4
$\begingroup$

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

  • 0
    @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.2012-10-03

1 Answers 1

4

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

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

  • 0
    @JoaoPereira You are welcome.2012-10-03