Integration by parts from a physics textbook

Question

I am trying to understand a formula derivation from a physics textbook, and could get the result up until the last step. The author mentions you just need to integrate by parts, however, I'm not able to get his results.

More specifically, I want to show that

$\int_l^d\int_r^dsu(s)ds dr = \int_l^dr(r-l)u(r)dr $,

and $u(s)$ is a potential such that

$u(s)=\int_s^df(s')ds' \ , \ \ \displaystyle \frac{du}{ds}=-f(s) \ , \ \ u(s\ge d)=0 \ , \ \ f(s\ge d)=0 $.

Can anyone help me in understanding that?

Answer 1

Another point-of-view is to see this as a change in integration order à la Fubini. Then $$ \iint_{l<r<s<d}su(s)\,ds\,dr=\iint_{l<r<s<d}su(s)\,dr\,ds =\int_l^d\int_l^s su(s)dr\,ds=\int_l^d(s-l)su(s)\,ds $$ In the second form the integrand is constant for the inner integal, so resolves to the difference of the endpoints. There is no difference if the fineal integration variable is $r$ or $s$.

Answer 2

Let $$ U(r) = \int_r^d su(s)\,ds. $$ Then \begin{align*} \int_l^d \int_r^d su(s)\,ds\,dr &= \int_l^d U(r)\,dr\\ &= dU(d) - lU(l) - \int_l^d rU'(r)\,dr\\ &= -l\int_l^d su(s)\,ds + \int_l^d r^2u(r)\,dl\\ &= \int_l^d r(r-l)u(r)\,dr \end{align*}