Lebesgue Integral of $\frac{\sin x}{x}$ ?

Note that $\left|\frac{\sin x}{x}\right| \; \ge \; \frac{1}{2(n+\frac56)\pi} \; > \; \frac{1}{2\pi(n+1)} \hspace{2cm} (n+\tfrac16)\pi \le x \le (n + \tfrac56)\pi$ for any integer $n \ge 0$ . Thus we deduce that $\int_{n\pi}^{(n+1)\pi}\left|\frac{\sin x}{x}\right|\,dx \; \ge \; \frac{1}{2(n+1)\pi} \times \tfrac23\pi \; =\; \frac{1}{3(n+1)}$ for any integer $n \ge 0$ .

If $\frac{\sin x}{x}$ is Lebesgue integrable over $\mathbb{R}$ , then so is $\left|\frac{\sin x}{x}\right|$ , and hence $\sum_{n=0}^{N-1} \frac{1}{3(n+1)} \; \le \; \sum_{n=0}^{N-1} \int_{n\pi}^{(n+1)\pi} \left|\frac{\sin x}{x} \right|\,dx \; = \; \int_0^{N\pi} \left|\frac{\sin x}{x}\right|\,dx \; \le \; \int_0^\infty \left|\frac{\sin x}{x}\right|\,dx$ for all integers $N \ge 1$ . But this means that $\sum_{n=1}^N n^{-1} \; \le \; 3\int_0^\infty \left|\frac{\sin x}{x}\right|\,dx$ for all positive integers $N$ , which would imply that the infinite series $\sum_{n=1}^\infty n^{-1}$ converged, which is does not. Thus the function $\frac{\sin x}{x}$ is not Lebesgue integrable on $\mathbb{R}$ .