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.

Table of Contents

## 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: What is the inverse Laplace of 1.**

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

**Q2: What is the inverse Laplace transform of 2.**

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

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.