Evaluate $\ \int_0^3 x^3\,d[\frac{x}{2}]$

Question

I know the result must be $1,$ I also know that:

$ n\leq x/2 < n+1 \Leftrightarrow$ $ 2n \leq x < 2n+2$

For $n=0$ we have that

$ 0 \leq x <2 $

$ 0\leq x/2 <1 \Rightarrow$ $ [x/2] = 0$

Similarly, for $n=1,$ $ [x/2] = 1$

How can I use these facts to evaluate the integral?

Answer 1

I think if $f(x)=\left\lfloor\dfrac{x}{2}\right\rfloor\!,$ then it's easier to work, on your domain, with $f(x)=U(x-2),$ the unit step function. Then $f'(x)=U'(x-2)=\delta(x-2)$ on your domain. You could use the identity $\int_a^b f(x)\,\delta(x-c)\,dx=f(c)$ if $a<c<b$.

Answer 2

Let's write the Riemann-Stieltjes sum $$ S_n = x_1^3([x_1/2]-[x_0/2])+\dots+x_i^3([x_i/2]-[x_{i-1}/2])+x_{i+1}^3([x_{i+1}/2]-[x_i/2])+\dots+x_n^3([x_n/2]-[x_{n-1}/2]) $$ Then we choose $i$ so that $x_{i-1}/2<1<x_{i+1}/2$, most terms will cancel out, and we will have $$ S_n =x_i^3([x_i/2]-[x_{i-1}/2])+x_{i+1}^3([x_{i+1}/2]-[x_i/2])= (x_i^3-x_{i+1}^3)[x_i/2]+x_{i+1}^3 $$ And by taking a limit $$ S_n \to 2^3 $$ So from the definition $$\int_0^3 x^3 \,\text{d}[\frac{x}{2}] = 2^3 $$