Multiple Convolution closed form.

Question

Define the convultion of two functions f and g as:

$$(f*g)(x)=\int_{-\infty}^{\infty}f(\tau)g(x-\tau)d\tau$$

Is there any closed form for a multiple convultion: $(f_1*f_2*...*f_p)(x)$?

Answer 1

\begin{eqnarray} (f_1 \star f_2 \star \cdots \star f_n)(x) &=& \int {\rm d} x_{n-1}{\rm d} x_n \cdots {\rm d }x_{1} ~f_n(x-x_{n-1}) f_{n-1}(x_{n-1} - x_{n-2})\cdots f_1(x_1) \\ &=& \int {\rm d}^{n-1}{\bf x} ~f_n(x - x_{n-1}) f_1(x_1)\prod_{k=2}^{n-1}f_k(x_{k} - x_{k-1}) \tag{1} \end{eqnarray}

To prove if use induction

Works for $n=2$

Indeed for $n=2$ the product in the integral is not evaluated and you recover

$$ (\color{blue}{f_1} \star \color{red}{f_2})(x) = \int {\rm d}x_1 \color{red}{f_2(x - x_1)} \color{blue}{f_1(x_1)} \tag{2} $$

(colors are there to help you follow the order of the operations)

Assume that works for $n - 1$

\begin{eqnarray} (f_1 \star f_2 \star \cdots \star f_{n-1})(x) &=& \int {\rm d}^{n-2}{\bf x} ~f_{n-1}(x - x_{n-2}) f_1(x_1)\prod_{k=2}^{n-2}f_k(x_{k} - x_{k-1}) \tag{3} \end{eqnarray}

Let's show that works for $n$

\begin{eqnarray} (f_1 \star f_2 \star \cdots \star f_n)(x) &=& (\color{blue}{(f_1 \star f_2 \star \cdots \star f_{n-1})}\star \color{red}{f_n})(x) \\ &\stackrel{(1)}{=}& \int {\rm d}x_{n-1}~ \color{red}{f_{n}(x - x_{n-1})} \color{blue}{ \left( f_1 \star f_2 \cdots \star f_{n-1} \right)(x_{n-1})} \\ &\stackrel{(2)}{=}& \int {\rm d}x_{n-1}~ \color{red}{f_{n}(x - x_{n-1})} \color{blue}{ \left(\int {\rm d}^{n-2}{\bf x} ~f_{n-1}(x_{n-1} - x_{n-2}) f_1(x_1)\prod_{k=2}^{n-2}f_k(x_{k} - x_{k-1}) \right)} \\ &=& \int {\rm d}x_{n-1}{\rm d}^{n-2}{\bf x}~ \color{red}{f_{n}(x - x_{n-1})} \color{blue}{ f_1(x_1)\prod_{k=2}^{n-1}f_k(x_{k} - x_{k-1}) } \\ &=& \int {\rm d}^{n-1}{\bf x}~ f_{n}(x - x_{n-1}) f_1(x_1)\prod_{k=2}^{n-1}f_k(x_{k} - x_{k-1}) \end{eqnarray}