Sine n x over Pi x Delta Sequence

Theorem
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:


 * $\ds \map {\delta_n} x := \frac {\map \sin {n x} }{\pi x}$

Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.

That is, in the distributional sense it holds that:


 * $\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$

or


 * $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$

where $\phi \in \map \DD \R$ is a test function, $\delta$ is the Dirac delta distribution, and $\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.

Proof
From Integral to Infinity of Sine p x over x for $n \in \N_{> 0}$ we have that:


 * $\ds \int_0^\infty \frac {\map \sin {n x} } x \rd x = \frac \pi 2$

Furthermore:

Hence:


 * $\ds \int_{-\infty}^0 \frac {\map \sin {n y} } y \rd y + \int_0^\infty \frac {\map \sin {n x} } x \rd x = \pi$

Let $a,b \in \R$.

Then:

Suppose $0 < a < b$.

Then:

Analogously, suppose $a < b < 0$.

Then:

Let $\epsilon \in \R_{> 0}$.

Then:

$\epsilon$ is an arbitrary positive real number.

Hence, for every $\epsilon \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_+}$ and $\map \phi {\xi_-}$ vanish.

Suppose $\xi_\epsilon \ne 0$.

By Real Numbers are Densely Ordered:


 * $\forall \epsilon \in \R_{> 0} : \exists \epsilon' \in \R_{> 0} : 0 < \epsilon' < \epsilon$

Then $\epsilon'$ we have that $\xi_\epsilon = \xi_{+'}$ or $\xi_\epsilon = \xi_{-'}$, where $\xi_{+'} \in \hointr {\epsilon'} \infty$ and $\xi_{-'} \in \hointl {-\infty} {-\epsilon'}$.

But from the result above, for every $\epsilon' \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_{+'}}$ and $\map \phi {\xi_{-'}}$ vanish.

Therefore, the only nonvanishing contribution can come from $\xi_\epsilon = 0$.