The Laplace transform of t is 1/s^2. In this post, we will find the Laplace transform of t by definition and by the Laplace transform formulas for integrals and derivatives.

Table of Contents

## Laplace transform of t formula

We know that the Laplace transform of integral powers of t is given by the formula:

L{t^{n}} = n!/s^{n+1}.

Putting n=1, we get the Laplace transform formula of t.

L{t} = 1!/s^{1+1} = 1/s^{2}.

## What is the Laplace transform of t?

**Answer:** The Laplace transform of t is 1/s^{2}.

*Proof:*

The definition of the Laplace transform of f(t) is given as follows:

L{f(t)} = $\int_0^\infty$ f(t) e^{-st} dt.

**Step 1:** Put f(t) = t in the above definition of Laplace transform.

So the Laplace transform of t by definition is equal to

L{t} = $\int_0^\infty$ te^{-st} dt

**Step 2:** We use the integrating by parts formula, we obtain that

L{t} = $[$ t ∫e^{-st }dt – $\int \{\frac{d}{dt}(t) \int e^{-st} dt\} dt$ $]_0^\infty$

= $\left[ t \cdot \dfrac{e^{-st}}{-s} – \int \dfrac{e^{-st}}{-s} dt \right]_0^\infty$

= $\left[ \dfrac{te^{-st}}{-s} – \dfrac{e^{-st}}{s^2} \right]_0^\infty$

= $\lim\limits_{t \to \infty}\left[ \dfrac{te^{-st}}{-s} – \dfrac{e^{-st}}{s^2} \right]$ – (0 – 1/s^{2})

= 0-0 + 1/s^{2}

= 1/s^{2}.

Thus, the Laplace transform of t is 1/s^{2} and this is obtained by the definition of the Laplace transform.

**Also Read:**

Laplace transform of sin at: | a/(s^{2}+a^{2}) |

Laplace transform of cos at: | s/(s^{2}+a^{2}) |

Laplace transform of e^{-t}: | 1/(s+1) |

Laplace transform of 1: | 1/s |

## Find The Laplace Transform of t

We know that L{t} =1/s^{2}.

**Method 1:** Using the multiplication by t^{n} Laplace transform formula, we will now find the Laplace transform of t. This formula says that

L{t^{n}f(t)} = (-1)^{n} $\dfrac{d^n}{ds^n}$[F(s)], where F(s)=L{f(t)}.

Put n=1 and f(t)=1. Then F(s)=L{f(t)}=1/s.

So L{t} = (-1)^{1} $\dfrac{d}{ds}$(1/s)

= – (-1/s^{2})

= 1/s^{2}.

**Method 2:** Using the Laplace transform formula for integrals, we will now find the Laplace transform of t. The formula says that

$L\{\int_0^t f(u) du \}=\frac{1}{s}$F(s), where F(s)=L{f(t)}.

Put f(u)=1. Then F(s)=L{f(u)}=1/s.

Therefore, $L\{\int_0^t du \}$ = 1/s × 1/s

⇒ L{t} = 1/s^{2}.

## FAQs

**Q1: What is the Laplace transform of 0?**

Answer: The Laplace transform of 0 is 0.

**Q2: What is the Laplace transform of t?**

Answer: 1/s^{2} is the Laplace transform of t.

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.