Integration - change of integrands within the integral I came across this equation in a paper:
$$\int_\pi^{p_*}\left(\int_p^{p_*}h(u)\;\mathrm du\right)S'(p)\;\mathrm dp=\int_\pi^{p_*}[S(u)-S(\pi)]h(u)\;\mathrm du, 0\leq\pi\leq p_*$$
I am not sure if this is correct, as this is not the usual change of integration order for a double integral. Can someone please explain why this makes sense?
 A: It is the usual change of order of integration. Draw a picture. When you change the order, you note that $\pi\le u\le p^*$ and, for fixed $u$, we have $\pi\le p\le u$. Then they apply the Fundamental Theorem of Calculus.
A: At first this also appeared to me as a case of fubini's theorem. When that didn't make sense, I tried substituting inner integral with H(p*)- H(p) but wasn't able to solve until Jean-Claude Arbaut commented above suggesting using integration by parts. Using integration by parts and treating each part separately leads to straightforward result.
$\int_\pi^{p_*}\left(\int_p^{p_*}h(u)\;\mathrm du\right)S'(p)\;\mathrm dp $
$\\=\int_\pi^{p_*}[H(p*)-H(p)]S'(p)\;\mathrm dp$   $\\=\int_\pi^{p_*}H(p*)S'(p)\;\mathrm dp - \int_\pi^{p_*}H(p)S'(p)\;\mathrm dp $         $\\ =H(p*)*(S(p)-S(\pi))\;\mathrm  - \int_\pi^{p_*}H(p)S'(p)\;\mathrm dp $
$\\=H(p*)*(S(p)-S(\pi))\;\mathrm  - \int_\pi^{p_*}d(H(p)S(p))\;\mathrm  - \int_\pi^{p_*}h(p)S(p)\;\mathrm dp $
$=H(p*)*(S(p)-S(\pi))\;\mathrm  - [H(p*)S(p*)- H(\pi)S(\pi)]\mathrm  - \int_\pi^{p_*}h(p)S(p)\;\mathrm dp $
$=S(\pi)*(H(p*)-H(\pi))\;\mathrm  - \int_\pi^{p_*}h(p)S(p)\;\mathrm dp $
  =RHS  
