Thank you for your time.


My basic question is whether the following change of variables allowed 


$$\int\_0^a \int\_0^b f(a-b)g(b-c)h(c)\,dc\,db = \int\_0^a \int\_0^b f(c)g(b-c)h(a-b)\,dc\,db$$
I fail to find a substitutiuon that does this, but as far as I see am I having in both case a combination of positive numbers $a-b$, $b-c$ and $c$ that sums up to $a$ where all the three numbers range between $0$ and $a$.


So it was tempting for me to assume it is correct, but I fail to find a rigorous argument for this to be true.



Change of variables for double integral
Yes, the equality is true (if e.g. $f,g,h,$ are locally integrable and $a\ge0$). Indeed, the left-hand side is $f\*(g\*h)$ and the right-hand side is $(f\*g)\*h$, where $(f\*g)(x):=\int\_0^x f(x-y)g(y)dy$. 


You can also rewrite your equality as 
$$\int\_{-\infty}^\infty \int\_{-\infty}^\infty f\_1(a-b)g\_1(b-c)h\_1(c)\,dc\,db=((f\_1\*g\_1)\*h\_1)(a) $$
$$=((f\_1\*g\_1)\*h\_1))(a)= \int\_{-\infty}^\infty \int\_{-\infty}^\infty f\_1(c)g\_1(b-c)h\_1(a-b)\,dc\,db,$$
where $f\_1(x):=f(x)$ for $x\ge0$ and $f\_1(x):=0$ for $x<0$, $g\_1$ and $h\_1$ are defined similarly, and $(f\_1\*g\_1)(x):=\int\_{-\infty}^\infty f\_1(x-y)g\_1(y)dy$. 


