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

## Theorem

The 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:

$\ds \map {\delta_n} x := \frac n \pi \frac 1 {1 + n^2 x^2}$

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, 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 1

 $\ds \int_0^\infty \frac n \pi \frac 1 {1 + n^2 x^2} \rd x$ $=$ $\ds \int_0^\infty \frac 1 \pi \frac 1 {1 + n^2 x^2} \rd \paren {n x}$ $\ds$ $=$ $\ds \int_0^\infty \frac 1 \pi \frac 1 {1 + y^2} \rd y$ $n x = y$, Integration by Substitution $\ds$ $=$ $\ds \frac 1 2$ Definite Integral to Infinity of Reciprocal of x Squared plus a Squared/Corollary

Furthermore:

 $\ds \int_{-\infty}^\infty \frac n \pi \frac 1 {1 + n^2 x^2} \rd x$ $=$ $\ds 1$

Let $a,b \in \R$.

Then:

 $\ds \int_a^b \frac n \pi \frac 1 {1 + n^2 x^2} \rd x$ $=$ $\ds \int_a^b \frac n \pi \frac 1 {1 + \paren {n x}^2} \rd \paren {n x}$ $\ds$ $=$ $\ds \int_{n a}^{n b} \frac 1 \pi \frac 1 {1 + y^2} \rd y$ $n x = y$, Integration by Substitution

Suppose $0 < a < b$.

Then:

 $\ds \lim_{n \mathop \to \infty} \int_{a n}^{b n} \frac 1 \pi \frac 1 {1 + y^2} \rd y$ $=$ $\ds \lim_{n \mathop \to \infty} \int_0^{b n} \frac 1 \pi \frac 1 {1 + y^2} \rd y - \lim_{n \mathop \to \infty} \int_0^{a n} \frac 1 \pi \frac 1 {1 + y^2} \rd y$ Sum of Integrals on Adjacent Intervals for Integrable Functions $\ds$ $=$ $\ds \int_0^\infty \frac 1 \pi \frac 1 {1 + y^2} \rd y - \int_0^\infty \frac 1 \pi \frac 1 {1 + y^2} \rd y$ $\ds$ $=$ $\ds \frac 1 2 - \frac 1 2$ $\ds$ $=$ $\ds 0$

Analogously, suppose $a < b < 0$.

Then:

 $\ds \lim_{n \mathop \to \infty} \int_{a n}^{b n} \frac 1 \pi \frac 1 {1 + y^2} \rd y$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{a n}^0 \frac 1 \pi \frac 1 {1 + y^2} \rd y - \lim_{n \mathop \to \infty} \int_{b n}^0 \frac 1 \pi \frac 1 {1 + y^2} \rd y$ Sum of Integrals on Adjacent Intervals for Integrable Functions $\ds$ $=$ $\ds \int_{-\infty}^0 \frac 1 \pi \frac 1 {1 + y^2} \rd y - \int_{-\infty}^0 \frac 1 \pi \frac 1 {1 + y^2} \rd y$ $\ds$ $=$ $\ds \frac 1 2 - \frac 1 2$ $\ds$ $=$ $\ds 0$

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

Then:

 $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \frac n \pi \frac 1 {1 + n^2 x^2} \rd x$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^{-\epsilon} \map \phi x \frac n \pi \frac 1 {1 + n^2 x^2} \rd x + \lim_{n \mathop \to \infty} \int_{-\epsilon}^\epsilon \map \phi x \frac n \pi \frac 1 {1 + n^2 x^2} \rd x + \lim_{n \mathop \to \infty} \int_\epsilon^\infty \map \phi x \frac n \pi \frac 1 {1 + n^2 x^2} \rd x$ $\ds$ $=$ $\ds \map \phi {\xi_-} \lim_{n \mathop \to \infty} \int_{-\infty}^{-\epsilon} \frac n \pi \frac 1 {1 + n^2 x^2} \rd x + \map \phi {\xi_\epsilon} \lim_{n \mathop \to \infty} \int_{-\epsilon}^\epsilon \frac n \pi \frac 1 {1 + n^2 x^2} \rd x + \map \phi {\xi_+} \lim_{n \mathop \to \infty} \int_{\epsilon}^\infty \frac n \pi \frac 1 {1 + n^2 x^2} \rd x$ Mean value theorem for integrals, $\xi_\epsilon \in \closedint {-\epsilon} \epsilon$, $\xi_- \in \hointl {-\infty} {-\epsilon}$, $\xi_+ \in \hointr \epsilon \infty$ $\ds$ $=$ $\ds 0 + \map \phi {\xi_\epsilon} \lim_{n \mathop \to \infty} \int_{-n \epsilon}^{n \epsilon} \frac 1 \pi \frac 1 {1 + y^2} \rd y + 0$ $\ds$ $=$ $\ds \map \phi {\xi_\epsilon}$

$\epsilon$ 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 $\xi_\epsilon \ne 0$.

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

Then with respect to $\epsilon'$ 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 $\xi_\epsilon = 0$.

$\blacksquare$

## Proof 2

Let $\map g x = \map \phi x - \map \phi 0$.

