# Proof of the convolution theorem for laplace transform

1. Dec 28, 2007

### O.J.

My textbook provides a proof but there's one thing about the proof i do not understand

it starts assuming L{f(t)} = the laplace integral with the f(t) changed to f(a)

same goes with L{g(t)} as it changes it to g(b)

i understand the big picture>>starting from a product of 2 L transforms and working ur way back to an expression in terms of the two functions transformed, but how can u change the variable for f and g when we started off stating f of T and g of T.???

2. Dec 28, 2007

### O.J.

Am I not presenting my questions clearly?

3. Dec 28, 2007

### Rainbow Child

Let

$$F(s)=\mathcal{L}\{f(t)\}=\int_0^{+\infty}e^{-s\,t}\,d\,t\,f(t) \quad \text{and} \quad G(s)=\mathcal{L}\{g(t)\}=\int_0^{+\infty}e^{-s\,t}\,g(t)\,d\,t$$

be the two Laplace transormations and let's denote

$$(f\star g)(t)=\int_0^{+\infty}f(\tau)\,g(t-\tau)\,d\,\tau$$

the convolution of $$f,\,g$$.

Thus the Laplace transform for the convolution would be

$$\mathcal{L}\{(f\star g)(t)\}=\int_0^{+\infty}e^{-s\,t}\left( \int_0^{+\infty}f(\tau)\,g(t-\tau)\,d\,\tau \right)\,d\,t =\int_0^{+\infty}\int_0^{+\infty}e^{-s\,t}f(\tau)\,g(t-\tau)\,d\,\tau \,d\,t \quad (1)$$

The above double integral is to be evaluated in the domain $$\mathcal{D}=\{(t,\tau): t \in (0,+\infty),\tau \in (0,+\infty)\}$$.

Performing the change of variables
$$(t,\tau)\rightarrow (u,v): t=u+v, \, \tau=v$$ with $$u \in (0,+\infty),v \in (0,+\infty)$$ we have $$d\,\tau\,d\,t=d\,u\,d\,v$$ since the Jacobian equals to 1. Thus (1) becomes

$$\mathcal{L}\{(f\star g)(t)\}=\int_0^{+\infty}\int_0^{+\infty}e^{-s\,(u+v)}f(u)\,g(v)\,d\,u\,d\,v= \int_0^{+\infty}e^{-s\,u}f(u)\,d\,u \cdot \int_0^{+\infty}e^{-s\,v}\,g(v)\,d\,v$$

yielding to

$$\mathcal{L}\{(f\star g)(t)\}=\mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\}$$

Nothing but double integrals!

4. May 10, 2010

### xenosicotte

Isn't the laplace convolution between 0 and t rather than 0 and infinity?

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