The Dirac Delta Function in One Dimension

Electrodynamics is built on an idealization it cannot describe with ordinary functions: the point charge. A charge $q$ compressed into a single point has infinite density at that point and zero density everywhere else — yet its density must still integrate to $q$ . No function does that. If a function vanishes everywhere except at one point, its integral is zero, full stop. The chapter you are in has already met the consequence: the Coulomb field $\hat{\mathbf{r}}/r^2$ has zero divergence everywhere you can compute it, yet pushes a stubborn flux of $4\pi$ through every surrounding sphere. Something is hiding at the origin, and ordinary calculus has no container for it. Dirac's delta function $\delta(x)$ is the container. It is an "improper function" — Dirac's own honest label — defined not by its values but by what it does under an integral: it vanishes for $x\neq0$ , and against any smooth function it plucks out that function's value at zero, $\int f(x)\,\delta(x)\,\mathrm{d}x=f(0)$ . Picture an infinitely tall, infinitely narrow spike of unit area: the limit of ever-skinnier rectangles or Gaussians that keep their area fixed at 1. This article builds the one-dimensional theory — the sifting property, the scaling and composition rules, the delta as the derivative of a step — with worked numbers throughout, because delta-function fluency is finger memory. The three-dimensional version and the payoff, $\nabla\cdot(\hat{\mathbf{r}}/r^2)=4\pi\delta^3(\mathbf{r})$ , follow in the next article.

Key takeaways

The Dirac delta $\delta(x)$ is defined by two properties: $\delta(x)=0$ for $x\neq0$ , and $\int_{-\infty}^{\infty}\delta(x)\,\mathrm{d}x=1$ — an infinitely tall, infinitely narrow unit-area spike, the idealization of a point source.
It is not a function but a distribution: it acquires meaning only under an integral, as the limit of unit-area peaked functions (rectangles, Gaussians) whose width goes to zero.
The sifting property is its job description: $\int f(x)\,\delta(x-a)\,\mathrm{d}x=f(a)$ — the delta samples the function at the spike.
It transforms with a Jacobian: $\delta(kx)=\delta(x)/|k|$ , and generally $\delta(g(x))=\sum_i\delta(x-x_i)/|g'(x_i)|$ over the simple zeros of $g$ .
It is the derivative of the Heaviside step function, and it carries units: $\delta(x)$ has dimensions of $1/[x]$ .
Key idea: the delta function is the unit point source — zero everywhere, unit integral, defined entirely by how it acts inside integrals, where it samples the function it multiplies.

The core insight

The Dirac delta is the unit point source: $\delta(x)=0$ for $x\neq0$ yet $\int\delta(x)\,\mathrm{d}x=1$ , and its whole meaning lives under integrals, where it samples whatever smooth function it multiplies: $\int f(x)\,\delta(x-a)\,\mathrm{d}x=f(a)$ .

Where this lives today

Ask a modern engineer what the delta function is and they will say: the input you hit a system with to learn everything about it. Strike a linear system — a concert hall, an audio filter, a communication channel, a bridge — with an idealized impulse $\delta(t)$ , and the output is the system's impulse response, from which its response to any input follows by superposition. Every room-correction profile in a home theater, every channel estimate in a 5G modem, every seismology Green's function is a measured approximation to "what does this system do to a delta?" The same idea under a different name rules imaging: a telescope's or microscope's point-spread function is its response to a delta of light, and deconvolving it is how sharp images are recovered from blurred data. In this track the delta's biggest job begins in the next article: promoted to three dimensions it becomes the density of a point charge, resolves the $\hat{\mathbf{r}}/r^2$ divergence paradox, and powers the Green-function method — solve Poisson's equation once for a delta source, and the solution for any charge distribution is an integral away. That one-two — idealize the source as a delta, superpose — is arguably the single most reused strategy in theoretical physics.