Then:

$\ds \int_{- \infty}^\infty \map \phi x \map {\delta_n} x \rd x = \map \phi 0 + \int_{- \infty}^\infty \map g x \map {\delta_n} x \rd x$

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

Then:

 $\ds \int_{- \infty}^\infty \map g x \map {\delta_n} x \rd x$ $=$ $\ds \int_{- \infty}^{- A} \map g x \map {\delta_n} x \rd x + \int_A^\infty \map g x \map {\delta_n} x \rd x + \int_{- A}^A \map g x \map {\delta_n} x \rd x$ Sum of Integrals on Adjacent Intervals for Integrable Functions $\ds$ $=:$ $\ds I_1 + I_2 + I_3$

Let:

$\ds \max_{x \mathop \in \closedint {-A} A} \size {\map g x} := \map M A$

Then:

 $\ds I_3$ $\le$ $\ds \int_{-A}^A \size {\map g x} \map {\delta_n} x \rd x$ $\ds$ $\le$ $\ds \map M A \int_{-A}^A \frac n {\pi \paren {1 + n^2 x^2} } \rd x$ $\ds$ $=$ $\ds \map M A \frac 2 \pi \map \arctan {n A}$ Primitive of Reciprocal of x squared plus a squared $\ds$ $\le$ $\ds \map M A$ $\ds \forall x \in \R_{\ge 0} : 0 \le \map \arctan x \le \frac \pi 2$

We have that $\map g 0 = 0$.

By definition, $\phi$ is a smooth real function on $\R$.

By Differentiable Function is Continuous, $\map g x$ is continuous at $x = 0$.

Furthermore:

 $\ds \lim_{A \mathop \to 0} \map M A$ $=$ $\ds \lim_{A \mathop \to 0} \max_{x \mathop \in \closedint {-A} A} \size {\map g x}$ $\ds$ $=$ $\ds \size {\map g 0}$ $\ds$ $=$ $\ds 0$

By definition of the limit of a real function:

$\forall \epsilon' \in \R_{>0} : \exists \delta \in \R_{>0}: \forall A \in \R_{> 0}: 0 < A < \delta \implies \map M A < \epsilon'$

Let $\ds \epsilon' = \frac \epsilon 2$.

It follows that:

$\ds \forall \epsilon \in \R_{> 0} : \exists A \in \R_{> 0} : I_3 \le \map M A < \frac \epsilon 2$

Suppose $A$ is such that the above inequality holds.

By definition, $\map \phi x$ is bounded.

Then:

 $\ds \size {\map g x}$ $=$ $\ds \size {\map \phi x - \map \phi 0}$ $\ds$ $<$ $\ds \size {\map \phi x} + \size {\map \phi 0}$ Triangle Inequality for Real Numbers

It follows that:

$\exists b \in \R_{> 0} : \forall x \in \R : \size {\map g x} < b$

Then:

 $\ds \size {I_1 + I_2}$ $=$ $\ds \size { \int_{- \infty}^{-A} \map g x \map {\delta_n} x \rd x + \int_A^\infty \map g x \map {\delta_n} x \rd x}$ $\ds$ $\le$ $\ds \size {\int_{- \infty}^{-A} \size {\map g x} \map {\delta_n} x \rd x + \int_A^\infty \size {\map g x} \map {\delta_n} x \rd x}$ $\ds$ $\le$ $\ds \size { b \int_{- \infty}^{-A} \map {\delta_n} x \rd x + b \int_A^\infty \map {\delta_n} x \rd x}$ $\ds$ $\le$ $\ds b \size {\int_{- \infty}^{-A} \map {\delta_n} x \rd x + \int_A^\infty \map {\delta_n} x \rd x}$ $\ds$ $=$ $\ds \frac b \pi \size {\bigintlimits {\map \arctan {n x} } {- \infty} {-A} + \bigintlimits {\map \arctan {n x} } A \infty }$ Primitive of Reciprocal of x squared plus a squared $\ds$ $=$ $\ds \frac b \pi \size {- \map \arctan {n A} + \frac \pi 2 + \frac \pi 2 - \map \arctan {n A} }$ $\ds$ $=$ $\ds b \size {1 - \frac 2 \pi \map \arctan {n A} }$

With the number $A$ fixed:

$\ds \lim_{n \mathop \to \infty} \frac 2 \pi \map \arctan {n A} = 1$.

By Squeeze Theorem and for a fixed $A$ we have:

$\ds \lim_{n \mathop \to \infty} \size {I_1 + I_2} = 0$

By definition of the limit of a real sequence:

$\ds \forall \overline \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \size {I_1 + I_2} \le b \size {1 - \frac 2 \pi \map \arctan {n A} } < \overline \epsilon$.

Let $\ds \overline \epsilon = \frac \epsilon 2$.

Then:

$\ds \forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \size {I_1 + I_2} < \frac \epsilon 2$.

Let $A$ and $N$ be such that the above inequalities for $I_3$ and $I_1 + I_2$ hold.

Then:

 $\ds \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x}$ $\le$ $\ds \size {I_1 + I_2 + I_3}$ $\ds$ $\le$ $\ds \size {I_1 + I_2} + \size {I_3}$ Triangle Inequality for Real Numbers $\ds$ $<$ $\ds \epsilon$

To sum up:

$\ds \forall \epsilon \in \R_{>0} : \exists N \in \R_{>0} : \forall n \in \N_{>0} : \forall n > N \implies \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x} < \epsilon$

By definition of the limit of a real sequence:

$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x = 0$

However:

 $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x - \map \phi 0 \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \rd x$ Sum Rule for Limits of Real Functions $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x - \map \phi 0$ $\ds$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x$ $=$ $\ds \map \phi 0$

$\blacksquare$