Motion blur integral

Question

I was going through a biophysics paper (Berglund, Phys Rev E Stat Nonlin Soft Matter Phys. 2010 Jul) and derived the main pieces of interest but ran into this difference of two integrals:$$R=\frac{1}{\Delta t}\left[\int \limits _0^{\Delta t}dt \int \limits _0^{\Delta t}dt's(t)s(t')\min (t,t')-\int \limits _0^{\Delta t}ts(t)\,dt\right],$$where $\Delta t$ is positive and $s(t)$ is a non-negative function that integrates to one over $[0,\Delta t]$ (i.e. $\displaystyle \int \limits _0^{\Delta t}s(t)\,dt=1$).

The question is, this is supposed to be equal to$$R=\frac{1}{\Delta t}\int \limits _0^{\Delta t}S(t)[1-S(t)]\,dt,$$where $\displaystyle S(t)=\int \limits _0^ts(t')\,dt'$.

Why is this? Why is

$$ \frac{1}{\Delta t} \left[ \int \limits _0^{\Delta t}dt\int \limits _0^{\Delta t}dt's(t)s(t')\min (t,t')-\int \limits _0^{\Delta t}ts(t)\,dt \right]=\frac{1}{\Delta t} \int \limits _0^{\Delta t}S(t)[1-S(t)]\,dt$$?

Any hints appreciated. Even though I studied math in college I don't do integrals very often anymore so any leads are helpful.

Answer 1

With slight relabelling for clarity, plus correcting an overall sign error:

Theorem Let $S(t):=\int_0^ts\left(u\right)du$ with $s\ge0,\,S(\Delta)=1$. Then$$\int_0^\Delta us(u)du-\int_0^\Delta du\int_0^\Delta dvs(u)s(v)\min(u,\,v)=\int_0^\Delta S(t)[1-S(t)]dt.$$Proof Since $S(\Delta)=1$, the left-hand side with a factor of $S(\Delta)$ inserted into one term is$$\int_0^\Delta\int_0^\Delta us(u)s(v)dudv-\int_0^\Delta du\int_0^\Delta dvs(u)s(v)\min(u,\,v)=\int_{0\le v\le u\le\Delta}dudvs(u)s(v)(u-v),$$while the right-hand side with $1-S$ rewritten as $S(\Delta)-S$ is$$\int_0^\Delta dt\int_0^ts(v)dv\int_t^\Delta s(u)du=\int_{0\le v\le u\le\Delta}dudvs(u)s(v)\int_v^udt=\int_{0\le v\le u\le\Delta}dudvs(u)s(v)(u-v)$$by Fubini's theorem, since the triple integral's range is $0\le v\le t\le u\le\Delta$.