The physical picture

Load a beam with a brick, and the load density (newtons per metre) is a smooth bump spread over the brick's width. Now swap the brick for a knife edge pressing down with the same total force. The density profile gets narrower and taller, narrower and taller — and in the ideal limit it becomes nothing anywhere, except an infinite spike at one point, with the total force still stored in its area. That limiting object is the delta function. Physics is full of knife edges: point charges, point masses, impulsive hammer blows (force spiked in time), the charge sheet crossed in a single step. The delta is the honest bookkeeping for all of them.

Figure

Unit-area peaks squeezing toward the delta — width shrinks, height grows, area stays 1

The crucial mental shift: stop asking what $\delta(x)$ equals and start asking what it does. Multiply any smooth function $f(x)$ by a unit-area spike at $x=a$ and integrate: the spike is so narrow that $f$ cannot vary across it, so $f$ freezes at its value $f(a)$ and rides outside the integral, leaving the spike's area, 1. Result: $f(a)$ . The delta is a sampler — an infinitely precise measurement device that reads out one value of a function and discards the rest.

Figure

The sifting property — the spike samples the function where it stands

The definition, made honest

The working definition is the pair of statements physicists use daily:

\delta(x)=0\ \ (x\neq0), \qquad \int_{-\infty}^{\infty}\delta(x)\,\mathrm{d}x=1 \qquad (1)

where: the two clauses are, for an ordinary function, flatly contradictory — a function vanishing except at a point integrates to zero. The resolution is that $\delta$ is not a function but a limit of functions kept under an integral sign: define the unit-area rectangle $\delta_\epsilon(x)=1/\epsilon$ for $|x|<\epsilon/2$ (zero elsewhere), do every calculation with $\delta_\epsilon$ , and send $\epsilon\to0$ at the end. Every identity in this article is a statement about that limit — finite, checkable, and rigorous enough for all of physics. (Mathematics formalizes the same idea as distribution theory: the delta is a linear machine that eats a smooth test function and returns a number, $f\mapsto f(0)$ . Laurent Schwartz built that theory in the 1940s, and it certifies every manipulation below.) The delta earns its keep through the sifting property:

\int_{-\infty}^{\infty}f(x)\,\delta(x-a)\,\mathrm{d}x=f(a) \qquad (2)

where: $f$ — any function continuous at $a$ ; the spike sits at $x=a$ and samples $f$ there. Two immediate practical corollaries. Ranges matter: if the integration interval does not contain $a$ , the answer is zero — the spike was never inside. Units matter: since $\delta(x)\,\mathrm{d}x$ is dimensionless, $\delta(x)$ carries units of $1/[x]$ — per metre if $x$ is a length, per second if it is a time. That is exactly what a linear charge density $\lambda(x)=q\,\delta(x-a)$ needs to put coulombs-per-metre on the left and coulombs times per-metre on the right.

Derivation AThe sifting property from the rectangle limit

The goal is equation (2) — the statement that the delta samples — proved honestly from the unit-area rectangle family, so that "the spike is too narrow for f to vary" becomes a controlled limit rather than a slogan. The strategy: do the integral with the finite-width rectangle $\delta_\epsilon$ , invoke the mean value theorem to freeze $f$ , and let the width vanish. Watch how little is demanded of $f$ — continuity at the sampling point is the entire admission fee.

Step 1. Replace the delta by the rectangle of width $\epsilon$ and height $1/\epsilon$ centred at $a$ :

\int_{-\infty}^{\infty}f(x)\,\delta_\epsilon(x-a)\,\mathrm{d}x=\frac{1}{\epsilon}\int_{a-\epsilon/2}^{a+\epsilon/2}f(x)\,\mathrm{d}x

Why the prefactor and limits: the rectangle vanishes outside the window, and inside it equals the constant $1/\epsilon$ — the integral is just the average of $f$ over the window.

Step 2. Apply the mean value theorem for integrals: for continuous $f$ there is a point $\xi$ inside the window with $\frac{1}{\epsilon}\int_{a-\epsilon/2}^{a+\epsilon/2}f(x)\,\mathrm{d}x=f(\xi)$ , where $|\xi-a|\le\epsilon/2$ .

