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Question Number 143786 by mohammad17 last updated on 18/Jun/21

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{iax}}{1+x^2}dxhowcanitsolvethis$$

Commented by mohammad17 last updated on 18/Jun/21

$$?????$$

Answered by mathmax by abdo last updated on 18/Jun/21

$$\mathrm{\Psi }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{iax}}}{{\mathrm{x}}^2+1}\mathrm{dx}\mathrm{considere}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iaz}}}{{\mathrm{z}}^2+1}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iaz}}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}$$$$\mathrm{residus}\mathrm{tbeorem}\mathrm{give}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})$$$$=2\mathrm{i}\pi \times \frac{{\mathrm{e}}^{\mathrm{ia}\left(\mathrm{i}\right)}}{2\mathrm{i}}=\pi {\mathrm{e}}^{-\mathrm{a}}\Rightarrow \mathrm{\Psi }=\pi {\mathrm{e}}^{-\mathrm{a}}$$

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