8
$\begingroup$

Evaluate: $$\int _0^2 (x^2+1) \; d \lfloor x\rfloor$$

Here $[x]$ denotes the greatest integer function of $x$.

I know this has to be done by parts as:

$$\int _0^2 (x^2+1) \, d [x]= {|(1+x^2)[x]|}_0^2- \int_0^2 [x] \, d(1+x^2)$$

Note:- This integral can be quite easily evaluated.I don't need the method for this.

But if we split the given integral into the sum of $2$ integrals as:-

$$\int _0^2 (x^2+1) \, d [x]=\int _0^1 (x^2+1) d [x]+\int _1^2 (x^2+1) \, d [x]$$

My question is as $[x]$ is constant in each of the intervals $[0,1)$ and $[1,2)$ in each of these $2$ integrals $d[x] = 0$

So, the value of the given integral should be $0$.

This seems contradictory !!

Kindly correct my reasoning for the part $d[x]=0$.

$\endgroup$
8
  • $\begingroup$ Quite nonstandard notation. You should check its meaning but this might refer to $$\int_0^2u(x)d[x]=\tfrac12u(0)+u(1)+\tfrac12u(2).$$ $\endgroup$
    – Did
    Oct 5, 2015 at 14:04
  • $\begingroup$ @Did, what has this to do with my question? $\endgroup$
    – miyagi_do
    Oct 5, 2015 at 15:15
  • 3
    $\begingroup$ @Did, why can't you answer my question rather than commenting down to close the question?Answer it and i will close it myself. I believe this group is for sharing knowledge.So please stop acting so rudely. $\endgroup$
    – miyagi_do
    Oct 6, 2015 at 11:14
  • 1
    $\begingroup$ Would you be so nice as to point at examples of rudeness in my two comments? Please be specific. If you are alluding to my second comment, I suggest that you engage in a little thinking about your attitude in your first comment. "why can't you answer my question" Well, I did. $\endgroup$
    – Did
    Oct 6, 2015 at 11:18
  • 1
    $\begingroup$ It's Stieltjes again. $\endgroup$ Oct 6, 2015 at 15:21

1 Answer 1

1
$\begingroup$

'Zero' contradicts the correct answer of '7' because the method of 'Integration by Parts' is invalid in this context since there is a discontinuity at x=1 in $\lfloor {x} \rfloor $

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .