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.

## 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 – $\int \{\frac{d}{dt}(t) \int e^{-st} dt\} dt$ $]_0^\infty$

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

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

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

= $ 0-0 + \dfrac{1}{s^2}$

= $\dfrac{1}{s^2}$.

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} $\dfrac{d}{ds}[\dfrac{1}{s}]$

= $- \dfrac{-1}{s^2}$

= 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, $L\{\int_0^t du \}=\dfrac{1}{s} \cdot \dfrac{1}{s}$

⇒ 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.