# Riemann Stieltjes Integral

I've been trying to get knowledge with Riemann Steltjes integral and came across to some assignments in the web about the subject.In doing my practice I could´t achieve to the solution of a particular example that states as follows $$\int_0^6 (x^2+[x])d(|3-x|) =$$ According to the assignment the solution is supposed to be 63.

I tried to get to the solution using two different ways but none solution coincide.

I developed the example like this $$\int_0 ^3(x^2+[x])d(3-x) + \int_3^6 (x^2+[x])d(x-3)$$ which cancels the absolute value, and then $$\int_0 ^3(x^2\cdot d(3-x)) +\int_0 ^3([x]\cdot d(3-x) +\int_3 ^6(x^2\cdot d(x-3) +\int_3 ^6([x]\cdot d(x-3)=$$ The problem is that I can't work out the integral that involves the$[x]$ floor function. I searched your archives but couldn't get any hint. Doesn't seem that complicated but in fact I'stuck.

Can you give some help? Tks in advance

Joao Pereira

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Write the floor function as a piecewise function, broken up at each integer. –  Shaun Ault Oct 2 '12 at 18:56
Think about what the graph of the floor function looks like, and what the area under the graph would be. –  Alex J Best Oct 2 '12 at 18:57
Hints: e.g. $u=3-x$ $$\left\lfloor 3-u\right\rfloor =3+\left\lfloor -u\right\rfloor$$ $$\left\lfloor -u\right\rfloor =\left\{ \begin{array}{ccc} -1 & \text{if} & 0<x<1 \\ -2 & \text{if} & 1<x<2 \\ -3 & \text{if} & 2<x<3 \end{array} \right.$$ –  Américo Tavares Oct 2 '12 at 19:49
@Shaun: Yes, as long as the integrator $|3-x|$ has no discontinuity at an integer, the way to do this is to subdivide at the integers into 6 integrals. –  GEdgar Oct 3 '12 at 14:25

## 1 Answer

Make the substitution $u=3-x$. Then

\begin{align} I =&\int_{0}^{6}(x^{2}+\left\lfloor x\right\rfloor )d(\left\vert 3-x\right\vert )\\ =&\int_{3}^{-3}(\left( 3-u\right) ^{2}+\left\lfloor 3-u\right\rfloor )d(\left\vert u\right\vert ) \\ =&\int_{3}^{-3}(\left( 3-u\right) ^{2}+3+\left\lfloor -u\right\rfloor )d(\left\vert u\right\vert ), \end{align} because $$\left\lfloor 3-u\right\rfloor =3+\left\lfloor -u\right\rfloor,$$ since for $x$ real and $n$ integer, $\lfloor x+n\rfloor=\lfloor x\rfloor +n$. Hence

\begin{align} I=&-\int_{-3}^{0}(\left( 3-u\right) ^{2}+3+\left\lfloor -u\right\rfloor )d\left\vert u\right\vert-\int_{0}^{3}(\left( 3-u\right)^{2}+3+\left\lfloor -u\right\rfloor )d\left\vert u\right\vert\end{align}

and

\begin{align} I=&\int_{-3}^{0}(\left( 3-u\right) ^{2}+3+\left\lfloor -u\right\rfloor )du-\int_{0}^{3}(\left( 3-u\right) ^{2}+3+\left\lfloor -u\right\rfloor )du \end{align}

\begin{align} I=&\int_{-3}^{0}(\left( 3-u\right) ^{2}+3)du-\int_{0}^{3}(\left( 3-u\right) ^{2}+3)du \\&+\int_{-3}^{0}\left\lfloor -u\right\rfloor du-\int_{0}^{3}\left\lfloor -u\right\rfloor du.\end{align}

The first two integrals are $72$ and $18$. As for the last two their evaluation follows from the definition of the floor function of $-u$

$$\left\lfloor -u\right\rfloor =\left\{ \begin{array}{ccc} 2 & \text{if} & -3<x\le -2 \\ 1 & \text{if} & -2<x\le -1 \\ 0 & \text{if} & -1<x\le 0 \\ -1 & \text{if} & 0<x\le 1 \\ -2 & \text{if} & 1<x\le 2 \\ -3 & \text{if} & 2<x\le 3. \end{array} \right.$$

So \begin{align} I=&72-18+\left( \int_{-3}^{-2}2du+\int_{-2}^{-1}1du+\int_{-1}^{0}0du\right)\\&-\left( \int_{0}^{1}-1du+\int_{1}^{2}-2du+\int_{2}^{3}-3du\right) \\ =&72-18+3-\left( -6\right) \\ =&63. \end{align}

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Tank you very much for working out the problem.It was very kind of you.Best regards. –  Joao Pereira Oct 3 '12 at 9:01
@JoaoPereira You are welcome. –  Américo Tavares Oct 3 '12 at 10:45