Determine limits of integration in double integral change of variables

  • #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

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

My question is: how do I find the limits of integration in (6)? What is a good strategy for reasoning about such a task?
 
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  • #2
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

1711773980024.png


$$\int_0^\infty\int_0^t g(\tau)f(t-\tau)e^{-st}d\tau dt$$
 
  • #3
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##.
 
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