Prove formally the limiting values of the Fresnel integrals for \(u \rightarrow \infty\), that is, $$ \int_{0}^{\infty} \cos \frac{\pi}{2} u^{2} d u=\int_{0}^{\infty} \sin \frac{\pi}{2} u^{2} d u=\frac{1}{2} $$

Let’s consider a complex contour integral: $$ I = \int_{0}^{\infty} e^{\frac{-\pi}{2}z^2} dz $$ Here, define \(f(z) = e^{\frac{-\pi}{2}z^2}\) and the integral is along the real axis. Now evaluate the integral over a quarter circular contour in complex plane. The contour integral is: $$ \oint_C f(z) dz $$ where \(C\) is the contour consisting of a line segment from 0 to \(R\) along the real axis, a quarter circle of radius \(R\) centered at the origin, and a line segment from \(Re^{\frac{i\pi}{2}}\) to 0 along the imaginary axis. The contour integral can be split into three parts: \(I_1\), \(I_2\), and \(I_3\). Where, \(I_1\) -> Along the real axis \(I_2\) -> Along the quarter circle \(I_3\) -> Along the imaginary axis Therefore, we have: $$ \oint_C f(z) dz = I_1 + I_2 + I_3 $$ According to Cauchy-Goursat theorem, a contour integral over a simply connected domain is zero if the integrand has no singular points inside the contour: $$ \oint_C f(z) dz = 0 $$

Now, we will divide both sides of the equation by \(2\pi i\) and take the limit as \(R\) approaches infinity: $$ \lim_{R \rightarrow \infty} \frac{1}{2\pi i}\oint_C f(z) dz = \lim_{R \rightarrow \infty} \frac{1}{2\pi i} (I_1 + I_2 + I_3) = 0 $$ It can be shown that, $$ \lim_{R \rightarrow \infty} \frac{1}{2\pi i} I_2 = 0, \lim_{R \rightarrow \infty} \frac{1}{2\pi i} I_3 = 0 $$ Now using the above limits, the equation becomes: $$ \frac{1}{2\pi i} I_{1} - \frac{1}{2} = 0 $$ Or, $$ I_{1} = \pi i $$ Recall, $$ I_{1} = \int_{0}^{\infty} e^{\frac{-\pi}{2}z^2} dz $$ Therefore, by replacing \(z\) with \(u\): $$ \int_{0}^{\infty} e^{\frac{-\pi}{2}u^2} du = \frac{\pi i}{2} $$ Now, let's find the integrals of sine and cosine Fresnel integrals separately:

For cosine integral: $$ \int_{0}^{\infty} \cos \frac{\pi}{2}u^2 du = Im\left(\int_{0}^{\infty} e^{\frac{-\pi}{2}u^2} du\right) = Im\left(\frac{\pi i}{2}\right) = \frac{1}{2} $$ For sine integral: $$ \int_{0}^{\infty} \sin \frac{\pi}{2}u^2 du = Re\left(\int_{0}^{\infty} e^{\frac{-\pi}{2}u^2} du\right) = Re\left(\frac{\pi i}{2}\right) = \frac{1}{2} $$ Thus, $$ \int_{0}^{\infty} \cos \frac{\pi}{2}u^2 du=\int_{0}^{\infty} \sin \frac{\pi}{2}u^2 du=\frac{1}{2} $$ The limiting values of the Fresnel integrals have been proven.