# sinc is not $L^{1}$

The main results used in the proof will be that $f\in L^{1}(A)\iff|f|\in L^{1}(A)$ and the dominated convergence theorem.

Let $f(x)=|\operatorname{sinc}(x)|$ and suppose it’s Lebesgue integrable in $\mathbb{R}^{+}$.

Consider the intervals $I_{k}=[k\pi,(k+1)\pi]$ and $U_{k}=\bigcup_{i=0}^{k}I_{k}=[0,(k+1)\pi]$.

and the succession of functions $f_{n}(x)=f(x)\chi_{U_{n}}(x)$, where $\chi_{U_{n}}$ is the characteristic function of the set $U_{n}$.

Each $f_{n}$ is a continuous function of compact support and will thus be integrable in $\mathbb{R}^{+}$. Furthermore $f_{n}(x)\nearrow f(x)$ (pointwise)

in each $I_{k}$, $f(x)\geq\frac{|\sin(x)|}{(k+1)\pi}$.

So

$\displaystyle\int_{\mathbb{R}^{+}}f_{n}=\sum_{k=0}^{n}\int_{k\pi}^{(k+1)\pi}% \frac{|\sin(x)|}{x}dx\geq\sum_{k=0}^{n}\int_{k\pi}^{(k+1)\pi}\frac{|sin(x)|}{(% k+1)\pi}=\sum_{k=0}^{n}\frac{2}{(k+1)\pi}$.

Suppose $f$ is integrable in $\mathbb{R}^{+}$. Then by the dominated convergence theorem $\lim_{n\to\infty}\int_{\mathbb{R}^{+}}f_{n}=\int_{\mathbb{R}^{+}}f$.

But $\lim_{n\to\infty}\int_{\mathbb{R}^{+}}f_{n}\geq\lim_{n\to\infty}\sum_{k=0}^{n}% \frac{2}{(k+1)\pi}=+\infty$ and we get the contradiction $\int_{\mathbb{R}^{+}}f\geq+\infty$.

So $f$ cannot be integrable in $\mathbb{R}^{+}$. This implies that $f$ cannot be integrable in $\mathbb{R}$ and since a function is integrable in a set iff its absolute value is

$\operatorname{sinc}(x)\notin L^{1}(\mathbb{R})$

Title sinc is not $L^{1}$ SincIsNotL1 2013-03-22 15:44:32 2013-03-22 15:44:32 cvalente (11260) cvalente (11260) 14 cvalente (11260) Result msc 26A06