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

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?

• – user119228 Apr 2 '14 at 20:13
• Tu as supprimé ta question ? – user119228 Apr 4 '14 at 17:02
• @Julien elle n'avait aucun sens en fait (formulée comme telle). En fait je cherchais le min de la fonction $x \rightarrow \frac{1}{ |1-e^{inx}|}$ à $n$ fixé. – Gabriel Romon Apr 4 '14 at 17:04
• Ah, d'accord..  – user119228 Apr 4 '14 at 17:07
• Oui tu as raison, cela dit $\left|1 - e^{inx}\right|=2\left|\sin\frac{nx}{2}\right|$, là tu devrais pouvoir conclure ;) – user119228 Apr 5 '14 at 17:13

$$\int_{0}^{x}\frac{\sin t}{t}\,dt \stackrel{\text{IBP}}{=} \left[\frac{1-\cos t}{t}\right]_{0}^{x}+\int_{0}^{x}\frac{1-\cos t}{t^2}\,dt$$ 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 $$\mathbb{R}$$.
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.