How to show positivity of Fresnel C integral?

Question

The Fresnel $C$-integral is defined as follows. $$C(x) = \int_0^x \cos(t^2) \, dt $$ From the plot found on Wikipedia it seems to be non-negative for all $x \geq 0$ however it is not obvious to me why this is. For example, if you make the change of variables $t = u^\frac{1}{2}$ then you find $$C(x) = \frac{1}{2} \int_0^{x^2} u^{-\frac{1}{2}} \cos(u) \, du$$ My first thought was to split it into positive and negative regions and make a crude bound. For example, $\cos(u)$ will be positive on $[0,\pi/2]$ and negative on $[\pi/2,3\pi/2]$. A crude lower bound on the first region would be $\left(\frac{\pi}{2}\right)^{-\frac{1}{2}} \int_0^\frac{\pi}{2} \cos(u) \, du$ and on the second region would be $\left(\frac{\pi}{2}\right)^{-\frac{1}{2}} \int_\frac{\pi}{2}^\frac{3\pi}{2} \cos(u) \, du$. However adding these two bounds together yields something negative and so it doesn't work (however, this approach does work to show the non-negativity of the Fresnel $S$-integral).

Answer 1

As DionelJamie states, the claim is true so long as you can show $\int_0^\frac{3\pi}{2} u^{-\frac{1}{2}} \cos(u) \,du \geq 0$. I will take this as obvious.

To show this let us split up the integral $J = \int_0^\frac{3\pi}{2} u^{-\frac{1}{2}} \cos(u) \,du$ into four regions. $J=J_1 + J_2 + J_3 + J_4$. \begin{align} J_1&= \int_0^\frac{\pi}{4} u^{-\frac{1}{2}} \cos(u) \,du \geq \cos\left(\frac{\pi}{4}\right)\int_0^\frac{\pi}{4} u^{-\frac{1}{2}} \,du = \sqrt{\frac{\pi}{2}} \\ J_2&= \int_\frac{\pi}{4}^\frac{\pi}{2} u^{-\frac{1}{2}} \cos(u) \,du \geq \left( \frac{\pi}{2} \right)^{-\frac{1}{2}}\int_\frac{\pi}{4}^\frac{\pi}{2} \cos(u) \,du = \left( \frac{\pi}{2} \right)^{-\frac{1}{2}} \left( 1 - \frac{1}{\sqrt{2}}\right)\\ J_3&= \int_\frac{\pi}{2}^\pi u^{-\frac{1}{2}} \cos(u) \,du \geq \left(\frac{\pi}{2}\right)^{-\frac{1}{2}}\int_\frac{\pi}{2}^\pi \cos(u) \,du = -\left(\frac{\pi}{2}\right)^{-\frac{1}{2}} \\ J_4&= \int_\pi^\frac{3\pi}{2} u^{-\frac{1}{2}} \cos(u) \,du \geq \pi^{-\frac{1}{2}}\int_\pi^\frac{3\pi}{2} \cos(u) \,du = - \pi^{-\frac{1}{2}} \end{align} Adding up these bounds we find that $J \geq \sqrt{\frac{\pi}{2}}- \frac{2}{\sqrt{\pi}} = \frac{1}{\sqrt{2\pi}}\left( \pi - 2 \sqrt{2} \right)$. Numerically one finds that $\pi - 2 \sqrt{2} > 0$ though I haven't tried to think of a proof of this.