Convolution of convolution Let us write a convolution 
$\int_{0}^{t} A(t-\tau) \mathrm{d}x(\tau)$   as
$A \star \mathrm{d}x$
I would like to write down the expression for the double convolution
$A \star  \mathrm{d}x \star \mathrm{d}x $
Following the definition I obtain
$ \int_{0}^{t} \int_{0} ^{t-\tau} A(t-\tau-s) \mathrm{d}x(s) \mathrm{d}x(\tau)$
Can this be given a more compact form, especially in reference to the upper limit of integration in the inner integral? 
I would like to perform the change of variable $t-\tau = w$ but unsure as tyo how to proceed, any hint would be the most appreciated, thanks
 A: When the function is differentiable and you can write the operation as a regular convolution, you can use the fact that $\dot x\ast \dot x $ makes sense, differently from $dx\star dx$, which is not defined.
In this case you would have $A\star dx\star dx = A\ast \dot x \ast \dot x = A\ast (\dot x \ast \dot x)$: $$\int_0^t A(t-u) \int_0^u \dot x(u-s)\dot x(s)\,ds\,du.$$
If you want to rewrite it as before: $$\int_0^t A(t-u) \int_0^u \dot x(u-s)\,dx(s)\,du.$$
Otherwise, you can change the limits, but at the cost of defining another function $x^t(w)=x(t-w)$ when changing $t-\tau=w$, in this case $dx^t(w)=-dx(t-w)$:
$$\int_0^t \int_0^{t-\tau} A(t-\tau-s)\,dx(s)\,dx(\tau)=\int_0^t \int_0^{w} A(w-s)\,dx(s)\,dx^t(w)$$
A: The result above can be obtained directly by a straight forward calculation. Suppose you have three Schwartz function $f,g,h$ (in particular: continuous, smooth and decaying fast enough at infinity). Then
$$
m(t) = (f*g)(t) = \int_0^t f(t-s_1) g(s_1) ds_1
$$
and 
$$
(f*g*h)(t) = (f*g)*h = (m*h)(t) = \int_0^t m(t-s_2) h(s_2) ds_2
$$
putting the explicit expression for $m(t-s_2)$ we obtain
$$
= \int_0^t ds_2 \int_0^{t-s_2} f(t-s_2-s_1) g(s_1) h(s_2) ds_2
$$
so
$$
(f*g*h)(t) = \int_0^t \int_0^{t-s_2} f(t-(s_1+s_2))g(s_1)h(s_2)ds_1 ds_2
$$
and without further assumptions this as far as you can go.
See also: http://mathworld.wolfram.com/Convolution.html
