# How to prove that $|\int_{x}^{x+1} \sin(t^2)dt| \leq \frac{1}{x}$?

I have $F(x) = \int_{x}^{x+1} \sin(t^2)dt$, how can i proof that $|F(x)| \leq \frac{1}{x}$ for all $x > 0$? I have no idea how to do it.

• So, whati have you tried? What happens if you substitute in $u=t^2$ in the integral, for example? (Hint: If you can't see the way clearly to the end goal, that is no reason not to start the journey. Play around and see what can be discovered.) Commented Apr 20, 2017 at 11:38

Observe that $$\left|\int_x^{x+1}\sin(t^2)dt\right|=\left|\int_x^{x+1}\frac{t}{t}\sin(t^2)dt\right|.$$ Using integration by parts, we have \begin{align*} u&=\frac{1}{t}&dv&=t\sin(t^2)dt\\ du=&-\frac{1}{t^2}&v&=-\frac{1}{2}\cos(t^2). \end{align*} Therefore, \begin{align*} \left|\int_x^{x+1}\frac{t}{t}\sin(t^2)dt\right| &=\left|\left.-\frac{1}{2t}\cos(t^2)\right|_x^{x+1}-\int_x^{x+1}\frac{1}{2t^2}\cos(t^2)dt\right|\\ &\leq\left|\left.\frac{1}{2t}\cos(t^2)\right|_x^{x+1}\right|+\left|\int_x^{x+1}\frac{1}{2t^2}\cos(t^2)dt\right|\\ &\leq\frac{1}{2(x+1)}+\frac{1}{2x}+\int_x^{x+1}\frac{1}{2t^2}dt \end{align*} since $|\cos(t^2)|$ is at most $1$. Then, we can just integrate to get $$=\frac{1}{2(x+1)}+\frac{1}{2x}+\left(\left.-\frac{1}{2t}\right|_{x}^{x+1}\right) =\frac{1}{2(x+1)}+\frac{1}{2x}-\frac{1}{2(x+1)}+\frac{1}{2x}=\frac{1}{x}$$
• Why did Rudin's hint say to put $t^2=u$? Commented Oct 30, 2023 at 5:35