Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods? I have seen the Fresnel integral
$$\int_0^\infty \sin x^2\, dx = \sqrt{\frac{\pi}{8}}$$
evaluated by contour integration and other complex analysis methods, and I have found these methods to be the standard way to evaluate this integral.  I was wondering, however, does anyone know a real analysis method to evaluate this integral?

First, use the substitution $t = x^2$ to obtain
$$ I := \int_{0}^{\infty}\sin\left(x^2\right)\;dx=\int_{0}^{\infty}\frac{\sin t}{2\sqrt{t}}\;dt. $$
This integral converges conditionally, thus we make an integration by parts to obtain an absolutely convergent integral as follows:
$$I
= \left[\frac{1-\cos t}{2\sqrt{t}}\right]_{0}^{\infty}-\int_{0}^{\infty}\left(\frac{d}{dt}\frac{1}{2\sqrt{t}}\right)(1 - \cos t)\;dt
=\int_{0}^{\infty}\frac{1 - \cos t}{4t^{3/2}}\;dt.$$
Now, from gamma integral,
$$ \begin{align*}
I
&=\frac{1}{4\Gamma(3/2)}\int_{0}^{\infty}\left(\frac{\Gamma(3/2)}{t^{3/2}}\right)(1 - \cos t)\;dt \\
&=\frac{1}{4\Gamma(3/2)}\int_{0}^{\infty}\left(\int_{0}^{\infty}u^{1/2}e^{-tu}\;du\right)(1 - \cos t)\;dt\\
&=\frac{1}{4\Gamma(3/2)}\int_{0}^{\infty}\int_{0}^{\infty}u^{1/2}e^{-tu}(1 - \cos t)\;dudt \\
&=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\int_{0}^{\infty}u^{1/2}e^{-tu}(1 - \cos t)\;dudt,
\end{align*} $$
where in the last line we have used the fact that $\Gamma(3/2) = \frac{1}{2}\sqrt{\pi}$, which is a direct consequence of the Gaussian integral. (Of course, this integral can be evaluated by a famous real analysis technique.) By Tonelli's theorem we can change the order of integration, and with the substitution $u = v^2$ we obtain
$$\begin{align*}I
&=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\int_{0}^{\infty}u^{1/2}e^{-tu}(1 - \cos t)\;dtdu \\
&= \frac{1}{2\sqrt{\pi}} \int_{0}^{\infty}\frac{u^{1/2}}{u(1+u^2)}\;du
= \frac{1}{2\sqrt{\pi}} \int_{0}^{\infty} \frac{2}{1+v^4} \; dv.
\end{align*}$$
To evaluate the last integral, we use the following decomposition
$$ \frac{2}{1+v^4} = \frac{1+v^{-2}}{(v-v^{-1})^2-2} - \frac{1-v^{-2}}{(v+v^{-1})^2-2}. $$
Thus with the substitution $z = v - v^{-1}$ and $w = v + v^{-1}$, the integral becmes
$$\begin{align*}I
&= \frac{1}{2\sqrt{\pi}} \left( \int_{0}^{\infty} \frac{d(v-v^{-1})}{(v-v^{-1})^2+2} -  \int_{0}^{\infty} \frac{d(v+v^{-1})}{(v+v^{-1})^2-2} \right) \\
&= \frac{1}{2\sqrt{\pi}} \left( \int_{-\infty}^{\infty} \frac{dz}{z^2+2} - \int_{\infty}^{\infty} \frac{dw}{w^2-2} \right)
= \frac{1}{2\sqrt{\pi}} \left( \frac{\pi}{\sqrt{2}} - 0 \right)
= \sqrt{\frac{\pi}{8}}.
\end{align*}$$
Only a slight modification of this argument immediately yields
$$ \int_{0}^{\infty} \cos\left(x^2\right) \; dx = \sqrt{\frac{\pi}{8}}. $$
