An Inequality involving integration

Question

Show that $$\int_{0}^{\pi} \left|\frac{\sin nx}{x}\right| dx \ge \frac{2}{\pi}\left( 1 + \frac{1}{2} + \cdots + \frac{1}{n} \right)$$

How do I go about proving this inequality ?

Answer 1

Hint. Make the substitution $u = nx$. Then bound the resulting integral separately on each interval $[(k-1)\pi,k\pi]$.

Answer 2

This is my solution :

Let us substitute $u=nx$.

Then we have , $$\int_{0}^{\pi} \left| \frac{\sin nx}{x} \right| dx = \int_{0}^{n\pi} \left| \frac{\sin u}{u} \right| du$$

Now, splitting up the limits, we get,

$$\int_{0}^{n\pi} \left| \frac{\sin u}{u} \right| du = \int_{0}^{\pi} \left| \frac{\sin u}{u} \right| du + \int_{\pi}^{2\pi} \left| \frac{\sin u}{u} \right| du + .... + \int_{(n-1)\pi}^{n\pi} \left| \frac{\sin u}{u} \right| du$$

Now, by the Weighted Mean-Value Theorem for Integrals, we have

$$\int_{(n-1)\pi}^{n\pi} \left| \frac{\sin u}{u} \right| du \ge \min \left|\frac{1}{u}\right| \int_{(n-1)\pi}^{n\pi} \left|\sin u \right| du = 2\ln n\pi \ge \frac{2}{n\pi}$$

Now, adding all inequalities of the following form, $$\int_{(k-1)\pi}^{k\pi} \left| \frac{\sin u}{u} \right| du \ge \frac{2}{k\pi}$$ such that $1 \le k \le n.$ We get,

$$\int_{0}^{\pi} \left|\frac{\sin nx}{x}\right| dx \ge \frac{2}{\pi}\left( 1 + \frac{1}{2} + \cdots + \frac{1}{n} \right)$$

Please point out any flaws in the above solution, if you can find one.