Lebesgue Integral of $\sin(x)/x$ using the Monotone convergence theorem [closed]

Recall that $f$ is (Lebesgue) integrable if and only if $|f|$ is Lebesgue integrable. Now, we may write $$f_n(x) = \sum_{k=0}^n\frac{|\sin x |}{x}\chi_{[k \pi , (k+1)\pi]}$$ so that $f_{n} \leq f_{n+1}$ and $f_n \to f$ pointwise, where $f(x) = \frac{|\sin x |}{x}$. We may then estimate the integral of each term in the sum via: $$\frac{2}{(k+1)\pi} = \int_{k \pi}^{(k+1)\pi} \frac{|\sin x|}{(k+1)\pi}dx \leq \int_{k \pi}^{(k+1)\pi} \frac{|\sin x|}{x}dx$$

So that by the monotone convergence theorem, $$\begin{align*} \int_0^\infty \frac{|\sin x|}{x}dx &= \lim_{n\to \infty} \int_0^\infty f_n dx \\ &\geq \sum_{k=0}^\infty \frac{2}{(k+1) \pi} \\ &= \infty \end{align*}$$ Which implies that the Lebesgue integral of $\frac{\sin x}{x}$ does not exist.