The inverse Laplace transform of 1 is the Dirac delta function δ(t). The inverse Laplace transform of a constant is equal to constant times δ(t). In this post, we will learn how to find the inverse Laplace transform of 1, and more generally of a constant.

## Dirac delta function

How is Dirac delta function defined? The Dirac delta function is defined as follows:

$\delta(t) = \begin{cases}

0 & \text{ if } t \neq 0 \\

\infty & \text{ if } t = 0.

\end{cases}$

Note that Dirac delta function is also known as the unit impulse function. It satisfies the following property:

$\int_{-\infty}^\infty$ δ(t-a) f(t) dt = f(a) for a≥0 **…(I)**

We will show that the inverse Laplace transform of 1 is the Dirac delta function.

## What is the inverse Laplace transform of 1?

**Answer:** The inverse Laplace transform of 1 is δ(t) where δ(t) denotes the Dirac delta function.

*Proof:*

Recall, the definition of the inverse Laplace transform: Let F(s) be the Laplace transform of f(t), that is,

F(s) = L{f(t)} = $\int_0^\infty$ f(t) e^{-st} dt. **…(II)**

Then the inverse Laplace transform of F(s) is f(t). In other words,

L^{-1}{F(s)} = f(t).

Now, to find the inverse Laplace transform of 1, we will put f(t)=e^{-st} in the above Property **(I)** that Dirac delta function satisfies.

∴ We have that

$\int_{-\infty}^\infty$ δ(t-a) f(t) dt = f(a) = e^{-as}

⇒ $\int_{-\infty}^0$ δ(t-a) f(t) dt + $\int_{0}^\infty$ δ(t-a) f(t) dt= e^{-as}

⇒ $\int_{-\infty}^0$ 0⋅f(t) dt + $\int_{0}^\infty$ δ(t-a) f(t) dt= e^{-as}. Because, δ(t-a) = 0 for all t ∈ (0, ∞) as a≥0. It follows from the definition of the Dirac delta function.

⇒ 0 + $\int_{0}^\infty$ δ(t-a) f(t) dt= e^{-as}

⇒ $\int_{0}^\infty$ δ(t-a) f(t) dt= e^{-as}

Thus, the definition **(II)** of the Laplace transform implies that

L{δ(t-a)} = e^{-as} **…(*)**

Put a=0

∴ L{δ(t)} = e^{0 }= 1.

This shows that the Laplace transform of δ(t) is 1. Hence,

L^{-1}{1} = δ(t).

Thus, we have proven that the Laplace transform of 1 is δ(t).

Find the inverse Laplace transform of 1.Summary:The inverse Laplace transform of 1 is the Dirac delta function δ(t). |

## What is the inverse Laplace transform of constants?

**Answer:** The inverse Laplace transform of a constant c is cδ(t).

*Proof:*

We have

L{c δ(t-a)}

= c L{δ(t-a)}

= c e^{-as} by **(*)**

Therefore, L{c δ(t-a)} = c e^{-as}.

Put a=0

∴ L{c δ(t)} = c e^{0 }= c.

This shows that the Laplace transform of cδ(t) is c. Hence,

L^{-1}{c} = c δ(t).

Thus, the inverse Laplace transform of a constant is equal to constant times the Dirac delta function.

Find the inverse Laplace transform of c, where c is a constant.Summary:The inverse Laplace transform of c is c δ(t), where δ(t) is the Dirac delta function. |

**Also Read: **

**Laplace transform of constants**

**Question:** Find the inverse Laplace transform of 5.

*Solution:*

From above we have that the inverse Laplace transform of c is cδ(t), that is L{c}=cδ(t). Putting c=5, we obtain the inverse Laplace transform of 5 is 5δ(t) where δ(t) is the unit impulse function.

## FAQs

**Q1: Find the inverse Laplace transform of 1.**

Answer: The inverse Laplace transform of 1 is the Dirac delta function δ(t).

**Q2: Find the inverse Laplace transform of 2.**

Answer: The inverse Laplace transform of 2 is 2δ(t) where δ(t) is the unit impulse function.