Estimate of Fresnel-like integral Does it correct that if $x\in\mathbb{R}_+$ then
$$\left | \int_{x^2}^{(x+1)^2}\frac{\sin t}{\sqrt{t}}dt \right |\le \frac{2}{x}?$$

Given $x\in\mathbb{R}^+$,
$$\int_{x^2}^{(x+1)^2}\frac{\sin t}{\sqrt{t}}\,dt = \left.\frac{1-\cos t}{\sqrt{t}}\right|_{x^2}^{(x+1)^2}+\frac{1}{2}\int_{x^2}^{(x+1)^2}\frac{1-\cos t}{t\sqrt{t}}\,dt $$
hence:

$$\left|\int_{x^2}^{(x+1)^2}\frac{\sin t}{\sqrt{t}}\,dt\right|\leq \frac{2}{x}+\int_{x}^{x+1}\frac{1-\cos(u^2)}{u^2}\,du\leq \frac{2}{x}+\frac{2}{x^2}. $$

This is quite close to the best approximation we can hope for. 
