I am trying to proof that $\displaystyle\left|\int_a^b \frac{\sin(x)}{x}\mathrm{d}x\right|\leqslant C$ for some $C>0$ and arbitrary $a,b$.
I am unsure whether I can work with $\mathrm{Si}(x)$, as we never discussed it in class. So my idea is to define my function $f(x)$ as the integral above and calculate the maximum values of $f(x)$ in $(0, \infty)$. Now
$f^\prime(x)=\dfrac{\sin(x)}{x}$. As we know, we first have to show that there exists an $x_0$ for which $f'(x)=0$. Because $\sin(x)=0\quad\forall x\in\pi\Bbb{Z}$, those are our values.
With $f''(x)$ I can show that for $x_0=\pm\pi$ we indeed have a maximal/minimal value (also the global extrema). Now to my question:
I am pulling the absolute value into the integral and then using the sum formula for $\sin(x)$ and integrating the summand which leaves me with:
$$\sum_{n=0}^\infty(-1)^n \frac{x^{2n+1}}{(2n+1)! \cdot (2n+1)} \leq \sum_{n=0}^\infty(-1)^n \frac{x^{2n+1}}{(2n+1)!}=\sin(x)$$
Is this upper bound too optimistic? I would then state, that $\sin(x)\in[-1,1] \implies\max\big\{|\sin(a)-\sin(b)|\big\}=2$, and therefore my integral is less or equal to $4$.
Please do not hesitate to correct my $\mathbf{laTeX}$, as I am new. Thanks in advance.
First question:
This follows immediately from the fact that $\left|\int_{-\infty}^\infty \operatorname{sinc}(x)\mathrm dx\right|<\infty$. Now finding that constant $C$ is a bit more difficult. But it is possible, with the following method.
Let us first focus on the case when $0\leq a<b<\infty$. We will prove that, under these conditions, the maximum possible value of $\left|\int_a^b \operatorname{sinc}(x)\mathrm dx\right|$ is achieved when $a=0$ and $b=\pi$. Let us begin. First, since $\left|\int_0^\infty \operatorname{sinc}(x)\mathrm dx\right|<\infty$, we can already observe that
$$\left|\int_{a'}^{b'}\operatorname{sinc}(x)\mathrm dx\right|<\left|\int_{a}^{b}\operatorname{sinc}(x)\mathrm dx\right| \\ \text{if}~a'>a~\text{and}~b'>b\tag 1$$
This is pretty easy to justify if you write $\left|\int_{a}^{b}\operatorname{sinc}(x)\mathrm dx\right|=|\operatorname{Si}(b)-\operatorname{Si}(a)|$, and recall that $\lim_{x\to\infty}\operatorname{Si}(x)$ exists and is finite.
However, another key point is that due to the behavior of $\operatorname{sinc}$, we know that
$$\left|\int_{n\pi}^{(n+1)\pi+s}\operatorname{sinc}(x)\mathrm dx\right|<\left|\int_{n\pi}^{(n+1)\pi}\operatorname{sinc}(x)\mathrm dx\right|\tag{2A}$$ For all $n\in\mathbb N\cup\{0\}$, and all $s>0$. Similarly, $$\left|\int_{n\pi-s}^{(n+1)\pi}\operatorname{sinc}(x)\mathrm dx\right|<\left|\int_{n\pi-s}^{(n+1)\pi}\operatorname{sinc}(x)\mathrm dx\right|\tag{2B}$$ For all $n\in\mathbb N$, and all $s>0$. $(2A),(2B)$ allow us to conclude that the optimal choice for $a,b$ must be of the form $n\pi,(n+1)\pi$. And, $(1)$ allows us to conclude that this optimal choice must be $a=0,b=\pi$.
By symmetry, this shows us that the optimal choice for the entire real line is $a=-\pi,b=\pi$, and that the optimal constant $C$ is $$C=\int_{-\pi}^\pi \operatorname{sinc}(x)\mathrm dx=\operatorname{Si}(\pi)-\operatorname{Si}(-\pi)=2\operatorname{Si}(\pi)\approx 3.704 $$ Which is indeed less than $4$, as claimed by Grafakos.