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Delta Function — Definition, Formula & Examples

The delta function (also called the Dirac delta function) is a generalized function that is zero everywhere except at a single point, where it is infinitely tall, yet integrates to exactly 1 over the entire real line. It captures the idea of a perfect impulse or point concentration of mass.

The Dirac delta δ(x)\delta(x) is not a function in the classical sense but a distribution (linear functional) defined by its action on test functions: for any smooth function ff that is continuous at x=ax = a, ∫−∞∞f(x) δ(x−a) dx=f(a)\int_{-\infty}^{\infty} f(x)\,\delta(x - a)\,dx = f(a). This is called the sifting property. The delta distribution satisfies δ(x)=0\delta(x) = 0 for x≠0x \neq 0 and ∫−∞∞δ(x) dx=1\int_{-\infty}^{\infty} \delta(x)\,dx = 1.

Key Formula

∫−∞∞f(x) δ(x−a) dx=f(a)\int_{-\infty}^{\infty} f(x)\,\delta(x - a)\,dx = f(a)
Where:
  • δ\delta = The Dirac delta function (distribution)
  • f(x)f(x) = Any continuous function being "sampled"
  • aa = The point at which the delta is centered

How It Works

The delta function acts as a "selector" inside an integral: when you multiply a function f(x)f(x) by δ(x−a)\delta(x - a) and integrate, the result is simply f(a)f(a). You can think of it as concentrating all its weight at one point. In practice, you never evaluate δ(x)\delta(x) by itself — you always use it inside an integral or as part of a larger expression. Physicists and engineers often model point charges, impulse forces, or instantaneous signals using δ(x)\delta(x). Mathematically, it is rigorously defined through distribution theory, where it maps test functions to real numbers rather than having pointwise values.

Worked Example

Problem: Evaluate the integral ∫−∞∞(3x2+5) δ(x−2) dx\int_{-\infty}^{\infty} (3x^2 + 5)\,\delta(x - 2)\,dx.
Identify the sifting property: The integrand contains δ(x−a)\delta(x - a) with a=2a = 2 and f(x)=3x2+5f(x) = 3x^2 + 5. The sifting property states the integral equals f(a)f(a).
∫−∞∞f(x) δ(x−2) dx=f(2)\int_{-\infty}^{\infty} f(x)\,\delta(x - 2)\,dx = f(2)
Substitute $x = 2$ into $f(x)$: Replace xx with 2 in f(x)=3x2+5f(x) = 3x^2 + 5.
f(2)=3(2)2+5=3(4)+5=17f(2) = 3(2)^2 + 5 = 3(4) + 5 = 17
State the result: The delta function "picks out" the value of the function at x=2x = 2.
∫−∞∞(3x2+5) δ(x−2) dx=17\int_{-\infty}^{\infty} (3x^2 + 5)\,\delta(x - 2)\,dx = 17
Answer: The integral evaluates to 1717.

Another Example

Problem: Evaluate ∫05e−x δ(x−3) dx\int_{0}^{5} e^{-x}\,\delta(x - 3)\,dx.
Check that $a$ lies within the integration limits: Here a=3a = 3, and the limits are [0,5][0, 5]. Since 0≤3≤50 \leq 3 \leq 5, the delta contributes to the integral.
Apply the sifting property: Because a=3a = 3 is inside the interval, the integral equals f(3)=e−3f(3) = e^{-3}.
∫05e−x δ(x−3) dx=e−3\int_{0}^{5} e^{-x}\,\delta(x - 3)\,dx = e^{-3}
Compute the numerical value: Evaluate the exponential.
e−3≈0.0498e^{-3} \approx 0.0498
Answer: The integral equals e−3≈0.0498e^{-3} \approx 0.0498.

Why It Matters

The delta function is essential in courses like signals and systems, partial differential equations, and quantum mechanics. Electrical engineers use it to model impulse responses of circuits and filters, which is the foundation of linear systems theory. In physics, Green's functions — which solve PDEs for point sources — are built directly from the delta function.

Common Mistakes

Mistake: Treating δ(0)\delta(0) as a finite number (like 1 or ∞\infty) and trying to multiply by it.
Correction: The delta function has no meaningful pointwise value at x=0x = 0. It only produces well-defined results inside an integral via the sifting property.
Mistake: Applying the sifting property when the center aa lies outside the interval of integration.
Correction: If aa is not in the interval [b,c][b, c], then ∫bcf(x) δ(x−a) dx=0\int_b^c f(x)\,\delta(x - a)\,dx = 0, not f(a)f(a). Always check that the delta's center falls within the limits.

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