$\int_{0}^{\infty}{{\mathrm{e}}^{-{\mathrm{i}}\,x^2}dx}$
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$\int_{0}^{\infty}{{\mathrm{e}}^{-{\mathrm{i}}\,x^2}dx}$
Calculate the integral:
$$\int_{0}^{\infty}{{\mathrm{e}}^{-{\mathrm{i}}\,x^2}dx}\,.$$
$$\int_{0}^{\infty}{{\mathrm{e}}^{-{\mathrm{i}}\,x^2}dx}\,.$$
Grigorios Kostakos
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Re: $\int_{0}^{\infty}{{\mathrm{e}}^{-{\mathrm{i}}\,x^2}dx}$
We will use contour integration on a wedge shaped contour of angle $\widehat{\omega} =\pi/4$ as our contour and an infinite large radius $R$. We are integrating on this contour the function $f(z)=e^{-iz^2}$. The contour is shown at the image below:
[/centre]
Clearly $f$ has no poles inside the contour , hence from Cauchy Residue Theorem we have that:
$$\oint_{\gamma} f(z) \, {\rm d}z =0 $$
Splitting the contour apart (into the two segments and the arc that is consisted of and at the same time taking care of the paramatrization) we have that:
$$\oint_{\gamma}f(z)\, {\rm d}z=\int_{0}^{R}e^{-x^2}\,{\rm d}x+\int_{0}^{\pi/4}iRe^{i\phi}e^{-R^2 e^{2i\phi}}\,{\rm d} \phi+\int_{R}^{0}e^{i\pi /4}e^{-ir^2}\,{\rm d}r$$
However the integral over the arc , as $R \rightarrow +\infty$ , goes to $0$ since:
$$\begin{aligned}\left | \int_{0}^{\pi/4}iRe^{i\phi}e^{-R^2e^{2i\phi}} \,{\rm d} \phi \right | &\leq \int_{0}^{\pi/4}\left | iRe^{i\phi}e^{-R^2 e^{2i\phi}} \right |\,{\rm d} \phi\\&\leq \int_{0}^{\pi/4}R\left | e^{-R^2 e^{2i\phi}} \right |\,{\rm d} \phi =\int_{0}^{\pi/4}Re^{-R^2 \cos 2\phi}\, {\rm d} \phi\\
&\overset{2\phi =t}{=\! =\! =\!}\int_{0}^{\pi/2}\frac{R}{2}e^{-R^2 \sin t }\,{\rm d}t \\&\leq \int_{0}^{\pi/2}\frac{R}{2}e^{-2R^2 t /\pi}\,{\rm d}t \\&\leq \frac{\pi}{4R}\left ( -e^{-R^2}+1 \right )\overset{R\rightarrow +\infty}{\longrightarrow }0 \end{aligned}$$
Hence:
$$\int_{0}^{\infty}e^{-x^2}\,{\rm d}x=e^{i\pi/4}\int_{0}^{\infty}e^{-ir^2}\,{\rm d}r \Rightarrow e^{-i\pi/4}\int_{0}^{\infty}e^{-x^2}\,{\rm d}x=\int_{0}^{\infty}e^{-ix^2}\,{\rm d}r$$
So combining Euler's Formula ($e^{ix}=\cos x +i \sin x$) and the Gaussian Integral we finally get that,
$$\int_0^\infty e^{-ix^2}\, {\rm d}x = \frac{1}{2}\sqrt{\frac{\pi}{2}} - i \frac{1}{2}\sqrt{\frac{\pi}{2}}$$
[/centre]
Clearly $f$ has no poles inside the contour , hence from Cauchy Residue Theorem we have that:
$$\oint_{\gamma} f(z) \, {\rm d}z =0 $$
Splitting the contour apart (into the two segments and the arc that is consisted of and at the same time taking care of the paramatrization) we have that:
$$\oint_{\gamma}f(z)\, {\rm d}z=\int_{0}^{R}e^{-x^2}\,{\rm d}x+\int_{0}^{\pi/4}iRe^{i\phi}e^{-R^2 e^{2i\phi}}\,{\rm d} \phi+\int_{R}^{0}e^{i\pi /4}e^{-ir^2}\,{\rm d}r$$
However the integral over the arc , as $R \rightarrow +\infty$ , goes to $0$ since:
$$\begin{aligned}\left | \int_{0}^{\pi/4}iRe^{i\phi}e^{-R^2e^{2i\phi}} \,{\rm d} \phi \right | &\leq \int_{0}^{\pi/4}\left | iRe^{i\phi}e^{-R^2 e^{2i\phi}} \right |\,{\rm d} \phi\\&\leq \int_{0}^{\pi/4}R\left | e^{-R^2 e^{2i\phi}} \right |\,{\rm d} \phi =\int_{0}^{\pi/4}Re^{-R^2 \cos 2\phi}\, {\rm d} \phi\\
&\overset{2\phi =t}{=\! =\! =\!}\int_{0}^{\pi/2}\frac{R}{2}e^{-R^2 \sin t }\,{\rm d}t \\&\leq \int_{0}^{\pi/2}\frac{R}{2}e^{-2R^2 t /\pi}\,{\rm d}t \\&\leq \frac{\pi}{4R}\left ( -e^{-R^2}+1 \right )\overset{R\rightarrow +\infty}{\longrightarrow }0 \end{aligned}$$
Hence:
$$\int_{0}^{\infty}e^{-x^2}\,{\rm d}x=e^{i\pi/4}\int_{0}^{\infty}e^{-ir^2}\,{\rm d}r \Rightarrow e^{-i\pi/4}\int_{0}^{\infty}e^{-x^2}\,{\rm d}x=\int_{0}^{\infty}e^{-ix^2}\,{\rm d}r$$
So combining Euler's Formula ($e^{ix}=\cos x +i \sin x$) and the Gaussian Integral we finally get that,
$$\int_0^\infty e^{-ix^2}\, {\rm d}x = \frac{1}{2}\sqrt{\frac{\pi}{2}} - i \frac{1}{2}\sqrt{\frac{\pi}{2}}$$
Imagination is much more important than knowledge.
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