# Substituting integral into an integral

Let $\displaystyle \ \ r(t) := \int_0^t \alpha(u,t)du\,\,\,\,$ be a deterministic, finite process.

What is $\displaystyle \ \ \int_0^t r(s)ds\,\,\,$, and could you please give an explanation for your answer?

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To integrate $r(s)$, we should write down $r(s)$: $$r(s) := \int_0^s \alpha(u,s)\,du$$ Now put this inside $\int_0^t r(s)\,ds$:
$$\int_0^t \int_0^s \alpha(u,s)\,du\,ds$$ The domain of integration is described by inequalities $0\le u\le s\le t$. So, if we begin integration with $s$, the inner limits are $u$ to $t$: $$\int_0^t \int_u^t \alpha(u,s)\,ds\,du$$ which matches your slides.
$\int_0^t r(s)ds=\int_0^t \int_0^s \alpha(u,t)duds$. You can use integration by parts to rewrite it as: $$\int_0^t r(s)ds=\int_0^t \int_0^s \alpha(u,t)duds=[s\int_0^s \alpha(u,t)du]^{s=t}_{s=0}-\int_0^t s\alpha(u,s)duds$$
My lecture slides say that the answer is $\displaystyle \ \ \int_0^t \int_u^t \alpha(u,s)dsdu$ –  Jase Nov 25 '12 at 9:15