Estimate of Fresnel-like integral

Question

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

Answer 1

Assume $x>0$.

One may integrate by parts: $$ \begin{align} \left| \int_{x^2}^{(x+1)^2}\frac{\sin t}{\sqrt{t}}dt\right|&=\left|\left[-\frac{1}{\sqrt{t}}\cos t\right]_{x^2}^{(x+1)^2}-\frac{1}{2}\int_{x^2}^{(x+1)^2}\frac{1}{t\sqrt{t}}\cos t\:dt\right| \\\\&= \left|\frac{\cos (x^2)}x-\frac{\cos ((x+1)^2)}{x+1}-\frac{1}{2}\int_{x^2}^{(x+1)^2}\frac{1}{t\sqrt{t}}\cos t\:dt\right| \\\\&\leq\left|\frac{\cos (x^2)}x-\frac{\cos ((x+1)^2)}{x+1}\right|+\frac{1}{2}\int_{x^2}^{(x+1)^2}\left|\frac{1}{t\sqrt{t}}\cos t\right|\:dt \\\\&\leq\left|\frac{(x+1)\cos (x^2)-x\cos ((x+1)^2)}{x(x+1)}\right|+\frac{1}{2}\int_{x^2}^{(x+1)^2}\frac{1}{t\sqrt{t}}\:dt \\\\&\leq\frac{2x+1}{x(x+1)}+\frac{1}{x(x+1)} \\\\&\leq\frac{2}{x}. \end{align} $$

Answer 2

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.