Example 12. Calculate the Fresnel integrals

$$
\begin{aligned}
& I_{1}=\int_{0}^{\infty} \cos x^{2} d x \\
& I_{2}=\int_{0}^{\infty} \sin x^{2} d x
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-100.jpg?height=349&width=552&top_left_y=293&top_left_x=655)

Fig. 10

$$
\int_{0}^{\infty} e^{-x^{2}} d x=\frac{\sqrt{\pi}}{2}
$$
Solution. Consider the auxiliary function $f(z)=e^{i z^{2}}$ and the contour shown in Fig. 10 (a circular sector $O B A O$, where $O A=O B=R$ and $\angle B O A=\frac{\pi}{4}$). Inside this contour, $f(z)$ is analytic, and by Cauchy's theorem,

$$
\int_{O B A O} e^{i z^{2}} d z=\int_{0}^{R} e^{i z^{2}} d x+\int_{C_{R}} e^{i z^{2}} d z+\int_{A O} e^{i z^{2}} d z=0
$$

We will show that

$$
\lim _{R \rightarrow \infty} \int_{C_{R}} e^{i z^{2}} d z=0
$$

Indeed, setting $z^{2}=\xi$, we get $d z=\frac{d \xi}{2 \sqrt{\xi}}$ and

$$
\int_{c_{R}} e^{i z^{2}} d z=\int_{\Gamma_{R^{2}}} \frac{e^{i \xi}}{2 \sqrt{\xi}} d \xi
$$

where $\Gamma_{R^{2}}$ is a quarter arc of a circle of radius $R^{2}$.

The function $g(\xi)=\frac{1}{2 \sqrt{\xi}}$ satisfies the conditions of Jordan's lemma, hence,

$$
\lim _{R \rightarrow \infty} \int_{\Gamma_{R^{2}}} \frac{e^{i \xi}}{2 \sqrt{\xi}} d \xi=\lim _{R \rightarrow \infty} \int_{C_{R}} e^{i z^{2}} d z=0
$$

On the segment $A O: z=\rho e^{i \pi / 4}, z^{2}=\rho^{2} e^{i \pi / 2}=\rho^{2} i, 0 \leqslant \rho \leqslant R$. Therefore,

$$
\int_{A O} e^{i z^{2}} d z=\int_{R}^{0} e^{-\rho^{2}} e^{i \pi / 4} d \rho=-e^{i \pi / 4} \int_{0}^{R} e^{-\rho^{2}} d \rho
$$

Taking the limit as $R \rightarrow \infty$ in (14), and considering (15), (16), and (13), we have

$$
\int_{0}^{\infty} e^{i \pi^{2}} d x=e^{i \pi / 4} \frac{\sqrt{\pi}}{2}
$$

or

$$
\int_{0}^{\infty} \cos x^{2} d x+i \int_{0}^{\infty} \sin x^{2} d x=\frac{1}{2} \sqrt{\frac{\pi}{2}}+i \frac{1}{2} \sqrt{\frac{\pi}{2}}
$$

from which we obtain

$$
\int_{0}^{\infty} \cos x^{2} d x=\frac{1}{2} \sqrt{\frac{\pi}{2}}, \quad \int_{0}^{\infty} \sin x^{2} d x=\frac{1}{2} \sqrt{\frac{\pi}{2}}
$$