- Prove that
\[ \int_ {0} ^ {\infty} \sin (x ^ {2}) d x = \int_ {0} ^ {\infty} \cos (x ^ {2}) d x = \frac {\sqrt {2 \pi}}{4}. \]
These are the Fresnel integrals. Here, \(\int _ { 0 } ^ { \infty }\) is interpreted as \(\lim_{R \to \infty} \int _ { 0 } ^ { R }\)
[Hint: Integrate the function \(e ^ { - z ^ { 2 } }\) over the path in Figure 14. Recall that \(\textstyle \int _ { - \infty } ^ { \infty } e ^ { - x ^ { 2 } } d x = { \sqrt { \pi } } . ]\)
Solution
For \(R>0\), integrate the entire function \(e^{-z^2}\) around the positively oriented boundary of the sector \[ 0\le |z|\le R, \qquad 0\le \arg z\le \frac{\pi}{4}. \] Cauchy’s theorem gives \[ \int_0^R e^{-x^2}\,dx +\int_{\Gamma_R}e^{-z^2}\,dz -e^{i\pi/4}\int_0^R e^{-ir^2}\,dr=0, \tag{1} \] where \(\Gamma_R\) is the circular arc \(z=Re^{i\theta}\), \(0\le\theta\le\pi/4\).
We first show that the arc integral tends to \(0\). Since \[ |e^{-z^2}|=e^{-R^2\cos(2\theta)}, \] we have, after putting \(u=\pi/4-\theta\), \[ \left|\int_{\Gamma_R}e^{-z^2}\,dz\right| \le R\int_0^{\pi/4}e^{-R^2\sin(2u)}\,du. \] For \(0\le u\le\pi/4\), concavity of \(\sin\) on \([0,\pi/2]\) gives \[ \sin(2u)\ge \frac{4u}{\pi}. \] Therefore \[ \left|\int_{\Gamma_R}e^{-z^2}\,dz\right| \le R\int_0^{\pi/4}e^{-4R^2u/\pi}\,du \le \frac{\pi}{4R}\longrightarrow0. \tag{2} \]
Letting \(R\to\infty\) in (1), using \[ \int_0^\infty e^{-x^2}\,dx=\frac{\sqrt\pi}{2}, \] and (2), yields \[ \frac{\sqrt\pi}{2} =e^{i\pi/4}\int_0^\infty e^{-ir^2}\,dr. \] Hence \[ \int_0^\infty e^{-ir^2}\,dr =e^{-i\pi/4}\frac{\sqrt\pi}{2} =\frac{\sqrt{2\pi}}{4}(1-i). \] Taking real and imaginary parts gives \[ \int_0^\infty \cos(x^2)\,dx=\frac{\sqrt{2\pi}}{4}, \qquad -\int_0^\infty \sin(x^2)\,dx=-\frac{\sqrt{2\pi}}{4}. \] Thus \[ \boxed{ \int_0^\infty \sin(x^2)\,dx = \int_0^\infty \cos(x^2)\,dx =\frac{\sqrt{2\pi}}{4}}. \]