calculus - Prove that the integral $int_{0}^{infty}bigg|frac{sin(pi x)}{x}bigg|dx$ diverges

I've been asked to prove that the following integral diverges $$\int_{0}^{\infty}\bigg|\frac{\sin(\pi x)}{x}\bigg|dx$$
The thing is, that the textbook explicitly suggests a way to prove that. It states the following:

Start by showing that the integral $\int_{0}^{b}\bigg|\frac{\sin(\pi x)}{x}\bigg|dx$ is bounded below by $\int_{0}^{n}\bigg|\frac{\sin(\pi x)}{x}\bigg|dx$ with a respective integer $n \in \mathbb Z$

I can't figure out what they mean by that - how can a function be bounded below by itself? It would be great if someone could clear that up for me.

Answer

Hint: For $n\leq x< n+1$, we have $$\int_{n}^{n+1}\left|\frac{\sin \pi x}{x}\right|\,dx>\frac{1}{n+1}\int_{n}^{n+1}|\sin \pi x|\,dx=\frac{1}{n+1}\left|\int_{n}^{n+1}\sin \pi x\,dx\right|$$ This follows just from properties of fractions; the integral that remains is easy to calculate. Now use the fact that $\sum \frac1n$ diverges.