The Laplace transform of te^{t} is equal to 1/(s-1)^{2}. In this article, we will learn how to find the Laplace transform of the product function te^{t}.

Table of Contents

## What is the Laplace Transform of te^{t}?

**Answer:** The Laplace transform of te^{t} is 1/(s-1)^{2} when s>1.

*Proof:*

We will find the Laplace transform of te^{t }by definition. Recall the definition of the Laplace transform of f(t) which is given below:

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

Put f(t) = te^{t}.

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

= $\lim\limits_{T\to \infty} \int_0^T$ t e^{-(1-s)t} dt

Using integrating by parts formula, the above is

= $\lim\limits_{T\to \infty} \Big( \left[ t \cdot \dfrac{e^{(1-s)t}}{1-s} \right]_{t=0}^T$ $- \int_0^T \dfrac{e^{(1-s)t}}{1-s} \, dt\Big)$

= $\dfrac{1}{1-s} \lim\limits_{T \to \infty} \Big(T e^{(1-s)T}$ $- \int_0^T e^{(1-s)t} \, dt\Big)$

= $\dfrac{1}{s-1} \lim\limits_{T \to \infty} \int_0^T e^{(1-s)t} \, dt$ [as s>1, we have e^{(1-s)T} **→**0 when T**→**∞]

= $\dfrac{1}{s-1} \lim\limits_{T \to \infty} \Big[ \dfrac{e^{(1-s)t}}{1-s} \Big]_{t=0}^T$

= $\dfrac{1}{s-1} \lim\limits_{T \to \infty} \Big[ \dfrac{e^{(1-s)T}}{1-s} – \dfrac{1}{1-s} \Big]$

= $\dfrac{1}{s-1} \Big[ 0 – \dfrac{1}{1-s} \Big]$

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

So the Laplace transform of te^{t} by definition is 1/(s-1)^{2}.

Find the Laplace transform of te^{t}.Summary:L{t e ^{t}} = 1/(s-1)^{2}. |

**Also Read:**

Laplace transform of sin t: | 1/(s^{2}+1) |

Laplace transform of cos t: | s/(s^{2}+1) |

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

Laplace transform of 1: | 1/s |

**Prove that Laplace of te ^{t} is 1/(s-1)^{2}.**

*Proof:*

Recall the multiplication by t^{n} Laplace transform formula, we have that:

$L\{t f(t)\} = – \dfrac{d}{ds}(F(s))$, where L{f(t)}=F(s)

Put f(t) = e^{t}. Note that F(s) = L{e^{t}}=1/(s-1).

Then L{te^{t}} = $- \dfrac{d}{ds}\left( \dfrac{1}{s-1} \right)$

= $- \left(- \dfrac{1}{(s-1)^2} \right)$

= $\dfrac{1}{(s-1)^2}$.

Thus, the Laplace of te^{t} is 1/(s-1)^{2}.

## Laplace of te^{t }by Differentiation

We prove that the Laplace of te^{t} is 1/(s-1)^{2}.

*Proof:*

We know that the Laplace transform of first derivative of f(t) is given as follows:

$L\{f'(t)\}=f(0)+s L\{f(t)\}$ **…(I)**

Let f(t) = te^{t}

Then we have from **(I)** that

L{e^{t}+te^{t}} = 0+sL{te^{t}}

By the linearity of Laplace transform, we get that

L{e^{t}} + L{te^{t}} = s L{te^{t}}

⇒ L{e^{t}} = (s-1) L{te^{t}}

⇒ 1/(s-1) = (s-1) L{te^{t}} as the Laplace of e^{t} is 1/(s-1).

⇒ L{te^{t}} = 1/(s-1)^{2}.

Thus, we prove that the Laplace transform of te^{t} by first derivative formula is equal to 1/(s-1)^{2}.

## FAQs

**Q1: What is Laplace of te**

^{t}?Answer: The Laplace of te^{t} is 1/(s-1)^{2}.

**Q2: What is Laplace of e**

^{t}?Answer: The Laplace of e^{t} is 1/(s-1).

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.