# The Improper Function and the Poetry of Proofs

Later the Delta Function was named after their founder. Dirac himself called it an improper function.

This time, the poem is not from repurposed snippets of his prose. These are just my own words to describe a proof:

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In the limit
the Lorentzian becomes
the improper function.

In the limit of tiny epsilons
it becomes infinitely thin.
Thus under the integral
it singles out
another function’s value at zero.

Yet the area covered
by this delicate needle
is finite and unity.
This is the property
we need to assert.

To integrate over the Lorentzian bell,
we consider the argument
a complex number.
Note the poles on the imaginary axis!

The spike peters out
at infinite distance,
regardless of angle.
We can safely add a tour
through the upper imaginary half plane.
to the integral.

The path has been closed
outlining a half moon of sorts.
It lassos the upper pole!

Shrink the infinite half moon
into a small loop around the pole.
Trace out the circular path.
Parameterize: Crank the phase angle!

Exponential functions cancel,
and the pole is defused.

A tiny radius and a tiny epsilon:
Both will become zero in the limit.

Then epsilons cancel out,
as well as the imaginary units.
We are left with a number of one half.

Integrating over the circle just gives
two times the Ludolphian number.
Numbers cancel.
We obtain unity.

~ $\displaystyle \delta(x) = \lim_{\varepsilon \to 0} \int_{-\infty}^{\infty} \frac{1}{\pi} \frac{\varepsilon}{x^2 + \varepsilon^2}$

~ $\displaystyle \int_{-\infty}^{\infty} \delta(x) = \oint_{\text{around }i\varepsilon} dz \frac{1}{\pi} \frac{\varepsilon}{(z + i\varepsilon)(z - i\varepsilon) } = \lim_{r \to 0} \int_0^{2\pi}d\varphi ire^{i\varphi} \frac{1}{\pi} \frac {\varepsilon}{( i\varepsilon + re^{i\varphi} + i\varepsilon)re^{i\varphi}} = \lim_{r \to 0}\int_0^{2\pi} d\varphi \frac{i}{\pi} \frac{\varepsilon}{re^{i\varphi} +2i\varepsilon} = \int_0^{2\pi} d\varphi \frac{i}{\pi} \frac{\varepsilon}{2i\varepsilon} = \frac{2\pi}{2\pi} = 1$

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