Question Number 200187 by sonukgindia last updated on 15/Nov/23
Answered by Mathspace last updated on 15/Nov/23
$${I}={Re}\left(\int_{−\infty} ^{+\infty} \frac{{e}^{{iax}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\right) \\ $$$${and}\:\int_{−\infty} ^{+\infty} \frac{{e}^{{iax}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\mathrm{2}{i}\pi{Re}\left({f},{i}\right)\:\:\:\:\:\left({f}\left({z}\right)=\frac{{e}^{{iaz}} }{{z}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${Res}\left({f},{i}\right)={lim}_{{z}\rightarrow{i}} \:\left({z}−{i}\right){f}\left({z}\right) \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\left({z}−{i}\right)\frac{{e}^{{iaz}} }{\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$$$=\frac{{e}^{{iai}} }{\mathrm{2}{i}}=\frac{{e}^{−{a}} }{\mathrm{2}{i}}\:\Rightarrow \\ $$$$\int_{{R}} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi.\frac{{e}^{−{a}} }{\mathrm{2}{i}}=\pi\:{e}^{−{a}} \\ $$$${I}={Re}\left(\int_{{R}} {f}\left({z}\right){dz}\right)=\pi{e}^{−{a}} \\ $$
Answered by Mathspace last updated on 15/Nov/23
$${J}=\int_{−\infty} ^{+\infty} \frac{{xsin}\left({ax}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$${J}={Im}\left(\int_{−\infty} ^{+\infty} \frac{{xe}^{{iax}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\right) \\ $$$${let}\:{f}\left({z}\right)=\frac{{ze}^{{iaz}} }{{z}^{\mathrm{2}} +\mathrm{1}}\left(=\frac{{ze}^{{iaz}} }{\left({z}−{i}\right)\left({z}+{i}\right)}\right) \\ $$$${poles}\:{are}\:{iand}\:−{i} \\ $$$${by}\:{residus}\:{theorem} \\ $$$$\int_{−\infty} ^{+\infty} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left({f},{i}\right) \\ $$$${Res}\left({f},{i}\right)={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right){f}\left({z}\right) \\ $$$$=\frac{{ie}^{{iai}} }{\mathrm{2}{i}}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{a}} \:\Rightarrow \\ $$$$\int_{{R}} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi.\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{a}} \\ $$$$={i}\pi\:{e}^{−{a}} \\ $$$${J}={Im}\left(\int….\right)\:\Rightarrow \\ $$$${J}=\pi\:{e}^{−{a}} \\ $$$$ \\ $$