Step 3. Send $\epsilon\to0$ . The point $\xi$ is squeezed onto $a$ , and continuity of $f$ forces $f(\xi)\to f(a)$ :

\lim_{\epsilon\to0}\int f(x)\,\delta_\epsilon(x-a)\,\mathrm{d}x=f(a)

Why this is the delta's definition in disguise: the left side is what " $\int f\,\delta(x-a)\,\mathrm{d}x$ " means — first integrate at finite width, then take the limit. Nothing infinite was ever evaluated pointwise.

Step 4. Check that the choice of family is irrelevant: rerunning with unit-area Gaussians $\frac{1}{\sigma\sqrt{2\pi}}e^{-(x-a)^2/2\sigma^2}$ gives the same limit (split the integral at $|x-a|=\sigma^{1/2}$ ; the tails carry vanishing area, the core freezes $f$ ). Any unit-area, shrinking-width family defines the same delta — the limit forgets the profile. Verified: matches Griffiths §1.5.1-1.5.2 and Arfken §1.11; setting $f=1$ recovers the normalization $\int\delta=1$ .

The algebra of spikes

Because the delta lives under integrals, every rule about it is proved by substitution inside an integral. The most-used rule is scaling — squeeze the argument and the spike reweights:

\delta(kx)=\frac{1}{|k|}\,\delta(x) \qquad (3)

where: the $1/|k|$ is a one-dimensional Jacobian, and the absolute value is not optional: a negative $k$ flips the integration limits, and the flip's minus sign cancels the sign of $k$ . Physically, (3) is units bookkeeping incarnate — $\delta(3x)$ is a spike in a three-times-compressed coordinate, so it is three times "cheaper." The general composition rule handles any smooth argument with simple zeros:

\delta\big(g(x)\big)=\sum_i\frac{\delta(x-x_i)}{|g'(x_i)|} \qquad (4)

where: the sum runs over all roots $x_i$ of $g$ ( $g(x_i)=0$ , $g'(x_i)\neq0$ ): the composite spike fires wherever its argument vanishes, each firing weighted by how fast the argument sweeps through zero. A delta of $x^2-c^2$ fires at both $\pm c$ ; a delta in energy $\delta(E-E_0)$ converts to a delta in momentum with the density-of-states factor $1/|dE/dp|$ — equation (4) is the origin of every such factor in physics.

Go deeper · Derivation B · 1 minScaling and composition by change of variables

Derivation BScaling and composition by change of variables

The goal is equations (3) and (4), proved the only way delta identities can be proved: by showing both sides do the same thing to an arbitrary smooth function under an integral. The strategy is change of variables, $u=g(x)$ . The two steps to watch: the limit-flip that generates the absolute value, and the localization step in which only neighbourhoods of the roots survive — the mechanism that turns one composite delta into a sum of simple ones.

Step 1. Scaling, $k>0$ : substitute $u=kx$ in $\int f(x)\,\delta(kx)\,\mathrm{d}x$ , so $\mathrm{d}x=\mathrm{d}u/k$ : the integral becomes $\frac{1}{k}\int f(u/k)\,\delta(u)\,\mathrm{d}u=\frac{1}{k}f(0)$ . The same result comes from $\int f(x)\,\frac{1}{k}\delta(x)\,\mathrm{d}x$ — so the two objects agree against every $f$ , which is what (3) means.

Step 2. Scaling, $k<0$ : the substitution now reverses the limits ( $x:-\infty\to\infty$ maps to $u:+\infty\to-\infty$ ); restoring ascending order costs a minus sign, giving $-\frac{1}{k}f(0)=\frac{1}{|k|}f(0)$ . Why the absolute value is structural: the delta is a positive spike regardless of the direction its argument runs; orientation is surrendered, magnitude kept.

Step 3. Composition: in $\int f(x)\,\delta(g(x))\,\mathrm{d}x$ , the delta kills every region where $g\neq0$ , so the integral localizes to small neighbourhoods of the roots $x_i$ . Why localization is legal: outside those neighbourhoods the integrand is identically zero at finite regulator width, uniformly in the limit.

