What would a 3-way convolution be? Convolution is defined as
$$
(f*g)(t) = \int^\infty_{-\infty} f(\tau) g(t-\tau)d\tau
$$
If I want to do a 3-way convolution like
$$
3way(f,g,h)(t) = \int^\infty_{-\infty}\int^\infty_{-\infty} f(\tau) g(t-\tau)h(t-\upsilon)d\tau
d\upsilon
$$
What is it normally known as?
Is it different from a normal convolution of 3 functions which is defined as
$$
(f*g*h)(t) = \int^\infty_{-\infty}\left(\int^\infty_{-\infty} f(\tau) g(\upsilon-\tau)d\tau\right)
h(t-\upsilon)d\upsilon
$$
 A: The usual way of writing a convolution is conducive to explicit calculations but it obscures the underlying symmetry of the definition. If you write it as
$$(f \ast g)(t) = \int_{\tau_1 + \tau_2 = t} f(\tau_1) g(\tau_2) \, d \mu$$
where $\mu$ refers to Lebesgue measure then it's clear how to generalize to any number of functions: the $n$-fold convolution is
$$(f_1 \ast \dots \ast f_n)(t) = \int_{\tau_1 + \dots + \tau_n = t} f_1(\tau_1) \dots \dots f_n(\tau_n) \, d \mu.$$
This definition preserves the full symmetry of convolution: you can arbitrarily permute the functions $f_i$ and you get the same result. Writing it down as an ordinary repeated integral requires breaking this symmetry, e.g. by writing $\tau_n = t - \tau_1 - \dots - \tau_{n-1}$ and writing it as an integral over $\tau_1, \dots \tau_{n-1}$. For $n = 3$, for example, this gives
$$(f \ast g \ast h)(t) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(\tau_1) g(\tau_2) h(t - \tau_1 - \tau_2) \, d \tau_1 \, d \tau_2.$$
What you describe as "a normal convolution" is the same as this, just with a different parameterization, namely $\tau_1 = \tau, \tau_2 = v - \tau$.
