# Laplace transform of convolution integral

If $f(t)$ an $g(t)$ are piecewise continuous functions on $[\ 0, \infty)$ then the convolution integral of $f(t)$ and $g(t)$ is,

$$(f*g)(t) = \int_{0}^{t}f(t-\tau)g(\tau) \text{d} \tau.$$

The text then gives a 'fact':

$\mathcal{L} \{f*g \} =F(s)G(s),$ where $\mathcal{L} \{ f(t) \} = F(s)$.

I tried to show this, but I'm not sure if it's correct. First

$$\mathcal{L} \{f*g \} = \int_{0}^{\infty} e^{-st} \left [\ \int_{0}^{t}f(t-\tau)g(\tau) \text{d} \tau \right]\ \text{d}t.$$

I'm weary of swapping integral signs because don't exactly know when it is valid, but

$$\mathcal{L} \{f*g \} = \int_{0}^{\infty} g(\tau) \left [\ \int_{\tau}^{\infty}f(t-\tau) e^{-st} \text{d} t \right]\ \text{d} \tau.$$ Using the substitution $u = t - \tau,$

$$\mathcal{L} \{f*g \} = \int_{0}^{\infty} g(\tau) e^{-s \tau} \left [\ \int_{0}^{\infty}f(u) e^{-su} \text{d} u \right]\ \text{d} \tau = \left [\ \int_{0}^{\infty}f(u) e^{-su} \text{d} u \right]\ \left [\ \int_{0}^{\infty} g(\tau) e^{-s \tau} \text{d} \tau \right]$$

Does this make any sense?

Edit: $e^{-st}$ replaced with $e^{-su}$ and corrections to errors

• What if I made both limits $\infty$ at the beginning? Jun 25, 2015 at 15:19
• integrating $\tau$ from $0$ to $t$ and then integating $t$ from $0$ to $\infty$ is the same as integrating $t$ from $\tau$ to $\infty$ and then $\tau$ from $0$ to $\infty$ Jun 25, 2015 at 15:22
• Ok I found the limits and they match the ones you have just given. So in the substitution $u = t - \tau$, the lower limit becomes $0$. Thanks. I feel bad because I should know this. Jun 25, 2015 at 15:25
• But now you do know it and that's what matters. Jun 25, 2015 at 15:33

As you know we have $$\mathcal{L} \{f*g \} = \int_{0}^{\infty} e^{-st} \left [\ \int_{0}^{t}f(t-\tau)g(\tau) \text{d} \tau \right]\ \text{d}t.$$ Note that the exponential has the following property $$e^{-st} = e^{-s(t - \tau )} e^{-s\tau}$$ We may exploit this by sneaking the exponential into the $\tau$ integral with Fubini's to obtain $$\int_0^\infty \left [\int_0^t e^{-s ( t- \tau)} f( t - \tau) e^{-s \tau} g ( \tau) d \tau \right ] dt$$ Now let's change variables, say $u = t-\tau$ and $v = \tau$. What happens to the integral under this change of variables?
• Thanks, but did you read my attempt? The problem was my limits were incorrect. With your substitution, $g(\tau)$ becomes $g(u - t)$ and the integral upper and lower limit become $t$ and $0$ respectively, but I'm not sure how that would help. Jun 25, 2015 at 15:35
• Your change of integrals was incorrect, I thought I'd show you how to justify the swap with the above. In addition to that, your change of variables wasn't quite right (since you forgot the limits). I thought I'd rewrite the change using $u$ and $v$ since that's what you've most likely seen. You're glancing over the Jacobian of the transformation and what not... Hence why I rewrote it the way I did. Also, your comment applies to your direct change of variables as well if you notice... you may want to have another look.
• I'm still a bit confused. I thought the problem was that when I swapped the integrals I didn't have the right limits. With the right limits and the substitution it looks like it comes out correctly. The reason I kept the lower limit as $0$ in the substitution originally is because $f(t) = 0$ for $t < 0$, so I thought it made no difference. Jun 25, 2015 at 16:15
• It's much more than that, consider when you wrote $$\int_0^\infty f(u) e^{-st} du = e^{-st} \int_0^\infty f(u) du \neq \mathcal{L}\{f\} [s]$$, it's still a function of $t$! That's not at all what you wanted to show. Have a look at what I've written, and perform the suggested change of variables according to the definition.
• I'm not sure when I wrote that. I thought that I had $$\int_{0}^{\infty} f(u)e^{-s(t + \tau)} du = e^{-s \tau } \int_{0}^{\infty} f(u)e^{-st} du$$ Jun 25, 2015 at 16:29