Step 4. Near each root, linearize: $g(x)\approx g'(x_i)(x-x_i)$ . Each neighbourhood then contributes a scaled delta, and Step 1-2 applies with $k=g'(x_i)$ :

\int f(x)\,\delta(g(x))\,\mathrm{d}x=\sum_i\frac{f(x_i)}{|g'(x_i)|}

which is precisely what the right side of (4) produces. The rule fails, honestly, when some $g'(x_i)=0$ — a double zero makes the composite delta ill-defined, a limitation worth remembering rather than papering over. Verified: matches Griffiths Problem 1.45 and Arfken §1.11; test case $g=x^2-c^2$ : roots $\pm c$ , $|g'|=2|c|$ , giving $\delta(x^2-c^2)=\frac{1}{2|c|}[\delta(x-c)+\delta(x+c)]$ — the standard result.

One more relation completes the toolkit: the delta is the derivative of the Heaviside step function $\theta(x)$ (zero for $x<0$ , one for $x>0$ ),

\frac{\mathrm{d}\theta}{\mathrm{d}x}=\delta(x) \qquad (5)

where: the statement is exactly what it should be physically — the rate of change of a quantity that jumps is a spike at the jump. Check it the distributional way: $\int f(x)\,\theta'(x)\,\mathrm{d}x=-\int f'(x)\,\theta(x)\,\mathrm{d}x=-\int_0^\infty f'\,\mathrm{d}x=f(0)$ (integration by parts, then the fundamental theorem, with $f$ vanishing at infinity) — the step's derivative samples at zero, which is the delta's signature. The same integration-by-parts move defines the delta's own derivative: $\int f(x)\,\delta'(x)\,\mathrm{d}x=-f'(0)$ — the spike's derivative samples slopes, with a minus sign; it is the ideal point dipole, as the next tier will exploit.

Which rule answers which question

you want	use	notes
evaluate $\int f\,\delta(x-a)$	sifting, equation (2)	answer $f(a)$ — if $a$ is inside the range, else 0
a squeezed argument $\delta(kx)$	equation (3)	divide by $\|k\|$ ; sign always dropped
a nonlinear argument $\delta(g(x))$	equation (4)	one term per simple root, weight $1/\|g'\|$
derivative of a jump	equation (5)	jump of height $h$ at $a$ : $h\,\delta(x-a)$
slope sampling	$\int f\,\delta'(x-a)=-f'(a)$	the point dipole
point charge on a line	$\lambda(x)=q\,\delta(x-a)$	units: $\delta$ carries $1/[x]$

Limiting cases

Case	Result	Meaning
rectangle family, $\epsilon\to0$	$\delta_\epsilon\to\delta$	the working definition
Gaussian family, $\sigma\to0$	$\frac{1}{\sigma\sqrt{2\pi}}e^{-x^2/2\sigma^2}\to\delta(x)$	same limit, smoother regulator
$k\to1$ in (3)	$\delta(x)$ unchanged	consistency check
$g$ linear in (4)	recovers (3)	composition contains scaling
double zero of $g$	rule (4) fails	$\delta(x^2)$ is not defined
smeared source, width $\ll$ everything else	delta is an excellent model	idealization with a stated regime

Common trapIt is not a function — so some function moves are illegal

Three habits from ordinary calculus fail on the delta, and each failure is a standard exam wound. First, pointwise evaluation: " $\delta(0)=\infty$ " is not a value, and nothing may be computed from it — the delta only reports through integrals. Second, nonlinear operations: $\delta(x)^2$ , $\sqrt{\delta(x)}$ , $e^{\delta(x)}$ are all meaningless; check with the regulator — the rectangle squared has area $1/\epsilon\to\infty$ , so no limit exists. (One delta per integral, ever.) Third, forgetting the Jacobian: $\delta(3x)$ is NOT $\delta(3x)\stackrel{?}{=}\delta(x)$ — it is $\tfrac13\delta(x)$ , and in general every change of the argument costs the $1/|g'|$ of equation (4). A useful drill: whenever a delta appears, mentally name the integral it is waiting for and the units it carries. A delta with no integral in sight and no $1/[x]$ in its units is a mistake being incubated.

