The question is to prove that $((f *g)*h) (x) = (f*(g*h)) (x)\\$
So far, I have the following:
\begin{align*}
((f \ast g) \ast h) (x) &= \int^x_0 (f \ast g)(t) h (x-t) dt\\
&=\int^x_0 \left[\int^t_0 f(h)g(t-h) dh\right] h(x-t) dt\\
&= \int^t_0 \int^x_0 g(t-h) h(x-t) f(h) dt \quad dh\\
&= \int^t_0 \int^{\alpha = x – h}_{\alpha = -h} g(\alpha) h(x – h – \alpha) f(h) d\alpha \quad dh, \quad \text{where } \alpha = t – h \Rightarrow d\alpha = dt\\
\end{align*}
I'm stuck here because somehow, I need:
$$\int^{\alpha = x – h}_{\alpha = 0} \text{instead of} \int^{\alpha = x – h}_{\alpha = -h}$$
I'm aware that there are different definitions of convolution that go from $(-\infty, +\infty)$ that might simply this, but I'm wondering how to do for the above bounds.
Is there something I did incorrectly? It's been a while since I've learned Fubini's theorem, so I may be missing up the bounds somehow. Thank you!
Best Answer
We obtain \begin{align*} \color{blue}{((f\ast g)\ast h)(x)}&=\int_0^x(f\ast g)(t)h(x-t)\,dt\\ &=\int_{t=0}^x\left(\int_{u=0}^tf(u)g(t-u)\,du\right)h(x-t)\,dt\tag{1}\\ &=\int\!\!\!\int_{0\leq u\leq t\leq x}f(u)g(t-u)h(x-t)\,du\,dt\tag{2}\\ &=\int_{u=0}^x\int_{t=u}^xf(u)g(t-u)h(x-t)\,dt\,du\tag{3}\\ &=\int_{u=0}^x\int_{t=0}^{x-u}f(u)g(t)h(x-t-u)\,dt\,du\tag{4}\\ &=\int_{u=0}^xf(u)\left(\int_{t=0}^{x-u}g(t)h(x-u-t)\,dt\right)\,du\\ &=\int_{u=0}^xf(u)(g\ast h)(x-u)\,du\\ &\,\,\color{blue}{=(f\ast (g\ast h))(x)} \end{align*}
Comment:
In (1) we introduce a new integration variable $u$. It is crucial to use different names for different objects to not mix up function names ($h$) with variable names.
In (2) we write the region of integration conveniently as preparation for changing the order of integration in the next step.
In (3) we change the order of integration.
In (4) we substitute $t\to t+u$ to start the region of integration with $t=0$. In this substitution $u$ is to be seen as constant, so that we have again $dt$.