Question Number 177665 by Ahmed777hamouda last updated on 07/Oct/22
Answered by Ar Brandon last updated on 07/Oct/22
$$\int_{−\infty} ^{\infty} \mathrm{sin}{x}^{\mathrm{2}} {dx}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{t}}{\:\sqrt{{t}}}{dt}=\frac{\pi}{\mathrm{2}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\right)}=\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$
Answered by Mathspace last updated on 07/Oct/22
$$\int_{−\infty} ^{+\infty} {cos}\left({x}^{\mathrm{2}} \right){dx}−{i}\int_{−\infty} ^{+\infty} {sin}\left({x}^{\mathrm{2}} \right){dx} \\ $$$$=\int_{−\infty} ^{+\infty} {e}^{−{ix}^{\mathrm{2}} } {dx}\:\:\left(\sqrt{{i}}{x}={t}\right) \\ $$$$=\int_{−\infty} ^{+\infty} {e}^{−{t}^{\mathrm{2}} } \frac{{dt}}{\:\sqrt{{i}}}=\frac{\mathrm{1}}{\:\sqrt{{i}}}\int_{−\infty} ^{+\infty} {e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$={e}^{−\frac{{i}\pi}{\mathrm{4}}} \sqrt{\pi}=\sqrt{\pi}\left\{{cos}\left(\frac{\pi}{\mathrm{4}}\right)−{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right\} \\ $$$$=\sqrt{\pi}\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{i}\right\} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} {cos}\left({x}^{\mathrm{2}} \right){dx}=\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$$${and}\:\int_{−\infty} ^{+\infty} {sin}\left({x}^{\mathrm{2}} \right){dx}=\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$$${another}\:{way}\:{with}\:{residus}\: \\ $$$${theorem}\:{but}\:{very}\:{long} \\ $$