Question Number 143786 by mohammad17 last updated on 18/Jun/21
$$\int_{−\infty} ^{\:\infty} \frac{{e}^{{iax}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:\:\:\:{how}\:{can}\:{it}\:{solve}\:{this} \\ $$
Commented by mohammad17 last updated on 18/Jun/21
$$????? \\ $$
Answered by mathmax by abdo last updated on 18/Jun/21
$$\Psi=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{iax}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}\:\:\:\mathrm{considere}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iaz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iaz}} }{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)} \\ $$$$\mathrm{residus}\:\mathrm{tbeorem}\:\mathrm{give}\:\int_{−\infty} ^{+\infty} \varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{i}\right) \\ $$$$=\mathrm{2i}\pi×\frac{\mathrm{e}^{\mathrm{ia}\left(\mathrm{i}\right)} }{\mathrm{2i}}=\pi\mathrm{e}^{−\mathrm{a}} \:\Rightarrow\Psi=\pi\:\mathrm{e}^{−\mathrm{a}} \\ $$