integration - Definite integral involving Fresnel integrals

I am seeking to evaluate

$\int_0^{\infty} f(x)/x^2 \, dx$

with

$f(x)=1-\sqrt{\pi/6} \left(\cos (x) C\left(\sqrt{\frac{6 x}{\pi }} \right)+S\left(\sqrt{\frac{6 x}{\pi }} \right) \sin (x)\right)/\sqrt{x}$.

$C(x)$ and $S(x)$ are the Fresnel integrals. Numerical integration suggests that the integral equals $2 \pi/(3 \sqrt{3})$, which would also be desirable within the (physical) context it arose. How can this be proved?

Answer

First of all,

$$C(x)=\int_{0}^{x}\cos\left(\frac{1}{2}\pi t^{2}\right)dt=\sqrt{\frac{2}{\pi}}\int_{0}^{\sqrt{\pi/2} \,\left (x\right )}\cos(z^{2})\, dz,$$
and
$$S(x)=\int_{0}^{x}\sin\left(\frac{1}{2}\pi t^{2}\right)dt=\sqrt{\frac{2}{\pi}}\int_{0}^{\sqrt{\pi/2} \,\left (x\right )}\sin(z^{2})\, dz.$$
So we have

$$C\left(\sqrt{\frac{6x}{\pi}}\right)=\sqrt{\frac{2}{\pi}}\int_{0}^{\sqrt{3x}}\cos(z^{2})\, dz,$$
and
$$S\left(\sqrt{\frac{6x}{\pi}}\right)=\sqrt{\frac{2}{\pi}}\int_{0}^{\sqrt{3x}}\sin(z^{2})\, dz.$$
Using these we can write
$$\begin{eqnarray*} f(x) & = & 1-\sqrt{\pi/6}\frac{\cos(x)C\left(\sqrt{\frac{6x}{\pi}}\right)+\sin(x)S\left(\sqrt{\frac{6x}{\pi}}\right)}{\sqrt{x}} \\ & = & \frac{\int_{0}^{\sqrt{3x}}(1-\cos(x-z^{2}))\, dz}{\sqrt{3x}} \end{eqnarray*}$$
and
$$\int_{0}^{\infty}\frac{f(x)}{x^{2}}\, dx=\frac{2}{\sqrt{3}}\int_{0}^{\infty}\int_{0}^{\sqrt{3x}}\frac{\sin^{2}((x-z^{2})/2)}{x^{5/2}}\, dz\, dx.$$
Let's introduce the new variable $z=t\sqrt{x}$. Then $dz=\sqrt{x}dt$ and
$$\begin{eqnarray*} \int_{0}^{\infty}\int_{0}^{\sqrt{3x}}\frac{\sin^{2}((x-z^{2})/2)}{x^{5/2}}\, dz\, dx & = & \int_{0}^{\infty}\int_{0}^{\sqrt{3}}\frac{\sin^{2}(x(1-t^{2})/2)}{x^{2}}\, dt\, dx\\ & = & \frac{1}{2}\int_{0}^{\infty}\int_{0}^{\sqrt{3}}\frac{\sin^{2}(x(1-t^{2}))}{x^{2}}\, dt\, dx\\ & = & \frac{1}{2}\int_{0}^{\infty}\int_{0}^{1}\frac{\sin^{2}(x(1-t^{2}))}{x^{2}}\, dt\, dx\\ & & +\frac{1}{2}\int_{0}^{\infty}\int_{1}^{\sqrt{3}}\frac{\sin^{2}(x(t^{2}-1))}{x^{2}}\, dt\, dx\\ & = & \frac{1}{2}\int_{0}^{1}\int_{0}^{\infty}\frac{\sin^{2}(x(1-t^{2}))}{x^{2}}\, dx\, dt\\ & & +\frac{1}{2}\int_{1}^{\sqrt{3}}\int_{0}^{\infty}\frac{\sin^{2}(x(t^{2}-1))}{x^{2}}\, dx\, dt\\ & = & \int_{0}^{1}\frac{\pi}{4}(1-t^{2})\, dt+\int_{1}^{\sqrt{3}}\frac{\pi}{4}(t^{2}-1)\, dt\\ & = & \frac{\pi}{3}, \end{eqnarray*}$$

where the formula
$$\int_0^{\infty}\frac{\sin^2(Ax)}{x^2}\,dx=\frac{1}{2}A\pi,\quad(A>0)$$
was applied. Your conjecture was excellent.