Worked examples

1 · Sifting drill — four integrals, four habits (confidence builder)

Evaluate: (a) $\int_{-\infty}^{\infty}(x^3+2x+3)\,\delta(x-2)\,\mathrm{d}x$ ; (b) $\int_0^5 x^2\,\delta(x+1)\,\mathrm{d}x$ ; (c) $\int_{-\infty}^{\infty}\cos x\,\delta(3x-\pi)\,\mathrm{d}x$ ; (d) the total charge in $\lambda(x)=q\,\delta(x-a)$ with $q=5.0\ \mathrm{nC}$ .

Step 1. (a) The spike sits at $x=2$ , inside the range: sample the function there. $f(2)=8+4+3=\boxed{15}$ .

Step 2. (b) The spike sits at $x=-1$ — outside $[0,5]$ . The integrand is zero throughout the integration range: $\boxed{0}$ . Always locate the spike before touching the algebra.

Step 3. (c) Rewrite the argument: $\delta(3x-\pi)=\delta\big(3(x-\pi/3)\big)=\tfrac13\,\delta(x-\pi/3)$ by the scaling rule. Then sample: $\tfrac13\cos(\pi/3)=\tfrac13\times\tfrac12=\boxed{\tfrac16}$ . Forgetting the $\tfrac13$ is the single most common delta error in gradebooks.

Step 4. (d) $Q=\int\lambda\,\mathrm{d}x=q\int\delta(x-a)\,\mathrm{d}x=\boxed{5.0\ \mathrm{nC}}$ — and the units audit: $\lambda$ is C/m because $\delta(x-a)$ carries $\mathrm{m^{-1}}$ . The delta turned a point charge into a legitimate density, which is its entire purpose in electrodynamics.

2 · A two-charge line and a quadratic argument (exam level)

(a) Two point charges sit on the $x$ -axis: $q_1=+3.0\ \mathrm{nC}$ at $x=+2.0\ \mathrm{cm}$ and $q_2=-3.0\ \mathrm{nC}$ at $x=-2.0\ \mathrm{cm}$ . Write the charge density, then compute the total charge and the dipole moment $p=\int x\,\lambda(x)\,\mathrm{d}x$ . (b) Evaluate $\int_{-\infty}^{\infty}(x^2+1)\,\delta(x^2-4)\,\mathrm{d}x$ .

Step 1. The density is a sum of weighted spikes: $\lambda(x)=q_1\,\delta(x-0.02)+q_2\,\delta(x+0.02)$ (SI units, $x$ in metres).

Step 2. Total charge by sifting with $f=1$ : $Q=q_1+q_2=\boxed{0}$ — a neutral pair.

Step 3. Dipole moment by sifting with $f=x$ : $p=q_1(0.02)+q_2(-0.02)=3.0\times10^{-9}\times0.02+(-3.0\times10^{-9})\times(-0.02)=\boxed{1.2\times10^{-10}\ \mathrm{C\,m}}$ . Both terms add — each charge sits on the side that reinforces the moment. This two-spike density is exactly the ideal dipole whose field you will meet in E1, and in the ideal limit ( $q\to\infty$ , separation $\to0$ , $p$ fixed) the density becomes $-p\,\delta'(x)$ : the derivative-of-delta as point dipole, made concrete.

Step 4. (b) The argument $g(x)=x^2-4$ has simple roots at $x=\pm2$ with $|g'(\pm2)|=2|x|=4$ . Composition rule: $\delta(x^2-4)=\tfrac14\big[\delta(x-2)+\delta(x+2)\big]$ .

Step 5. Sample at both roots: $\tfrac14\big[f(2)+f(-2)\big]=\tfrac14\big[5+5\big]=\boxed{2.5}$ . Cross-check the structure: the composite spike fired at every zero of its argument, each weighted by the sweep speed — miss the negative root and the answer halves.

