Integrating the Delta Function (Again) – Dirac Version

The Delta Function is, roughly speaking, shaped like an infinitely tall and infinitely thin needle. It’s discovery – or invention – is commonly attributed to Paul Dirac[*].

Dirac needed a function like this to work with integrals that are common on quantum mechanics, a generalization of a matrix that has 1’s in the diagonal and 0’s elsewhere. The argument of the Delta Function is the difference of two values, positions of a particle in space for example. So it has a peak when the two positions or momenta are equal. It is a bit like an infinitely large and ‘infinitely fine’ matrix whose rows and columns are tagged with the continuous values of these ‘positions’.

The Dirac Delta Function is not a proper function but rather a distribution as I learned from skimming mathematically rigorous articles. There will not be more rigor in this article though, rather a physicist’s sloppy ‘intuitive’ arguments, relying on functions that ‘do not fluctuate wildly’ (Dirac used these words!).

This function is defined by is its ability to ‘single out’ another function’s value at zero under an integral. As it is zero everywhere except where the Delta Function has its peak, another function multiplied with it can be replaced by this function’s value at the peak. Then the constant value can be pulled out of the integral:

$\displaystyle f(0) = \int_{-\infty}^{+\infty}dx f(x) \delta(x)$

If the function f is a constant of 1, this equality turns into the condition that the area under the Delta Function has to be 1. Despite being an infinitely thin and infinitely tall needle it has a well-defined area (or volume) nonetheless.

There are many ways to represent the Delta function: Needle-like functions, such as a rectangle whose height is the reciprocal of its width or the Lorentzian bell-shaped curve I featured in my first Delta Function poem and proof. Any needle-shaped function singles out another function’s value at zero. What remains to be proved for a Delta Function candidate is that value of the integral is equal to one – and stays so in the limit. Then some width typically called epsilon becomes zero and the candidate really becomes the Delta Function.

In my poem-proof I took the usual (?) route, by turning the integral over the Lorentzian into a contour integral in the plane of complex numbers – and also providing part of the proof of method of residues along the way. Fortunately, the tiny epsilons cancel – the integral is independent of the thickness of the peak. In the second poem and proof, the Delta Function Haiku, I presented the spectral representation of the Delta Function and showed why it is equivalent to the infinitely thin Lorentzian.

Paul Dirac proves it directly – without reference to a specific needle-like function. The spectral representation has the defining property to single out another function’s value at zero. Here is his proof, from the section The momentum representation in his textbook The Principles of Quantum Mechanics:

Representation of the Delta Function:

$\displaystyle \int_{-\infty}^{\infty}e^{iax}da = 2\pi\delta(x)$

Proof of the characteristic property, introducing a number g that tends to infinity in the limit:

$\displaystyle \int_{-\infty}^{\infty} f(a) da \int_{-g}^{g}e^{iax}dx = \int_{-\infty}^{\infty} 2f(a) da \frac{\sin ag}{a} = 2\pi f(0)$

First step, integrate over x: The exponential function is equivalent to a complex function whose real part is the cosine function and its imaginary part is the sine function. As the sine is asymmetric its integral is zero. What remains is the real-valued cosine function whose integral is the sine function.

Second step, integrate over a: The outer integral is tougher as the argument a shows up both in the trigonometric function and in the denominator. I do not know Dirac’s preferred method to solve this integral. The shortest way would be to use complex numbers and contours again, but perhaps he / scientists in his day considered it an integral you knew or looked up? Similar functions appear, for example, in the general theory of the diffraction of (sound, light…) waves.

~

I want to use rather elementary methods,
neither rely on too much theorems
nor summon complex numbers.
This was more involved than I thought.

$\sin a g / a$ looks like a needle,
the larger g, the sharper.
So we salvage $f(0)$,
and pull it out.

Substituting …

$x = ag$ , thus $a = x/g$ and $da = dx/g$.
so the factor g cancels
(even though this tends to infinity –
I hope you rigorous folk agree),
and we are left with:

$\displaystyle \int_{-\infty}^{+\infty} dx \frac {\sin x}{x}$

To prove: Is it really π?

As the ratio of two asymmetric functions
it is symmetric,
The integral from minus to plus infinity
is twice the one from zero to infinity.

Get rid of the x in the denominator:
Look at the derivative, when the integrand
is multiplied with an exponential function
(a trick you learn as a budding theoretical physicist):

$\displaystyle - \frac{d}{d\beta} \int_{0}^{\infty} dx e^{-\beta x} \frac{\sin x}{x} = \int_{0}^{\infty} dx e^{-\beta x} \sin x$

The exponential function becomes 1
when β tends to zero.

This is the idea:
/1/ Integrate this over β
/2/ Set β = 0 at the end.

It does not look simpler at first glance,
but it actually is:
Integrate by parts …

$\displaystyle \int_{0}^{\infty} dx e^{-\beta x} \sin x = \left . e^{-\beta x} (-\cos x) \right |_0^{\infty} - \int_{0}^{\infty} dx \frac{e^{-\beta x} }{-\beta} (- \cos x) = ...$

… twice …

$\displaystyle ... = 0 + 1 +\left . \frac{e^{-\beta x} }{\beta} ( - \sin x) \right |_0^{\infty} - \frac{1}{\beta^2} \int_{0}^{\infty} dx e^{-\beta x} \sin x$

… so that the integral to be solved
shows up again,
and an algebraic equation emerges:

$\displaystyle \int_{0}^{\infty} dx e^{-\beta x} \sin x = \frac{1}{1 + \beta^2}$

This expression is the derivative
(with respect to β)
of the thing we actually want.
So we need to integrate it.

The denominator might ring a bell:
You need to invoke trigonometric functions.

Substituting $\beta = \tan \phi$
and using  $\tan \phi = \sin \phi / \cos \phi$ and $\cos^2 x + \sin^2 x = 1$,
thus $d(\tan \phi) = d\phi / \cos^2 \phi$

and

$\displaystyle \int d\beta \frac{1}{1 + \beta^2} = \int d \phi \frac{\frac{1}{\cos^2 \phi}}{1 + \frac{\sin^2 \phi}{\cos^2 \phi} } = \int \frac {1}{\cos^2 \phi + \sin^2 \phi} d\phi = \phi + C = \arctan \beta + C$

In summary,
collecting all the factors of 2 and minus signs
(hopefully):

$\displaystyle \int_{-\infty}^{+\infty}dx \frac {\sin x}{x} = -2 \lim_{\beta \to 0} \int_{0}^{\infty} dx e^{-\beta x} \sin x = - 2 \lim_{\beta \to 0} ( \arctan \beta + C)$

What is the integration constant?
Looking at the original integral
(over exponential times sine):
If β tends to infinity,
the integral becomes 0.

arctan of infinity is π/2
(I always think of the tan function and mirror it at the diagonal).
Make the integral 0
for an infinite β
by a C of -π/2.

Finally, finally:
we can insert this C and make β zero:

$\displaystyle \int_{-\infty}^{+\infty}dx \frac {\sin x}{x} = -2 (\arctan(0) - \frac{\pi}{2}) = -2( 0 - \frac{\pi}{2} ) = \pi$

_____

The ‘poetry’ was unintended, but I could not resist to add line breaks at the end.

Chances are high that I got two minus signs wrong and pairs of errors cancel out.

_____

[*] Often when a thing in science is tagged with one specific scientist’s name, it – or precursors of it – can be traced back to other. In this case, the ‘Dirac’ Delta Function might better be called the Cauchy-Dirac Delta function. Roger Penrose attributes the Delta Function to Oliver Heaviside in his book The Road to Reality.

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