Theorem: N over Pi times Reciprocal of 1 Plus n Squared x Squared Delta Sequence

thumb600pxThe graph of the $\ds \frac n \pi \frac 1 {1 + n^2 x^2}$ delta sequence. As $n$ grows, the graph becomes thinner and taller. The area under each graph is equal to $1$.
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:$\map {\delta_n} x := \frac n \pi \frac 1 {1 + n^2 x^2}$
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
:$\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.

Proof:
{{begin-eqn}}
{{eqn | l = \int_0^\infty \frac n \pi \frac 1 {1 + n^2 x^2} \rd x
      | r = \int_0^\infty \frac 1 \pi \frac 1 {1 + n^2 x^2} \rd \paren {n x}
}}
{{eqn | r = \int_0^\infty \frac 1 \pi \frac 1 {1 + y^2} \rd y
      | c = $n x = y$, Integration by Substitution
}}
{{eqn | r = \frac 1 2
      | c = {{Corollary|Definite Integral to Infinity of Reciprocal of x Squared plus a Squared}}
}}
{{end-eqn}}
Furthermore:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty \frac n \pi \frac 1 {1 + n^2 x^2} \rd x
      | r = 1
}}
{{end-eqn}}
Let $a,b \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = \int_a^b \frac n \pi \frac 1 {1 + n^2 x^2} \rd x
      | r = \int_a^b \frac n \pi \frac 1 {1 + \paren {n x}^2} \rd \paren {n x}
}}
{{eqn | r = \int_{n a}^{n b} \frac 1 \pi \frac 1 {1 + y^2} \rd y
      | c = $n x = y$, Integration by Substitution
}}
{{end-eqn}}
Suppose $0 < a < b$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_{a n}^{b n} \frac 1 \pi \frac 1 {1 + y^2} \rd y
      | r = \lim_{n \mathop \to \infty} \int_0^{b n} \frac 1 \pi \frac 1 {1 + y^2} \rd y - \lim_{n \mathop \to \infty} \int_0^{a n} \frac 1 \pi \frac 1 {1 + y^2} \rd y
      | c = Sum of Integrals on Adjacent Intervals for Integrable Functions
}}
{{eqn | r = \int_0^\infty \frac 1 \pi \frac 1 {1 + y^2} \rd y - \int_0^\infty \frac 1 \pi \frac 1 {1 + y^2} \rd y
}}
{{eqn | r = \frac 1 2 - \frac 1 2
}}
{{eqn | r = 0
}}
{{end-eqn}}
Analogously, suppose $a < b < 0$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_{a n}^{b n} \frac 1 \pi \frac 1 {1 + y^2} \rd y 
      | r = \lim_{n \mathop \to \infty} \int_{a n}^0 \frac 1 \pi \frac 1 {1 + y^2} \rd y - \lim_{n \mathop \to \infty} \int_{b n}^0 \frac 1 \pi \frac 1 {1 + y^2} \rd y
      | c = Sum of Integrals on Adjacent Intervals for Integrable Functions
}}
{{eqn | r = \int_{-\infty}^0 \frac 1 \pi \frac 1 {1 + y^2} \rd y - \int_{-\infty}^0 \frac 1 \pi \frac 1 {1 + y^2} \rd y
}}
{{eqn | r = \frac 1 2 - \frac 1 2
}}
{{eqn | r = 0
}}
{{end-eqn}}
Let $\epsilon \in \R_{> 0}$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \frac n \pi \frac 1 {1 + n^2 x^2} \rd x
      | r = \lim_{n \mathop \to \infty} \int_{-\infty}^{-\epsilon} \map \phi x \frac n \pi \frac 1 {1 + n^2 x^2} \rd x + \lim_{n \mathop \to \infty} \int_{-\epsilon}^\epsilon \map \phi x \frac n \pi \frac 1 {1 + n^2 x^2} \rd x + \lim_{n \mathop \to \infty} \int_\epsilon^\infty \map \phi x \frac n \pi \frac 1 {1 + n^2 x^2} \rd x
}}
{{eqn | r = \map \phi {\xi_-} \lim_{n \mathop \to \infty} \int_{-\infty}^{-\epsilon} \frac n \pi \frac 1 {1 + n^2 x^2} \rd x + \map \phi {\xi_\epsilon} \lim_{n \mathop \to \infty} \int_{-\epsilon}^\epsilon \frac n \pi \frac 1 {1 + n^2 x^2} \rd x + \map \phi {\xi_+} \lim_{n \mathop \to \infty} \int_{\epsilon}^\infty \frac n \pi \frac 1 {1 + n^2 x^2} \rd x
      | c = Mean value theorem for integrals, $\xi_\epsilon \in \closedint {-\epsilon} \epsilon$, $\xi_- \in \hointl {-\infty} {-\epsilon}$, $\xi_+ \in \hointr \epsilon \infty$
}}
{{eqn | r = 0 + \map \phi {\xi_\epsilon} \lim_{n \mathop \to \infty} \int_{-n \epsilon}^{n \epsilon} \frac 1 \pi \frac 1 {1 + y^2} \rd y + 0
}}
{{eqn | r = \map \phi {\xi_\epsilon}
}}
{{end-eqn}}
We have {{hypothesis}} that $\epsilon$ 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 {{WRT}} $\epsilon'$ we have that:
:$\xi_\epsilon = \xi_{+'}$
or:
:$\xi_\epsilon = \xi_{-'}$
where:
:$\xi_{+'} \in \hointr {\epsilon'} \infty$
:$\xi_{-'} \in \hointl {-\infty} {-\epsilon'}$
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 $\xi_\epsilon = 0$.
{{Proofread|Check the rigour of the last few lines. Make the argument stricter.}}
{{qed}}