm{dx}=\mathrm{f}\left(1\right)=-\frac{\pi }{\mathrm{e}}\sin \left(1\right)$$

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