# Calculating Inverse Laplace Using Convolutions

1. Feb 5, 2008

### ColdFusion85

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

We are to use the convolution theorem to compute the inverse Laplace transform of the function

$$L=\frac{1}{s^2 + 16}e^{-2s}$$

2. Relevant equations
Using a table, I find that $$L^{-1}[\frac{1}{s^2 + 16}] = \frac{1}{4}sin(4t)$$
and
$$L^{-1}[e^{-2s}] = \delta(t-2)$$

3. The attempt at a solution
Using the convolution theorem,

$$L^{-1}[\frac{1}{s^2 + 16}e^{-2s}] = \int_{0}^{t} \frac{1}{4}sin(4\tau)\delta(\tau-2)d\tau$$

My question is, how do I evaluate that integral? I know that when you have the delta function, e.g. $$\delta(t-2)$$ times a function f(x), and the limits are negative infinity to infinity, it is just f(2). But what if the limits are 0 to t?

2. Feb 6, 2008

### ColdFusion85

Can anyone help?

3. Feb 6, 2008

### HallsofIvy

Staff Emeritus
$$\int f(x)\delta(x-a) dx= f(a)$$
as long as the interval of integration includes x= a and is 0 if it doesn't. It doesn't have to be from $-\infty$ to $\infty$

In particular, with a variable upper limit (and assuming a> x0),
$$\int_{x_0}^t f(x)\delta(x-a) dx$$
is the function that is 0 if t< a, f(a) if t$\ge$ a.