Show that there exists a constant $c$ such that $\left|\int_0^b\frac{\sin ax}{x}dx\right|\le c$ Show that there exists a constant $c$ such that $$\left|\int_0^b\frac{\sin ax}{x}dx\right|\le c$$ In fact, show that the smallest such number is $c=\int_0^\pi\frac{\sin x}xdx$.

Well, I'm thinking of a change of variable $y=ax$, and we want to show $\left|\int_0^B\frac{\sin x}xdx\right|$ is bounded. Then consider the intervals $[n\pi,(n+1)\pi)$ for natural number n.
Thank you.
 A: You have the right idea with the change of variable. Now show that $x\mapsto \int_0^x\frac{sin y}{y}dy$ is bounded, by using the FTC to differentiate it.
A: Start with a special case: a = 1, b > $\pi$.  Will generalize.
$\int_0^b \frac{sinx}{x}dx = \int_0^{\pi} \frac{sinx}{x}dx  + \int_{\pi}^b \frac{sinx}{x}dx$.  
For the second integral do integration by parts with u(x) = sinx, u'(x) = cosx, v'(x) = 1/x, v(x) = logx, giving
sin(b)log(b) - $\int_{\pi}^b cos(x)log(x)dx$ which gives us
$\int_0^b \frac{sinx}{x}dx = \int_0^{\pi} \frac{sinx}{x}dx + sin(b)log(b) - \int_{\pi}^b cos(x)log(x)dx$
|$\int_{\pi}^b cos(x)log(x)dx| \ge log(\pi) | \int_{\pi}^b cos(x)dx | =log(\pi) sin(b)$
$\int_0^b \frac{sinx}{x}dx \le \int_0^{\pi} \frac{sinx}{x}dx + |sin(b)[log(b) -  log(\pi)]|$ 
Since for some b sinb = 0, the smallest constant would be $\int_0^{\pi} \frac{sinx}{x}dx$.
If b = $\pi$ there is nothing to do. In the case that b < $\pi$ write 
$\int_0^b \frac{sinx}{x}dx = \int_0^{\pi} \frac{sinx}{x}dx  - \int_b^{\pi}\frac{sinx}{x}dx$  and proceed similarly.
For the sin(ax) case make the substitution y = ax and everything should fall out from there.