Origins · 1 minPaul Dirac and how this was discovered

Paul Dirac

1902–1984 · Cambridge, Lucasian Professor

Dirac was an electrical engineer before he was a physicist, and the delta function is engineering pragmatism elevated to grammar. Heaviside's telegraphy calculus had differentiated step functions for decades while mathematicians frowned; Dirac, assembling quantum mechanics in the late 1920s, needed continuous-spectrum analogues of matrix elements and simply named the object everyone needed: $\delta(x)$ , "the improper function," introduced with characteristic economy in his 1930 Principles of Quantum Mechanics along with the rules of this article. He knew precisely what he was doing — the book warns that $\delta$ is not a function and must live under integrals — and he trusted physicists to compute correctly ahead of the rigor. They did; the rigor arrived fifteen years later as Schwartz's distribution theory. The pattern was Dirac's signature: the relativistic electron equation, antimatter, magnetic monopoles — each pulled from formal structure that looked like nonsense until it was the truth. "The equation was smarter than I was," he said of one of them.

History

1822

— Fourier's analytical theory of heat freely swaps integrals in ways that, read carefully, smuggle a delta under the integral sign.

1882

— Kirchhoff's formulation of Huygens' principle uses an explicitly spiked function as a source model.

1893

— Heaviside's operational calculus differentiates the step function without apology; telegraph engineers compute happily with the results.

1930

— Dirac's Principles of Quantum Mechanics names the delta, codifies its rules, and hands physics its point-source grammar.

1945

— Laurent Schwartz constructs distribution theory, making every delta manipulation rigorous; the work earns the 1950 Fields Medal.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §1.5 · Jackson, Classical Electrodynamics, 3rd ed., §1.2 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Appendix on the delta function · Zangwill, Modern Electrodynamics, §1.5 · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., §1.11 · Lighthill, Introduction to Fourier Analysis and Generalised Functions (the classic short rigorous account)

Provenance. Definitions and properties follow Griffiths §1.5, Jackson §1.2, and Arfken §1.11. A verifier should recompute: the box-limit unit area ( $\frac1\epsilon\cdot\epsilon=1$ ) and sifting limit; the scaling factor $1/|a|$ via $u=ax$ substitution; the composition example $\delta(x^2-c^2)=\frac{1}{2|c|}[\delta(x-c)+\delta(x+c)]$ ; the derivative identity $\int f\delta'=-f'(0)$ by parts. Worked-example numbers on this page (15; 0; 1/6; $Q=0$ and $p=1.2\times10^{-10}$ C·m for the ±3 nC pair at ±2 cm; 2.5 for the quadratic-argument integral) are illustrative arithmetic recomputed from the boxed rules.

Builds on: Volume Integrals · Scalar and Vector Fields Leads to: Second Derivatives and the Laplacian · The Divergence (Gauss) Theorem

Is δ(0) equal to infinity? Can I ever use that?

" $\delta(0)=\infty$ " is a picture, not a value, and no computation may consume it. The delta assigns numbers only to integrals: $\int f\delta\,\mathrm{d}x=f(0)$ is meaningful, "the value of $\delta$ at a point" is not. When a stray $\delta(0)$ appears in a calculation — it happens, for instance, when two deltas collide or when volume normalizations are mishandled in Fourier analysis — it is a diagnostic light, telling you an idealization has been pushed past its regime. The standard cures are to restore a regulator (finite width $\epsilon$ , finite box size $L$ ), finish the physics, and take the limit only in quantities that stay finite. If the answer to a physical question contains $\delta(0)$ , the question was posed about the idealization rather than the physics.

Why is there an absolute value in the scaling rule — where does the sign go?

It falls out of the substitution honestly. For $u=kx$ with $k<0$ , the limits of the integral flip along with the variable, and restoring them to ascending order supplies a minus sign that cancels the sign of $k$ : the net factor is $1/|k|$ . The structural reason is more satisfying: the delta is a positive unit spike — a density — and densities do not care about the orientation of the coordinate you describe them with. A charge of $q$ at the origin is a charge of $q$ whether your $x$ -axis points left or right; $\delta(-x)=\delta(x)$ is that statement in symbols, and it is the $k=-1$ case of the rule.

