Definite integral of $\sin(x)/x$

Question

I'm trying to evaluate absolute value of:

$$\left | \int_{a}^{b} \frac{\sin(x)}{x} dx \right | \leq\frac{2}{a}, \ \text{where}\ 0<a<b$$

I tried taylor series, mean value theorems, and tried to divide interval on some like $\left[\pi k, \pi(k+1) \right]$ or $\left[2\pi k, 2\pi(k+1) \right]$, but it didn't help

Answer 1

Use $$ \int\frac{\sin x}xdx=-\int\frac1xd\cos x=-\frac{\cos x}{x}-\int\frac{\cos x}{x^2}dx.$$ one has \begin{eqnarray} &&\bigg|\int_a^b\frac{\sin x}xdx\bigg|\\ &=&\bigg|-\frac{\cos x}{x}\bigg|_a^b-\int_a^b\frac{\cos x}{x^2}dx\bigg|\\ &\le&\bigg|\frac{\cos a}{a}-\frac{\cos b}{b}\bigg|+\int_a^b\frac1{x^2}dx\\ &\le&\frac{1}{a}+\frac1b+\frac1a-\frac1b\\ &=&\frac{2}{a}. \end{eqnarray}

Answer 2

Integrate by parts and let $u=\frac{1}{x}$ and $dv=\sin(x)\,dx$ to get

$$\int_{a}^{b} \frac{\sin(x)}{x} \,dx=\frac{\cos(a)}{a}-\frac{\cos(b)}{b}-\int_a^b \frac{\cos(x)}{x^2}\,dx.$$

Then, by the triangle inequality,

\begin{align}\left|\int_{a}^{b} \frac{\sin(x)}{x} \,dx\right|&= \left|\frac{\cos(a)}{a}-\frac{\cos(b)}{b}-\int_a^b \frac{\cos(x)}{x^2}\,dx\right|\\&\le \frac{1}{a}+\frac{1}{b}+\int_a^b \left|\frac{\cos(x)}{x^2}\right|\,dx\\&\le \frac{1}{a}+\frac{1}{b}+\int_a^b \frac{1}{x^2}\,dx\\&= \frac{2}{a}. \end{align}