How to show that $\left|\int_0^x \frac{\sin u} u \, du\right|$ is bounded? How to show that $\left|\int_0^x \frac{\sin u} u \, du\right|$ is bounded?


Is it because $\left|\int_0^x \frac{\sin u} u \, du\right|\leq \left|\cos1-\int_0^x \frac{\cos u}{u^2} \, du\right|$, which is finite?
As is well known, $\lim_{x \to \infty}\int_0^x \frac{\sin u }{u } \, du  = \frac{\pi}{2}$  which proves boundedness, but without using this fact we can argue as follows.

Fixing $a &gt; 0$, by Bonnet's mean value theorem for integrals
 there exists $\xi_x \in (a,x)$ such that for all $x &gt; a$,

$$\left|\int_a^x \frac{\sin u}{u}\, du\right| = \left|\frac{1}{a}\int_a^{\xi_x} \sin u\, du\right| = \frac{|\cos \xi_x - \cos  a|}{a} \leqslant \frac{2}{a},$$

and $\int_0^a \frac{\sin u }{u} \, du$ is finite.
Prove that $\sin(x) \leq x$ for $x \geq 0$
For $x &lt; 0$, note that the integrand is even.