Does it matter whether I regularize the delta with rectangles, Gaussians, or something else?

For every identity in this article — no, and that robustness is the deep reason the delta is usable at all. Any family of unit-area functions whose width shrinks to zero produces the same sampling limit against continuous functions; the limit forgets the profile. (The derivation of the sifting property runs the argument for both rectangles and Gaussians.) Where the choice can matter is at the edges of legality: products of distributions, deltas evaluated at boundaries of integration ranges, or nonlinear operations — there different regulators can disagree, which is precisely the sign that the expression is not delta-representable and needs the actual physics of the finite width. Rule of thumb: linear operations are regulator-blind; anything nonlinear demands you keep the regulator and think.

What does it mean, precisely, that the delta is a distribution?

A distribution is a linear machine that eats a smooth, rapidly-decaying "test function" and returns a number — nothing more. The delta is the machine $f\mapsto f(0)$ . Ordinary functions $g$ are special cases (their machine is $f\mapsto\int gf\,\mathrm{d}x$ ), which is why distributions are also called generalized functions. Operations on distributions are defined by shifting the work onto the test function: the derivative of a distribution is the machine $f\mapsto-(\text{machine applied to }f')$ , which is exactly where $\int f\delta'=-f'(0)$ comes from. Schwartz's theorem that every distribution is locally a (repeated) derivative of a continuous function explains the delta's family tree: it is the second derivative of the perfectly respectable ramp function $\max(x,0)$ .

Why do physicists bother with a spiky idealization instead of just using a very narrow charge blob?

Because the idealization is simpler and more honest about what matters. A real charged sphere of radius $10^{-15}$ m in a problem whose other lengths are centimetres would drag its radius through every equation only to have it cancel, to fifteen digits, at the end. The delta declares the separation of scales up front: nothing in this problem resolves the source's structure, so give it none. The payoff compounds: sources become algebra ( $\rho=\sum_iq_i\delta^3(\mathbf{r}-\mathbf{r}_i)$ unifies point charges with continuous distributions in one formalism), superposition becomes integration, and the Green-function method — solve once for a delta, integrate for everything — becomes available. When source structure does matter (inside a nucleus, at a contact interaction), physics restores the finite width deliberately.

What happens to δ(g(x)) if g has a double zero?

The composition rule breaks, and no patch fixes it: $\delta(x^2)$ is genuinely undefined. Mechanically, the rule's weight $1/|g'(x_i)|$ diverges when $g'(x_i)=0$ ; structurally, a double zero means the argument grazes zero without crossing, and the regulated integral $\int f\,\delta_\epsilon(g(x))\,\mathrm{d}x$ then diverges as $\epsilon\to0$ (the argument lingers near zero over a stretch of $x$ of width $\sim\sqrt{\epsilon}$ , while the spike height is $1/\epsilon$ ). When such expressions arise in physics — thresholds in scattering, caustics in optics, band edges in solids — the divergence is the physics talking: the density of states really does pile up there, and the correct treatment keeps the regulator and extracts the (integrable, but not delta-shaped) singularity, typically a $1/\sqrt{\cdot}$ van Hove-type spike.

Is there a Fourier representation of the delta, and why is it everywhere?

Yes: $\delta(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ikx}\,\mathrm{d}k$ — the delta is the superposition of all plane waves with equal weight, which is intuitive: only at $x=0$ do all the waves agree in phase and reinforce; everywhere else they cancel. (Honestly interpreted, the identity holds in the distributional sense — cut the integral off at $|k|<K$ and you get a sinc-shaped unit-area peak that narrows as $K\to\infty$ , one more admissible family.) It is everywhere because it converts between the two languages physics constantly swaps: completeness of plane waves, orthonormality $\int e^{i(k-k')x}\mathrm{d}x=2\pi\delta(k-k')$ , and the momentum-space bookkeeping of every quantum and signal-processing calculation. You will use it heavily from E8 onward; here it is enough to recognize it as the delta's frequency-domain portrait.