# Questions About Textbook Proof of Convolution Theorem

I am reading a textbook on Laplace transforms.

In the proof of the convolution theorem, the author starts by writing the following:

$$\mathcal\{ f(t) * g(t) \} = \int_0^\infty e^{-st}\int_0^t f(\tau)g(t - \tau) \ d\tau dt \ \ \ \text{Using the definition of the Laplace transform}$$

The definition of the Laplace transform is

$$f(s) = \int_0^\infty F(t) e^{-st} \ dt$$

My questions are as follows:

1. Where does the $f(\tau)g(t - \tau)$ come from?
2. Where does the double integral and the limits $0$ and $t$ for the second integral come from?

I would greatly appreciate it if people could please take the time to clarify this.

• Isn't it for calculating Laplace transform of convolution of two functions $f(t)$ and $g(t)$, i.e., $\mathcal{L}\{f(t) * g(t)\}$? – Alla Tarighati Aug 30 '18 at 11:25
• @AllaTarighati Yes, but that part is obvious. I'm asking about where it came from in the context of the of convolution theorem proof. – The Pointer Aug 30 '18 at 11:29

As you said, we are looking for Laplace transform of a convolution. Let us at the moment assume $$h(t)=f(t)* g(t).$$ Then by definition we have $$h(t)=\int_0^t f(\tau)g(t-\tau)d\tau.$$

Now let us consider Laplace transform of $h(t)$ as

$$\mathcal{L}\{h(t)\}=\int_0^\infty e^{-st}h(t)dt$$

Now we plug $h(t)$ into equation above to get:

$$\mathcal{L}\{h(t)\}=\int_{t=0}^{t=\infty}e^{-st} \int_{\tau=0}^{\tau=t} f(\tau)g(t-\tau)d\tau dt .$$