Question Number 48170 by Abdo msup. last updated on 20/Nov/18
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({sin}\left({x}^{\mathrm{2}} \right)\right)}{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 24/Nov/18
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({sin}\left({x}^{\mathrm{2}} \right)\right)}{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}\:{dx}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({sin}\left({x}^{\mathrm{2}} \right)\right)}{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$={Re}\left(\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{isin}\left({x}^{\mathrm{2}} \right)} }{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\right)\:{let}\:\varphi\left({z}\right)=\frac{{e}^{{isin}\left({z}^{\mathrm{2}} \right)} }{\mathrm{2}{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\varphi\left({z}\right)=\frac{{e}^{{isin}\left({z}^{\mathrm{2}} \right)} }{\left(\sqrt{\mathrm{2}}{z}\:−{i}\right)\left(\sqrt{\mathrm{2}}{z}+{i}\right)}=\frac{{e}^{{isinz}^{\mathrm{2}} } }{\mathrm{2}\left({z}−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\left({z}+\frac{{i}}{\:\sqrt{\mathrm{2}}\:}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\:{but} \\ $$$${Res}\left(\varphi,\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)={lim}_{{z}\rightarrow\frac{{i}}{\:\sqrt{\mathrm{2}}}} \:\:\:\left({z}−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\varphi\left({z}\right)=\:\frac{{e}^{{isin}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)} }{\mathrm{4}\frac{{i}}{\:\sqrt{\mathrm{2}}}}\:=\frac{{e}^{−{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }{{i}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }{\mathrm{4}{i}}\:\sqrt{\mathrm{2}}=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:{e}^{−{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\Rightarrow\mathrm{2}{I}\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:{cos}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}{cos}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right). \\ $$
Answered by Abdulhafeez Abu qatada last updated on 24/Nov/18
