What is the Inverse Laplace transform of 1/s^3?

The inverse Laplace transform of 1/s^3 is t2/2. Here we learn how to find the inverse Laplace of 1/s3.

The inverse Laplace transform of 1/s3 is denoted by L-1(1/s3) and its formula is given by

L-1(1/s3) = t2/2.

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Inverse Laplace of 1/s3

Answer: The inverse Laplace of 1/s^3 is equal to t2/2.


We know that the inverse Laplace Transform of 1/sn is given by the formula:

L-1$\left(\dfrac{1}{s^{n+1}} \right)$ = $\dfrac{t^n}{n!}$

In order to get the inverse Laplace of 1/s3 we need to put n=2 in the above formula, so that we obtain

L-1$\left(\dfrac{1}{s^{2+1}} \right)$ = $\dfrac{t^2}{2!}$

This implies that

L-1$\left(\dfrac{1}{s^3} \right)$ = $\dfrac{t^2}{2}$.

So the inverse Laplace transform of 1 by s^3 is equal to t2/2.

Related Topics:

Table of Inverse Laplace Transformations

Inverse Laplace transform of 1

Inverse Laplace transform of 1/s

Inverse Laplace transform of 1/s2


Q1: What is the inverse Laplace of 1/s3?

Answer: The inverse Laplace transform of 1/s3 is t2/2, that is, L-1(1/s3) = t2/2.

Q2: Find L-1(1/s3).

Answer: L-1(1/s3) = t2/2.

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