# 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.

• Can you use the Taylor series? – Michael McGovern May 6 '18 at 17:32
• @MichaelMcGovern Haven't learned it yet – 0rka May 6 '18 at 17:33
• The upper limit is different: $b \neq n$ . I assume that $b > n$, so $\int_0^b = \int_0^n + \int_n^b > \int_0^n$ . – mr_e_man May 6 '18 at 17:33

## 2 Answers

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.

\begin{align*} &\int_{0}^{\infty}\bigg|\frac{sin(\pi x)}{x}\bigg|dx\\ &=\sum_{n=0}^\infty\int_n^{n+1}\bigg|\frac{sin(\pi x)}{x}\bigg|dx\\ &\geq\sum_{n=0}^\infty\int_{n+1/4}^{n+1-1/4}\bigg|\frac{sin(\pi x)}{x}\bigg|dx\\ &\geq\frac{\sqrt2}{2}{}\sum_{n=0}^\infty\int_{n+1/4}^{n+1-1/4}\frac1xdx\\ &=\frac{\sqrt2}{2}{}\sum_{n=0}^\infty\ln\left(\frac{n+3/4}{n+1/4}\right)\\ &=\frac{\sqrt2}{2}{}\sum_{n=0}^\infty\ln\left(1+\frac{1/2}{n+1/4}\right)\to\infty \end{align*}