How do I show that exists a constant $M>0$ such that, for all $0\leq a \leq b < \infty$, $$\left|\int_a^b \frac{\sin(x)}{x}dx\right| \leq M.$$
I just read on Richard Bass book's that is enough to prove the uniformly boundeness of the integral $\int_0^b \frac{\sin(x)}{x}dx$, but I can't see why...
Let $G(x)=\int^x_0\frac{\sin t}{t}\,dt$. If One proves that $|G(x)|\leq M$ for some $M>0$ and all $x\geq0$, then
$$|G(b)-G(a)|\leq |G(b)|+|G(a)|\leq 2M$$
Hint: $\lim_{x\rightarrow+\infty}G(x)$ exists (it is $\frac{\pi}{2}$ but the value for this exercise is not that important)
If you know this, then the problem is almost done. If you don't, then you may try to write $G(x)$ as sum of decreasing alternating terms just to show that indeed $\lim_{x\rightarrow\infty}G(x)$ exists. For instance for $n\pi\leq x<(n+1)\pi$,
$$G(x)=\int^x_{n\pi}\frac{\sin t}{t}\,dt +\sum^n_{k=1}\int^{n\pi}_{(n-1)\pi}\frac{\sin t}{t}\,dt $$
The first term is bounded by $\pi$, the second is a sum of terms $\sum^n_{k=1}a_k$ which alternate in sign but that in absolute value $|a_n|\xrightarrow{n\rightarrow\infty}0$.
The way I will approach it is to prove that $\lim_{x \to \infty}G(x)\le G(\pi)$, where $G(x) is \int_{0}^{x}\frac{sin(y)}y dy $. The reason being if $f(y)=\frac{sin(y)}y$, then $\int_{r \pi}^{(r+2)\pi}f(y)dy\leq0$ for odd $r$
The rest is trivial
