Integration - change of integrands within the integral

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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?

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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.

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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