Prove that:
$$\int_0^\infty \cos x^2 \, {\rm d}x = \int_0^\infty \sin x^2 \, {\rm d}x =\frac{1}{2}\sqrt{\frac{\pi}{2}}$$
Solution
These are classic integrals. 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:
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,
$$\int_0^\infty e^{-ix^2}\, {\rm d}x = \frac{1}{2}\sqrt{\frac{\pi}{2}} - i \frac{1}{2}\sqrt{\frac{\pi}{2}}$$
Now we can easily read the result of both integrals since we are equating real and imaginary parts.
$$\int_0^\infty \cos x^2 \, {\rm d}x = \int_0^\infty \sin x^2 \, {\rm d}x =\frac{1}{2}\sqrt{\frac{\pi}{2}}$$
Solution
These are classic integrals. 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:
$$\oint_{\gamma} f(z) \, {\rm d}z =0 $$
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,
$$\int_0^\infty e^{-ix^2}\, {\rm d}x = \frac{1}{2}\sqrt{\frac{\pi}{2}} - i \frac{1}{2}\sqrt{\frac{\pi}{2}}$$
Now we can easily read the result of both integrals since we are equating real and imaginary parts.
A real analysis method for evaluating this integral would be to invoke the Mellin Transformation of the $\sin$ function
ReplyDeleteIndeed we have that:
$$\begin{aligned}
\int_0^\infty x^{a-1}\sin(x)\,\mathrm{d}x
&=\frac1{2i}\left(\int_0^\infty x^{a-1}e^{ix}\,\mathrm{d}x-\int_0^\infty x^{a-1}e^{-ix}\,\mathrm{d}x\right)\\
&=\frac1{2i}\left(e^{ia\pi/2}\int_0^\infty x^{a-1}e^{-x}\,\mathrm{d}x-e^{-ia\pi/2}\int_0^\infty x^{a-1}e^{-x}\,\mathrm{d}x\right)\\
&=\sin\left(\frac{a\pi}{2}\right)\Gamma(a)
\end{aligned}$$
For $a=1/2$ we have that:
$$\int_{0}^{\infty}\frac{\sin x}{\sqrt{x}}\, {\rm d}x =\sqrt{\frac{\pi}{2}}$$
Now, for our initial integral by applying the sub $u=x^2$ we have that:
$$\int_{0}^{\infty}\sin x^2 \, {\rm d}x = \frac{1}{2}\int_{0}^{\infty} \frac{\sin x}{\sqrt{x}}\, {\rm d}x = \frac{1}{2}\sqrt{\frac{\pi}{2}}$$
A similar approach can be followed for evaluating the integral: $\displaystyle \int_0^\infty \cos x^2 \, {\rm d}x$.
ReplyDelete