Having Trouble Using Fubini's Theorem

Question

Calculate

$\int_{a}^{b}\int_{a}^{z}\int_{a}^{y}f(x)f(y)f(z)dxdydz$. I said let $F(x) = \int_{a}^{x}f(t)dt$ to simplify things a little bit, and I got to here:

$\int_{a}^{b}\int_{a}^{z}f(z)F(y)f(y)dydz$

I am stuck from there. Clearly $f(z)$ can just be treated as a constant, but when I use integration by parts, it just recursively gives me the same integral. I have been stuck for a long time, and I wonder if I am not using the right method. Change of coordinates does not seem to be the right approach either.

Any help would be greatly appreciated.

Thank you!

Answer 1

Hint: Given any point in the $[a,b]$ cube, it satisfies one of the following:
$ x \leq y \leq z $
$ x \leq z \leq y $
$ y \leq x \leq z $
$ y \leq z \leq x $
$ z \leq x \leq y $
$ z \leq y \leq x $

Show that the set of points where at least 2 coordinates are equal has measure 0.

Hence $6 \int_S f(x) f(y)f(z) \, dx \, dy \, dz = $ integrating over the cube $= \left( \int f(t) \, dt \right)^3$

Answer 2

I'll assume $f$ is continuous on $[a,b].$ Define $F(u)= \int_a^u f(v)\,dv.$ Our integral is

$$\int_a^b f(z) \int_a^z f(y)\int _a^y f(x)\,dx\,dy\,dz.$$

The $dx$ integral here equals $F(y).$ We now have

$$\int_a^b f(z) \int_a^z f(y)F(y)\,dy\,dz.$$

Since $F(y)^2/2$ is an antiderivative for $f(y)F(y),$ the $dy$ integral here equals $(F(z)^2/2).$ We're left with

$$\int_a^b f(z)(F(z)^2/2)\,dz.$$

More of the same leads to $F(b)^3/6$ for the final answer, which is what we want..