Consider the oscillatory integral $$ I(x)=\int_0^1 e^{ix t^2}dt. $$ Standard Van der Corput lemma tells us that $I(x)=O(|x|^{-1/2})$ for $|x|>1$.
I was looking for a lower bound like this: $|I(x)|\geq c|x|^{-1/2}$, but it turns out to be impossible, since someone tells me that $I(x)$ has infinitely many zeros $x_n\to \infty$. But is it true? If yes, how can I prove this?
As far as I can see, $I(x)$ does not show any zero at all. $$I(x)=\int_0^1 e^{ix t^2}\,dt=\frac{1-i}{2} \sqrt{\frac{\pi }{2}}\,\,\,\frac{\text{erfi}\left((1+i) \sqrt{\frac x2}\right)}{\sqrt{x}}$$
Developed as a series for large values of $x$, we have $$I(x)=\frac{1+i}{2} \sqrt{\frac{\pi }{2}} {\frac{1}{\sqrt x}}-e^{i x} \left(\frac{i}{2 x}+\frac{1}{4 x^2}\right)+\cdots$$
Computing for $x=100$, the exact value is $(0.0601125+ 0.0583671\, i)$ while the above approximation gives $(0.0601123+ 0.0583668\,i)$.