- #1

zenterix

- 524

- 74

- Homework Statement
- Show that if ##h(t)=f(t)*g(t)## then ##H(s)=F(s)G(s)##.

- Relevant Equations
- where capital letters represent the Laplace transform of the lowercase variables.

$$h(t)=f(t)*g(t)=\int_0^t f(\tau)g(t-\tau)d\tau=\int_0^t g(\tau)f(t-\tau)d\tau\tag{1}$$

The Laplace transform is

$$H(s)=\int_0^\infty h(t)e^{-st}dt=\int_0^\infty\left ( \int_0^t g(\tau)f(t-\tau)d\tau\right )e^{-st}dt\tag{2}$$

The Laplace transforms of $f$ and $g$ are

$$F(s)=\int_0^\infty f(x)e^{-sx}dx\tag{3}$$

$$G(s)=\int_0^\infty g(y)e^{-sy}dy\tag{4}$$

Now let's check what ##F(s)G(s)## is.

$$F(s)G(s)=\int_0^\infty\int_0^\infty f(x)e^{-sx}g(y)e^{-sy}dxdy\tag{5}$$

We now make a change of variables.

$$x=X(t,\tau)=t-\tau$$

$$y=Y(t,\tau)=\tau$$

The Jacobian of this mapping is

$$J(\tau,t)=\begin{vmatrix} -1&1\\1&0\end{vmatrix}=-1$$

and so

$$|J(\tau,t)|=1$$

The double integral (5) becomes

$$\int\int g(\tau)f(t-\tau)e^{-st}d\tau dt\tag{6}$$

The Laplace transform is

$$H(s)=\int_0^\infty h(t)e^{-st}dt=\int_0^\infty\left ( \int_0^t g(\tau)f(t-\tau)d\tau\right )e^{-st}dt\tag{2}$$

The Laplace transforms of $f$ and $g$ are

$$F(s)=\int_0^\infty f(x)e^{-sx}dx\tag{3}$$

$$G(s)=\int_0^\infty g(y)e^{-sy}dy\tag{4}$$

Now let's check what ##F(s)G(s)## is.

$$F(s)G(s)=\int_0^\infty\int_0^\infty f(x)e^{-sx}g(y)e^{-sy}dxdy\tag{5}$$

We now make a change of variables.

$$x=X(t,\tau)=t-\tau$$

$$y=Y(t,\tau)=\tau$$

The Jacobian of this mapping is

$$J(\tau,t)=\begin{vmatrix} -1&1\\1&0\end{vmatrix}=-1$$

and so

$$|J(\tau,t)|=1$$

The double integral (5) becomes

$$\int\int g(\tau)f(t-\tau)e^{-st}d\tau dt\tag{6}$$

**My question is:**how do I find the limits of integration in (6)? What is a good strategy for reasoning about such a task?