Substituting integral into an integral

Question

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?

Answer 1

$\int_0^t r(s)ds=\int_0^t \int_0^s \alpha(u,t)duds$. You can use integration by parts to rewrite it as: $$\int_0^t r(s)ds=\int_0^t \int_0^s \alpha(u,t)duds=[s\int_0^s \alpha(u,t)du]^{s=t}_{s=0}-\int_0^t s\alpha(u,s)duds$$