Dirac Delta Function

The Dirac delta, written as $\delta(x)$, 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:

$$\int_{-\infty}^{\infty} f(x)\,\delta(x-a)\,dx = f(a)$$

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 $x=a$.

Dirac delta definition and intuition

If $\delta(x-a)$ appears inside an integral, all of its effect is concentrated at $x=a$. That is why people picture it as a spike with total area $1$.

That picture is useful for intuition, but the reliable definition is still the integral rule above. Treat the spike image as a mnemonic, not a literal graph of an ordinary function.

Two consequences follow immediately:

$$\int_{-\infty}^{\infty} \delta(x-a)\,dx = 1$$

and if the interval does not contain $a$,

$$\int_{c}^{d} f(x)\,\delta(x-a)\,dx = 0$$

because the sampling point lies outside the interval.

Why the Dirac delta is not a regular function

For an ordinary function, you can usually discuss values like $f(0)$ and use standard algebra without much trouble. The Dirac delta does not fit that pattern.

In elementary problems, the safest approach is to define $\delta(x)$ by what it does under integration. The phrase "zero everywhere except at $0$ and infinite at $0$" is only a rough intuition, not a complete definition.

That distinction prevents common mistakes like trying to treat $\delta(0)$ as an ordinary number.

Worked example with the sifting property

Evaluate

$$\int_{-\infty}^{\infty} (x^2 + 1)\,\delta(x-3)\,dx$$

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:

$$\int_{-\infty}^{\infty} (x^2 + 1)\,\delta(x-3)\,dx = 3^2 + 1 = 10$$

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.

How to read the shift correctly

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

$$\delta(x-a) \text{ samples at } x=a$$

but

$$\delta(x+a) = \delta(x-(-a))$$

so it samples at $x=-a$.

For example,

$$\int_{-\infty}^{\infty} e^x\,\delta(x+2)\,dx = e^{-2}$$

not $e^2$.

Common mistakes with the Dirac delta

Treating $\delta(x)$ like a normal function

Its meaning comes from how it acts in integrals. If you try to handle it like a standard graphable function, you will usually make the wrong move.

Missing the sampling point

With $\delta(x-a)$, the sample is taken at $x=a$. With $\delta(x+a)$, it is taken at $x=-a$.

Ignoring the interval

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

Forgetting the condition on $f(x)$

The standard sampling rule is used when the other factor is well behaved at the sampling point. In many introductory settings, continuity at that point is enough.

Confusing Dirac delta with Kronecker delta

The Dirac delta is used in continuous settings. The Kronecker delta, written $\delta_{ij}$, is a discrete object that equals $1$ when $i=j$ and $0$ otherwise.

Where the Dirac delta is used

The Dirac delta appears when a model needs to represent something concentrated at one point in space or one instant in time.

Typical examples include an impulse force in mechanics, an idealized point charge or point mass, and instantaneous inputs in signal processing.

It also appears in Green's functions and in Fourier or Laplace methods, where it gives a compact way to describe an input that happens all at once.

Try a similar problem

Try

$$\int_{-\infty}^{\infty} (2x-5)\,\delta(x+1)\,dx$$

First find the sampling point, then substitute it into the linear expression. If you want one more check, compare it with the same integral over $[0,4]$ and see why the answer changes.