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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.

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

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

Back to your question:

Where does the f(τ)g(t−τ) come from?
- It comes from definition of convolution.
Where does the double integral and the limits 0 and t for the second integral come from?
- see the explanation above.

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  • $\begingroup$ Ahh, thanks for that. I understand now. $\endgroup$ – The Pointer Aug 30 '18 at 11:53
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By the way, it makes more sense to see where the convolution integral comes from if you start from the product of two Laplace transforms, and work out the inverse of that.

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