How prove this inequality $I=\int_{0}^{\sqrt{2\pi}}\sin{x^2}dx>0$

Question

show that $$I=\int_{0}^{\sqrt{2\pi}}\sin{x^2}dx>0$$

follow is my methods: let $$x^2=t$$ then $$I=\int_{0}^{\pi}\dfrac{\sin{t}}{2\sqrt{t}}dt+\int_{\pi}^{2\pi}\dfrac{\sin{t}}{2\sqrt{t}}dt=\int_{0}^{\pi}\dfrac{\sin{t}}{2}\left(\dfrac{1}{\sqrt{t}}-\dfrac{1}{\sqrt{\pi+t}}\right)dt$$ since $$\dfrac{1}{\sqrt{t}}-\dfrac{1}{\sqrt{\pi+t}}>0$$ so $$I>0$$

My quuestion: have other methods? Thank you

Answer 1

Lazy answer: the Fresnel sine is positive for any positive argument.