Prove that: $\left|\int\limits_a^b\frac{\sin x}{x}dx\right|\le \frac{2}{a}$

Question

Prove that: $$\left|\int\limits_a^b\frac{\sin x}{x}dx\right|\le \frac{2}{a}$$

$0<a<b$

I'm not asking for a complete solution to this, just a hint or a push in the right direction. I've been struggling with this and haven't really made any progress.

Answer 1

Integrating by parts,

$$\int_a^b\frac{\sin x}x\,dx = \frac{\cos a}a-\frac{\cos b}b -\int_a^b\frac{\cos x}{x^2}\,dx.$$ Then apply the triangle inequality (twice) to get $$\bigg|\int_a^b\frac{\sin x}{x}\,dx\bigg|\leqslant \frac1a+\frac1b+\int_a^b\frac{1}{x^2}\,dx = \frac{2}{a},$$ since $\int_a^b\frac 1{x^2}\,dx = \frac1a-\frac 1b$.