I am stuck on the following problem from an exercise in my analysis book:
Show that $$\int_0^4 x \mathrm d(x-[x])=-2$$ where $[x]$ is the greatest integer not exceeding $x$.
I think I have to partition the interval $[0,4]$ into some suitable subintervals and here I see $x-[x] \ge 0$.
But, I am not sure how to tackle it as no similar types of problems has been discussed in the book. Can someone explain? Thanks and regards to all.
Using integration by parts for the Riemann-Stieltjes integral, we get:
$$\int\limits_0^4xd(x-\lfloor x\rfloor)=\left.x(x-\lfloor x\rfloor)\right|_{x=4}-\left.x(x-\lfloor x\rfloor)\right|_{x=0}-\int\limits_0^4(x-\lfloor x\rfloor)dx=$$
$$=4(4-4)-0(0-0)-\int\limits_0^4 x\,dx+\int\limits_0^4\lfloor x\rfloor dx=$$
$$=\left.-\frac12x^2\right|_0^4+\int\limits_0^1 0\cdot dx+\int\limits_1^21dx+\int\limits_2^32dx+\int\limits_3^43dx=-8+1+2+3=-2$$
You can define $u=x-[x]$, which will make the $d(x-[x])$ much more friendly. Then $x=u+[x]$ and it seems natural to split the integral into $[0,1],[1,2],[2,3],[3,4]$ On each interval you can see what to do with the $[x]$. That said, I am not getting $\frac 32$ for an answer.
If we let $f(x) = x -\lfloor x \rfloor$, then the Lebesgue Stieltjes measure corresponding to $f$ can be written as $\mu_f = m -\sum_n \delta_n$, where $m$ is the Lebesgue measure and $\delta_n$ is the Dirac measure concentrated at $n$.
Then $\int_0^4 x d \mu_f = \int_0^4 xdx - \int_0^4 x d(\sum_n \delta_n) = 8-(0+1+2+3+4) = -2$.
What happens if you try partitions like $$ 0,1-\delta,1,2-\delta,2,3-\delta,3,4-\delta,4 $$ where $\delta>0$ is very small?
