Calculating Inverse Laplace Using Convolutions

[ColdFusion85]

Homework Statement

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

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

Homework Equations

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

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?

 

[ColdFusion85]

Can anyone help?

 

[HallsofIvy]

$$\int f(x)\delta(x-a) dx= f(a)$$
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.