0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Related: math.stackexchange.com/questions/5248/… $\endgroup$ – user119228 Apr 2 '14 at 20:13
  • $\begingroup$ Tu as supprimé ta question ? $\endgroup$ – user119228 Apr 4 '14 at 17:02
  • $\begingroup$ @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. $\endgroup$ – Gabriel Romon Apr 4 '14 at 17:04
  • $\begingroup$ Ah, d'accord..$$ $$ $$ $\endgroup$ – user119228 Apr 4 '14 at 17:07
  • 1
    $\begingroup$ Oui tu as raison, cela dit $\left|1 - e^{inx}\right|=2\left|\sin\frac{nx}{2}\right|$, là tu devrais pouvoir conclure ;) $\endgroup$ – user119228 Apr 5 '14 at 17:13
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.