# Swapping the integral sign for $\int_0^1 (\int_0^s x(u) \, du ) ds$

In trying to prove the iterated formula for the Volterra operator, I get stuck with a basic integration problem. Let $x$ be any $L_2$ function. To what extent can we simplify, and in particular, swap the integral signs in the following expression: $$\int_0^1 \left(\int_0^s x(u) \ du\right) \ ds$$ A change of variable would remove the $s$ from the second integral but that doesn't really help. I would appreciate any suggestion.

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The geometric way to interpret the integral, as hinted at by DonAntonio, is the most elegant way to look at it. A mechanical method hinges on the use of the Iverson bracket $[p]$, which evaluates to $1$ if the condition $p$ is true and $0$ if the condition $p$ is false.

Now, one can easily transform the double integral being considered to the Iversonian format:

$$\iint [0\leq s\leq 1][0\leq u\leq s]x(u)\;\mathrm du\mathrm ds$$

and the double integral is taken to be doubly infinite; this works out since as already mentioned, the Iverson bracket is zero if the condition within it is false, thus suitably restricting the domain.

Now, the Iverson bracket has the neat property that $[p\text{ and }q]=[p][q]$; applying this property to the double integral gives

\begin{align*} \iint [0\leq s\leq 1][0\leq u\leq s]x(u)\;\mathrm du\mathrm ds&=\iint [(0\leq s\leq 1)\text{ and }(0\leq u\leq s)]x(u)\;\mathrm du\mathrm ds\\ &=\iint [0\leq u\leq s\leq 1]x(u)\;\mathrm du\mathrm ds\\ &=\iint [0\leq u\leq s\leq 1]x(u)\;\mathrm ds\mathrm du\\ &=\iint [0\leq u\leq 1][u\leq s\leq 1]x(u)\;\mathrm ds\mathrm du\\ \end{align*}

and we can now convert back to the more conventional notation:

$$\int_0^1 x(u)\int_u^1 \mathrm ds\mathrm du=\int_0^1 (1-u)x(u)\mathrm du$$

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If you're in the $s$-$u$ plane , the innermost integral is from zero to the straight line $\,u=s\,$, so changing the order of integration leaves us almost the same:

$$\int_0^1\left(\int_0^sx(u)du\right)ds=\int_0^1x(u)du\int_u^1ds=\int_0^1x(u)(1-u)du$$

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The integral in $s$ runs from $u$ to $1$, not from 0 to $u$. –  Byron Schmuland Aug 5 '12 at 4:28
Indeed so, thanks –  DonAntonio Aug 5 '12 at 5:32