How to show that $\int_0^x\cos(t^2)\ dt>0 $ for $x>0$? The Wikipedia article regarding the Fresnel integral seems to show that
$$
C(x):=\int_0^x\cos(t^2)\ dt>0
$$
for all $x>0$. 
But I can't find a reference. Could anyone give a proof or point me to some known reference? 

 A: Outline of a proof:
Note that since $C'(x) = \cos (x^2)$, $C'$ has zeros whenever $t^2 = \frac{\pi}{2} + k\pi$. Also, as was mentioned in the comments, we have 
$$ C(x) =  \int_0^{x^2}\frac{1}{2\sqrt{y}} \cos y dy.$$
So define
$$c_0 = \int_0^{\pi/2}\frac{1}{2\sqrt{y}} \cos y dy,$$
$$c_k = \int_{\pi/2+(k-1)\pi}^{\pi/2+k\pi}\frac{1}{2\sqrt{y}} \cos y dy,$$
for $k>0$.
Claim 1:$\displaystyle \sum_{k=0}^\infty c_k $ is an alternating series (with $|c_1|$ strictly less than $|c_0|$ and $c_0>0$). 
Claim 2: If $\pi/2+(n-1)\pi < x \leq \pi/2+n\pi$ then 
$$ \sum_{k=0}^{n-1} c_k < C(x) \leq \sum_{k=0}^{n} c_k.$$
Hence...
A: First, since $C'(x) = \cos(x^2)$, $C(x)$ alternates increasing and decreasing at $x = \sqrt{\pi (n+1/2)}$ for $n\in\mathbb{Z}$.
Next, using the substution Daniel suggests,
\begin{align*}
\int_{\sqrt{\pi (2n+1/2)}}^{\sqrt{\pi(2n+3/2)}} \cos(t^2)\,dt &= \int_{\pi (2n+1/2)}^{\pi(2n+3/2)} \cos(u)\frac{1}{2\sqrt{u}}\,du\\
&> \frac{1}{2\pi(2n+3/2)}\int_{\pi (2n+1/2)}^{\pi(2n+3/2)} \cos(u)\,du\\
&= \frac{1}{\pi(2n+3/2)}
\end{align*}
and
$$\int_{\sqrt{\pi (2n+3/2)}}^{\sqrt{\pi(2n+5/2)}} \cos(t^2)\,dt = \int_{\pi (2n+3/2)}^{\pi(2n+5/2)} \cos(u)\frac{1}{2\sqrt{u}}\,du > -\frac{1}{\pi(2n+3/2)}$$
since $\cos u$ is positive and negative in the entire domain of integration, respectively.
The last detail we need to check is that $C(\sqrt{3\pi/2})>0$; there must be a nice way of doing so analytically, though it is easy enough numerically to calculate that $C(\sqrt{3\pi/2})>0.4 > 0.$
Then we have $C(\sqrt{\pi(n+1/2)})>0$ for all $n$; and of course $C$ must also be positive between these critical points.
