Question

Prove that if f is continuous, then:

∫ (x>0) f(u)(x-u).du= ∫ (x>0) ( ∫ (u>0) f(t) dt) du.

Answer 1

let

$g(x)=\displaystyle{\int ^x_0}f(u)(x-u)du-\displaystyle{\int ^x_0}(\displaystyle{\int ^u_0}f(t)dt)du$

By the Fundamental Theorem of Calculus:

$\frac{d}{dx}(\displaystyle{\int ^x_0}uf(u)du)=xf(x)$

$\frac{d}{dx}(\displaystyle{\int ^x_0}\displaystyle{\int ^u_0}f(t)dtdu)=\displaystyle{\int ^x_0}f(t)dt$

Notice here we needed both that f and the indefinite integral of f are continuous functions. Now,

$g'(x)=\displaystyle{\int ^x_0}f(t)dt+xf(x)-xf(x)-\displaystyle{\int ^x_0}f(u)du=0$

So by the Mean Value Theorem, g is a constant. Further, observe that 

$g(0)=\displaystyle{\int ^0_0}f(u)(x-u)du-\displaystyle{\int ^0_0}(\displaystyle{\int ^u_0}f(t)dt)du=0-0=0$

Thus, g(x) = 0 for all x. It follows that

$\displaystyle{\int ^x_0}f(u)(x-u)du=\displaystyle{\int ^x_0}(\displaystyle{\int ^u_0}f(t)dt)du$