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

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## 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

 $\ds \int_0^\infty \frac n \pi \frac 1 {1 + n^2 x^2} \rd x$ $=$ $\ds \int_0^\infty \frac 1 \pi \frac 1 {1 + n^2 x^2} \rd \paren {n x}$ $\ds$ $=$ $\ds \int_0^\infty \frac 1 \pi \frac 1 {1 + y^2} \rd y$ $n x = y$, 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 $\xi_\epsilon = 0$.

$\blacksquare$

## Sources

2013: George ArfkenHans J. Weber and Frank E. Harris: Mathematical Methods for Physicists (7th ed.): Chapter $1$ Mathematical Preliminaries $1.11$ Dirac Delta Function