$x\rightarrow \int_{0}^{x} \frac{\operatorname{sin}(t)}{t}$ is a bounded function

Question

I've already proved that the improper integral $\int_{0}^{\infty}\frac{\operatorname{sin}(t)}{t}$ is convergent.

I don't know its limit though...

I'm asked to prove that $\begin{array}{ccccc} f & : & \mathbb R & \to & \mathbb R \\ & & x & \mapsto & \int_{0}^{x} \frac{\operatorname{sin}(t)}{t} \\ \end{array}$

is a bounded function.

I don't know how to proceed. We're not looking for what happens at $\infty$ so my reasoning for the convergence of the integral yields nothing.

What approach is suitable here?

Answer 1

I don't know why people voted for closing...

Anyway, the answer to this question is actually very simple: the function considered here is continuous, has limits at both $0$ and $\infty$ and is therefore bounded.