The Dirac delta function is a special tool used in math and physics, especially when dealing with sharp, sudden changes like a quick force or a point charge. Even though it's called a "function," it’s not a real function in the usual sense. Instead, it acts like a symbol or shortcut to make certain calculations easier.

It’s mostly used to represent things that happen at a single point—like an instant hit, a spark, or the position of an electron. For example, when a billiard ball is hit, the force is very quick and strong at one point in time. The Dirac delta is used to model this sudden effect.

This idea is helpful in physics and engineering, especially when studying how particles move or how waves travel. In this topic, you’ll learn the definition, key properties, and see some examples to understand how the Dirac delta function works in solving problems.

Let us study more about the chord of contact in this math article.

What is the Dirac delta function?

The Dirac delta function, written as δ(x), is not a regular function but a special mathematical tool called a distribution. It is used to model very sudden changes, like a sharp spike at a single point. This is why it is also called the impulse symbol or Dirac's delta.

This function is defined as zero everywhere except at x = 0, where it becomes infinitely large, but in such a way that the total area under the curve is equal to 1. That means:

• δ(x) = 0 when x is not equal to 0
• δ(x) = ∞ at x = 0
• The total area under δ(x) from just before to just after 0 is 1:
  ∫ from b to c δ(x) dx = 1, if 0 is between b and c

Also, multiplying x by δ(x) gives 0:
x · δ(x) = 0

This makes δ(x) very useful in physics and engineering to represent sudden events.

It can be easiest to think of the delta function as the limit of a sequence of steps, each of which is higher and narrower than the previous step, such that the area under the step is always one and it can be seen in the below Figure. The function\( δ(x) \) can be approximated by a series of steps that get progressively thinner and higher in such a way that the area under the curve is always equal to one.

Learn about Greatest Integer Function

There are many properties of the delta function which follow from the defining properties mentioned above. The following are the Some of these :

• \( \delta(x)=\delta(-x) \)
• \( \frac{d}{dx}\delta (x)= -\frac{d}{dx}\delta (-x) \)
• \( \int_{a}^{b}f(x)\delta'(x-a)dx =-f'(a) \)
• \( \delta(ax)=\frac{1}{|a|}\delta (x) \)
• \( \delta(g(x))= \sum_i^0 \frac{1}{|g'(x_{i})|}\delta (x-x_{i}) \)
• \( \delta(x^{2}-a^{2})= |a|^{-1}[\delta (x-a) +\delta (x+a)] \)
• \( \delta ((x-a)(x-b))= \frac{1}{|a-b|}[\delta (x-a) +\delta (x-b)] \)

Where, a= constant , \( g(x_{i})=0\) and \( g'(x_{i}) \#0\).

The first two properties of the Dirac Delta Function show that the delta function is even and its derivative is odd.

Summary of Dirac Delta Function

• The Dirac delta function is not a function but a symbol. Used by physicists and engineers to represent some calculations.
• The Dirac delta function defining properties are mentioned above.
• Some of the properties of The Dirac delta function are mentioned above.

Applications of Dirac Delta Function

1. Impulse Modeling
   The Dirac delta function is used to represent an instant or sudden force acting at a specific point in time or space. For example, it models a hammer strike or a quick electric pulse in circuits.
2. Electronics and Signals
   In signal processing, delta functions are used to analyze systems and their responses to sharp input signals, like in filters or systems receiving an impulse.
3. Physics (Point Sources)
   It helps describe physical quantities that are concentrated at a point — like a point charge in electrostatics or a point mass in mechanics.
4. Control Systems
   In control engineering, it is used to test how systems react to sudden changes using the impulse response method.
5. Green’s Function Methods
   In solving differential equations, delta functions help express the response of a system to an impulse, which can be expanded for general inputs.
6. Quantum Mechanics
   The delta function is often used to define position and momentum states or boundary conditions in quantum systems.
7. Probability Theory
   It is used to represent probability distributions that are concentrated at a single point (called degenerate distributions).

Dirac Delta Function Solved Examples

Question1: Prove that \(\int_{-1}^{1}x^{2}\delta (3x+1)dx=\frac{1}{27}\)

Answer 1: Assume, \( y=3x+1\)

\( \frac{dy}{dx}=3 \)

\( x=\frac{y-1}{3} \)

LHS= \(\int_{-1}^{1}x^{2}\delta (3x+1)dx\)

\( \int_{-2}^{4}(\frac{y-1}{3})^{2}\delta (y)\frac{dy}{3} \)

\(\frac{1}{3}\int_{-2}^{4}(\frac{y-1}{3})^{2}\delta (y) dy \)

= \(\frac{1}{3}\times\frac{1}{9} \)

=\(\frac{1}{27} \)

= RHS

LHS=RHS

Hence it is proved \(\int_{-1}^{1}x^{2}\delta (3x+1)dx=\frac{1}{27}\).

Question 2: Let the function
f(x) =
  x² + 3  for x < 0
  x·e⁻ˣ  for x > 0

We are asked to find the derivative of f(x), including the Dirac delta term at x = 0.

Answer 2:

Step 1: Check continuity at x = 0

• Left-hand limit as x → 0⁻:
   f(x) = x² + 3 → 0² + 3 = 3
• Right-hand limit as x → 0⁺:
   f(x) = x·e⁻ˣ → 0·1 = 0

There is a jump (discontinuity) at x = 0:
 Gap = 0 – 3 = –3

Step 2: Derivative on both sides (excluding x = 0)

• For x < 0: f(x) = x² + 3 ⇒ f′(x) = 2x
• For x > 0: f(x) = x·e⁻ˣ ⇒ f′(x) = e⁻ˣ – x·e⁻ˣ
    (using product rule)

Let this part be called φ(x):

φ(x) =
  2x     for x < 0
  e⁻ˣ – x·e⁻ˣ  for x > 0

Step 3: Include delta function
From the jump, we know:

f′(x) = φ(x) – 3·δ(x)

Final Answer:
The derivative of f(x) including the delta function is:
f′(x) =
  2x     for x < 0
  e⁻ˣ – x·e⁻ˣ  for x > 0
plus a delta term: –3·δ(x)

Or in one line:

f′(x) = φ(x) – 3·δ(x), where φ(x) = { 2x for x < 0, e⁻ˣ – x·e⁻ˣ for x > 0 }

Question 3: Evaluate the integral:
∫ from –∞ to ∞ of (3x + 2) · δ(5(x – 1)) dx

Answer 3:

We are given:
∫ (3x + 2) · δ(5(x – 1)) dx

Step 1: Use the property of the delta function:
δ(a(x – b)) = (1/|a|) · δ(x – b)

Here, a = 5 and b = 1, so:
δ(5(x – 1)) = (1/5) · δ(x – 1)

So the integral becomes:
= (1/5) ∫ from –∞ to ∞ of (3x + 2) · δ(x – 1) dx

Step 2: Apply the sifting property of delta function:
∫ f(x) · δ(x – a) dx = f(a)

Here, f(x) = 3x + 2 and a = 1
So:
= (1/5) · [3(1) + 2]
= (1/5) · (3 + 2)
= (1/5) · 5 = 1

Final Answer:
∫ from –∞ to ∞ of (3x + 2) · δ(5(x – 1)) dx = 1

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