Do nested integrals exist? I have a problem that involves evaluating (or at lest simplifying) the expression
$$\int_{0}^{x}\int_0^{x'}f(y)dy dx'.$$
Playing around with Riemann sums has lead me to believe that this is just $\int_0^x f(x)dx$ which does not make much sense to me.
Does that expression have a simpler form? Is is even meaningful (i.e., can I do that)?
 A: Under mild hypotheses on $f$ this is a case of Fubini's theorem. Consider the triangle
$$T:=\{(t,y)\>|\>0\leq t\leq x, \ 0\leq y\leq t\}$$
in the $(t,y)$-plane (draw a figure!). Then
$$\int_0^x\int_0^t f(y)\>dy\>dt=\int_T f(y)\>{\rm d}(t,y)=\int_0^x\int_y^x f(y)\>dt\>dy=\int_0^x f(y)(x-y)\>dy\ .$$
A: By the Cauchy's repeated integral formula:
$$\int_a^{x_1}\int_a^{x_2}\int_a^{x_3}\dots\int_a^{x_n}f(t)\ dt\ dx_n\dots\ dx_3\ dx_2\ dx_1=\frac1{(n-1)!}\int_a^{x_1}f(t)(x_1-t)^{n-1}\ dt$$
which I find analogous to the Cauchy integral formula from complex analysis for $n$th derivatives (instead of integrals):
$$f^{(n)}(z)=\frac{n!}{2\pi i}\oint_\gamma f(s)(s-z)^{-n-1}\ ds$$
where pretty much everything gets flipped upside down.
In the event of $a=0$ and $n=2$, we get

$$\int_0^{x_1}\int_0^{x_2}f(t)\ dt\ dx_2=\int_0^{x_1}f(t)(x_1-t)\ dt$$

A: If $f$ is continuous, if $F_{1}$ is a primitive of $f$, and if $F_{2}$ is a primitive of $F_{1}$, then
$$
\int_{x'= 0}^{x}\int_{y=0}^{x'}f(y)dydx'= \int_{x'=0}^{x}\left[F_{1}(x') - F_{1}(0)\right]dx' = F_{2}(x) - F_{2}(0) - F_{1}(0)x.
$$
A: Such things exist, but your conclusion is at least a little wrong. For instance, let f(y) = 1. Then
$$\int_0^x \int_0^{x'} 1 dy dx' = \int_0^x x' dx' = x^2/2$$
unless by the primes you intended some sort of differentiation.
