Exercise:
Compute, $\int_{0}^{\infty }{\cos \left( {{x}^{2}} \right)dx}$
Solution: we know that ${{e}^{i\theta }}=\cos \theta +i\sin \theta $ where $i\in \mathbb{C}$ such that ${{i}^{2}}=-1$
So ${{e}^{-i{{x}^{2}}}}=\cos {{x}^{2}}-i\sin {{x}^{2}}\Leftrightarrow \int_{0}^{\infty }{{{e}^{-i{{x}^{2}}}}}dx=\int_{0}^{\infty }{\cos \left( {{x}^{2}} \right)dx-i\int_{0}^{\infty }{\sin \left( {{x}^{2}} \right)dx}}$
Notice that, $I=\int_{0}^{\infty }{{{e}^{-i{{x}^{2}}}}dx}\Leftrightarrow {{I}^{2}}=\int_{0}^{\infty }{\int_{0}^{\infty }{{{e}^{-i\left( {{x}^{2}}+{{y}^{2}} \right)}}}dxdy}$
Using polar form we get ${{I}^{2}}=\int_{0}^{\infty }{\int_{0}^{\frac{\pi }{2}}{{{e}^{-i{{r}^{2}}}}rdrd\theta =\int_{0}^{\frac{\pi }{2}}{\left( \int_{0}^{\infty }{r{{e}^{-i{{r}^{2}}}}dr} \right)d\theta }}}$
Let $u=i{{r}^{2}}\Leftrightarrow du=2irdr\Leftrightarrow rdr=\frac{du}{2i}$
So $\int_{0}^{\infty }{r{{e}^{-i{{r}^{2}}}}dr}=\frac{1}{2i}\int_{0}^{\infty }{{{e}^{-u}}du}=\frac{1}{2i}\left( -{{e}^{-u}} \right)_{0}^{\infty }=\frac{1}{2i}$
Thus ${{I}^{2}}=\int_{0}^{\frac{\pi }{2}}{\frac{1}{2i}d\theta }=\frac{\pi }{4i}\Leftrightarrow \int_{0}^{\infty }{{{e}^{-i{{x}^{2}}}}dx}=\frac{\sqrt{\pi }}{2\sqrt{i}}$
Hence $\int_{0}^{\infty }{\cos \left( {{x}^{2}} \right)dx}-i\int_{0}^{\infty }{\sin \left( {{x}^{2}} \right)dx}=\frac{\sqrt{\pi }}{2\sqrt{i}}$
But $\frac{1}{\sqrt{i}}=\frac{1}{\sqrt{{{e}^{i\frac{\pi }{2}}}}}=\frac{1}{{{\left( {{e}^{i\frac{\pi }{2}}} \right)}^{1/2}}}\frac{1}{{{e}^{i\frac{\pi }{4}}}}={{e}^{-i\frac{\pi }{4}}}=\cos \frac{\pi }{4}-i\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}$
Thus $\frac{\sqrt{\pi }}{2\sqrt{i}}=\frac{\sqrt{\pi }}{2}\left( \frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}} \right)=\frac{\sqrt{\pi }}{2\sqrt{2}}-i\frac{\sqrt{\pi }}{2\sqrt{2}}=\int_{0}^{\infty }{\cos {{x}^{2}}dx}-i\int_{0}^{\infty }{\sin {{x}^{2}}dx}$
By comparison we get $\int_{0}^{\infty }{\cos {{x}^{2}}dx=\int_{0}^{\infty }{\sin {{x}^{2}}dx=\frac{\sqrt{\pi }}{2\sqrt{2}}}}$
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