Determine limits of integration in double integral change of variables

[zenterix]

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

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

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

 

[zenterix]

The region of integration of (5) is the first quadrant in the $xy$-plane.

My first thought is that since $y=\tau$ then $\tau$ goes from $0$ to $\infty$.

However, this doesn't work for $x$.

For a given $t$, since $x>0$ then $\tau$ goes from $0$ to $t$.

It seems that the mapping is

$$\int_0^\infty\int_0^t g(\tau)f(t-\tau)e^{-st}d\tau dt$$

 

[Orodruin]

The integration region for x and y is $x > 0$ and $y > 0$. Replace by the expressions with $t$ and $\tau$ and you get
$$
t-\tau > 0, \quad \tau > 0.
$$
Move $\tau$ to the rhs of the first: $t > \tau$. The integration region is therefore $t > \tau > 0$.