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?

Let $$F(t)=\int_0^{\infty} \sin(tx^2)\ dx$$ Taking the Laplace Transform of $F(t)$: $$\mathscr{L}\{F(t)\}=\int_0^{\infty}F(t)e^{-st}\ dt$$ By being slightly less than rigorous this can be rewritten as $$\int_0^{\infty}\mathscr{L}\{\sin (tx^2)\}\ dx=\int_0^{\infty}\frac{x^2}{s^2+x^4}\ dx$$ This can be put into partial fractions as $$\frac{x^2}{s^2+x^4}\to\frac{Ax+B}{(s+ix^2)}+\frac{Cx+D}{(s-ix^2)}$$ $$(Ax+B)(s-ix^2)+(Cx+D)(s+ix^2)=x^2$$ $$A=C=0,\ B=\frac{i}{2},\ D=-\frac{i}{2}$$ $$\begin{equation}\begin{aligned}
\int_0^{\infty}\frac{x^2}{s^2+x^4}\ dx&=\int_0^{\infty}\frac{i}{2(s+ix^2)}\ dx\ -\int_0^{\infty}\frac{i}{2(s-ix^2)}\ dx \\
&=\frac{1}{2}\int_0^{\infty}\frac{1}{x^2-is}\ dx\ +\frac{1}{2}\int_0^{\infty}\frac{1}{x^2+is}\ dx \\
&=\lim_{c\to\infty}\frac{1}{2}\bigg[\frac{1}{\sqrt{is}}\tan ^{-1} \bigg(\frac{x}{\sqrt{is}}\bigg)-\frac{1}{\sqrt{is}}\tanh ^{-1}\bigg(\frac{x}{\sqrt{is}}\bigg)\bigg]_0^c \\
&=\frac{\pi}{4}\bigg(\frac{1}{\sqrt{is}}+\frac{1}{\sqrt{-is}}\bigg)=\frac{\pi}{4}\bigg(\frac{\sqrt{2}}{2\sqrt{s}}(1-i)+\frac{\sqrt{2}}{2\sqrt{s}}(1+i)\bigg)=\frac{\pi\sqrt{2}}{4\sqrt{s}}
\end{aligned}\end{equation}$$
Now taking the inverse Laplace Transform:
$$\mathscr{L}^{-1}\bigg\{\frac{\pi\sqrt{2}}{4\sqrt{s}}\bigg\}=\frac{\pi\sqrt{2}}{4}\mathscr{L}^{-1}\bigg\{\frac{1}{\sqrt{s}}\bigg\}=\frac{\pi\sqrt{2}}{4\sqrt{\pi t}}$$
By noting that $$F(1)=\int_0^{\infty}\sin (x^2)\ dx$$ 
$$\int_0^{\infty}\sin (x^2)\ dx=\frac{\pi\sqrt{2}}{4\sqrt{\pi\times 1}}=\sqrt{\frac{\pi}{8}}$$
