The Laplace transform of t^3 is equal to 6/s^4. In this article, we will learn how to find the Laplace transform of t cube.

Table of Contents

## Laplace Transform of t Cube Formula

The Laplace transform formula of t cube, that is, the formula of L{t^{3}} is given by

L{t^{3}} = 6/s^{4}.

## What is the Laplace Transform of t^{3}?

**Answer:** The Laplace transform of t^{3} is 6/s^{4}.

*Proof:*

To find the Laplace transform of t cube by the definition of Laplace transform, let us recall the definition. The Laplace transform of f(t) is defined by

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

**Step 1:** Put f(t) = t^{3}.

Therefore,

L{t^{3}} = $\int_0^\infty$ t^{3} e^{-st} dt.** …(I)**

**Step 2:** We use the theory of the Gamma function Γ(x) = $\int_0^\infty$ t^{x-1} e^{-t} dx. Assume that

z=st

∴ dz=s dt ⇒ dt = dz/s. Also, t=z/s.

t | z |

0 | 0 |

∞ | ∞ |

**Step 3:** Therefore, from **(I)** we get that

L{t^{3}} = $\int_0^\infty \Big(\dfrac{z}{s} \Big)^3 e^{-z} \dfrac{dz}{s}$

= (1/s^{3+1}) $\int_0^\infty z^{3+1-1} e^{-z} dz$

= (1/s^{4}) $\Gamma(3+1)$, by the definition of the Gamma function.

= (1/s^{4}) × 3! as we know that Γ(n+1) = n!

= 3!/s^{4}

= 6/s^{4}.

Therefore, the Laplace transform of t^3 is equal to 6/s^{4} and this is proved by the definition of Laplace transforms.

**Read Also:**

Concept of Laplace Transform: Definition, Table, Formulas, Properties & Examples

Laplace Transform of e^{at}: The Laplace transform of e^{at} is 1/(s-a).

Laplace transform of constant: The Laplace transform of c is c/s.

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

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

Inverse Laplace transform of constant: The inverse Laplace transform of c is cδ(t), where δ(t) is the Dirac delta function.

## FAQs

**Q1: What is the Laplace transform of t cube?**

Answer: The Laplace transform of t cube is equal to 6/s^{4}.