$x\rightarrow \int_{0}^{x} \frac{\operatorname{sin}(t)}{t}$ 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

$$\int_{0}^{x}\frac{\sin t}{t}\,dt \stackrel{\text{IBP}}{=} \left[\frac{1-\cos t}{t}\right]_{0}^{x}+\int_{0}^{x}\frac{1-\cos t}{t^2}\,dt $$ and since $0\leq 1-\cos t\leq \min(1,t^2)$ we have $$ 0 \leq \int_{0}^{x}\frac{\sin t}{t}\,dt \leq \min\left(x,\frac{1}{x}\right)+2\leq 3 $$ for any $x>0$. The stationary points of $f(x)=\int_{0}^{x}\frac{\sin t}{t}\,dt$ occur at $\pi,2\pi,3\pi,\ldots$, so we may easily improve such bound: $$ \forall x>0,\qquad 0\leq \int_{0}^{x}\frac{\sin t}{t}\,dt \leq 1.851937\ldots = \int_{0}^{\pi}\frac{\sin t}{t}\,dt. $$ $f(x)$ is an odd function, so $|f(x)|<\frac{13}{7}$ holds over $\mathbb{R}$.

Answer 2

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.