Dirac Delta Function

The Dirac delta, written $\delta(x)$, is best understood as a distribution, not an ordinary function. It represents a unit amount concentrated at one point, and its key rule is the sifting property:

when the interval includes $a$ and $f$ is continuous, or at least well behaved, at that point. In plain language, $\delta(x-a)$ acts like a sampler: inside an integral, it picks out the value of the other factor at $x=a$.

The Formula And Its Symbols

If $\delta(x-a)$ appears inside an integral, all of its effect is concentrated at $x=a$, which is why people picture it as a spike with total area $1$. That picture is useful for intuition, but the reliable definition is the integral rule above. Treat the spike image as a mnemonic, not a literal graph. Two consequences follow at once:

and, if the interval does not contain $a$,

because the sampling point lies outside the interval.

Why The Sifting Property Holds

The spike picture explains the calculation. Since $\delta(x-a)$ is zero everywhere except at $x=a$ and carries total area $1$, the product $f(x)\delta(x-a)$ is also concentrated entirely at $x=a$. Near that point $f$ is essentially constant at the value $f(a)$, so it factors out of the concentrated mass, and what remains is $f(a)$ times the total area $1$. That is why the integral collapses to $f(a)$ rather than to a full antiderivative. The same reasoning shows why the result is $0$ when $a$ lies outside the interval: there is no mass inside the region to sample. This is a heuristic, not a proof, but it is the right mental model for elementary problems, where the safest stance is to define $\delta(x)$ by what it does under integration rather than by any pointwise value. The phrase "zero everywhere except at $0$ and infinite at $0$" is rough intuition, not a complete definition, which is why treating $\delta(0)$ as an ordinary number leads to errors.

Worked Example: The Sifting Property In Action

Evaluate

Step 1: find the sampling point. Since the delta is $\delta(x-3)$, it samples at $x=3$. Step 2: substitute $x=3$ into the other factor:

That is the whole calculation. You do not integrate $x^2+1$ in the usual way; you locate the sample point and evaluate the remaining expression there.

Reading The Shift Correctly

Sign errors are one of the most common sources of wrong answers:

but

so it samples at $x=-a$. For example,

not $e^2$.

Practice With A Self-Check

Evaluate

First find the sampling point, then substitute it into the linear expression. Self-check: $\delta(x+1)$ samples at $x=-1$, so the value is $2(-1)-5 = -7$. Now compare with the same integrand over $[0,4]$: since $-1$ is outside $[0,4]$, that integral is $0$, which shows how the interval changes the answer.

Calculation Traps

Treating $\delta(x)$ like a normal function. Its meaning comes from how it acts in integrals; handling it like a graphable function usually leads to the wrong move.

Missing the sampling point. With $\delta(x-a)$ the sample is at $x=a$; with $\delta(x+a)$ it is at $x=-a$.

Ignoring the interval. If the integration interval does not include the sampling point, the integral is $0$, often the fastest thing to check.

Forgetting the condition on $f(x)$. The standard rule applies when the other factor is well behaved at the sampling point; in introductory settings, continuity there is usually enough.

Confusing Dirac with Kronecker. The Dirac delta is continuous; the Kronecker delta $\delta_{ij}$ is discrete, equal to $1$ when $i=j$ and $0$ otherwise.

Where The Dirac Delta Is Used

It appears when a model needs something concentrated at one point in space or one instant in time: an impulse force in mechanics, an idealized point charge or point mass, and instantaneous inputs in signal processing. It also shows up in Green's functions and in Fourier or Laplace methods, where it compactly describes an input that happens all at once.