# Inverse laplace of 1

1. Nov 12, 2008

### aruna1

1. The problem statement, all variables and given/known data

anyone know how to find inverse laplace of 1?
that is
L-11=?

3. The attempt at a solution

can we use
L-11=s ?
L{s}=s.(1/s)=1

Thanks

2. Nov 12, 2008

### HallsofIvy

Staff Emeritus
Normally, the Laplace tranform of a function of x is written as a function of s. You seem to be confusing the two. The Laplace transform of f(x)= x is
$$\int_0^\infty xe^{-sx}dx= \frac{1}{s^2}$$
by integration by parts, not 1. And you certainly cannot just multiply a Laplace transform you already know by the variable to get another Laplace transform!

The inverse Laplace transform of the constant 1 is the Dirac delta function $\delta(x)$:
$$\int_0^\infty e^{-sx}\delta(x)dx= e^{-s(0)}= 1$$
since, by definition, $\int_S f(x)\delta(x) dx= f(0)$ as long as the region of integration, S, includes 0.

Here's a good table of Laplace and inverse Laplace transforms:
http://www.vibrationdata.com/Laplace.htm

3. Nov 12, 2008

### aruna1

thank you

4. Nov 12, 2008

### mhill

regarding this topic , using the Laplace transform properties would it be valid that

$$\mathcal L^{-1} (s^{k})= D^{k}\delta (t)$$ ??

where k >0 any real number (at least this property seems to work with Fourier transforms)

the case k <0 would involve integration , but the integrals of the Dirac delta (with k integer) are well defined for t >0 (except perhaps at the point t=0 )