Question Number 27620 by abdo imad last updated on 11/Jan/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{x}^{\mathrm{2}} } }{\mathrm{3}+{x}^{\mathrm{2}} }{dx}\:. \\ $$
Commented by abdo imad last updated on 12/Jan/18
$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{x}^{\mathrm{2}} } }{\mathrm{3}+{x}^{\mathrm{2}} }{dx}\:\:\:{due}\:{to}\:\left(−\mathrm{1}\right)={e}^{{i}\pi} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{{i}\pi{x}^{\mathrm{2}} } }{\mathrm{3}+{x}^{\mathrm{2}} }{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{R}} \:\:\frac{{e}^{{i}\pi{x}^{\mathrm{2}} } }{\mathrm{3}+{x}^{\mathrm{2}} }{dx}\:\:{let}\:{introduce}\:{the} \\ $$$${complex}\:{function}\:\:{f}\left({z}\right)=\:\frac{{e}^{{i}\pi{z}^{\mathrm{2}} } }{{z}^{\mathrm{2}} \:+\mathrm{3}}\:\:{we}\:{have} \\ $$$${f}\left({z}\right)=\:\frac{{e}^{{i}\pi{z}^{\mathrm{2}} } }{\left({z}\:−{i}\sqrt{\mathrm{3}}\right)\left({z}+{i}\sqrt{\left.\mathrm{3}\right)}\right.}\:\:{the}\:{poles}\:{of}\:{f}\:{are}\:{z}_{\mathrm{1}} =\:{i}\sqrt{\mathrm{3}}\:{and} \\ $$$${z}_{\mathrm{2}} =−{i}\sqrt{\mathrm{3}}\:\:\:{the}\:{residus}\:{theorem}\:{give} \\ $$$$\int_{{R}} {f}\left({z}\right){dz}=\:\mathrm{2}{i}\pi\:{Res}\left({f},{i}\sqrt{\mathrm{3}}\right)\:\:{but} \\ $$$${Res}\left({f},{i}\sqrt{\mathrm{3}}\right)=\:{lim}_{{z}−>{i}\sqrt{\mathrm{3}}} \left({z}\:−{i}\sqrt{\mathrm{3}}\right){f}\left({z}\right)=\:{lim}_{{z}−>{i}\sqrt{\mathrm{3}}} \:\:\frac{{e}^{{i}\pi{z}^{\mathrm{2}} } }{{z}+{i}\sqrt{\mathrm{3}}} \\ $$$$=\:\:\frac{{e}^{{i}\pi\left(−\mathrm{3}\right)} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}=\:\:\frac{\left(−\mathrm{1}\right)^{−\mathrm{3}} }{\mathrm{2}{i}\sqrt{\mathrm{3}}}=\:\frac{−\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}} \\ $$$$\int_{{R}\:} {f}\left({z}\right){dz}=\:\mathrm{2}{i}\pi.\frac{−\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\:=\:\frac{−\pi}{\sqrt{\mathrm{3}}}\:\:\:\:\:\:{and}\:\:\:{I}=\:\frac{−\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:. \\ $$