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I come back again only to confirm (or not) a generalization. In my post on yesterday the integral

$ \int_{0}^{6}(x^2+[x])d(|3-x|) $

was worked out based on a change of variable. I tried to get the same solution in another way - with integration by parts .

After cancelling the absolute value and considering the second integral(due only to simplicity purpose) we have :

$\int_{3}^{6}(x^2+[x])d(x-3)$ whose solution is 66

Working the integral....

$ =\int_{3}^{6}x^2d(x-3) + \int_{3}^{6}[x]d(x-3)$

$ =\int_{3}^{6} x^2 dx + ( f(6)a(6) - f(3)a(3)) - \int_{3}^{6} (x-3)d[x] $

(integration by parts with $f(x)=[x]$ and $a(x)=x$ )

$ = (72-9) + (18-0) - \int_{3}^{6} x d([x] ) + \int_{3}^{6} (-3)d([x] )$

$= (72-9) + (18-0) - (4+5+6) + \int{3}^{6} (-3)d([x] )=66 + 0 =66 $

It seems reasonable for the last integral to be 0, as floor function [x] doesn't have derivative. Hence, as a generalization we may say that

$\int_{a}^{b} K d([x-w]) = 0 $

for every k, w real numbers.

Am I right? And if so how can we prove that?

Thks for cooperation Regards João Pereira

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    I edited your post. See, if somet$h$ing has changed.2012-10-04

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Related problems: (I). Here is a theorem you can apply it to the problem,

Theorem: Suppose $f$ and $g$ are bounded functions with no common discontinuities on the interval $[a,b]$, and the Riemann-Stieltjes integral of $f$ with respect to $g$ exists. Then the Riemann-Stieltjes integral of $g$ with respect to $f$ exists, and $\int_{a}^{b} g(x)df(x) = f(b)g(b)-f(a)g(a)-\int_{a}^{b} f(x)dg(x)\,. $

Note that,

$\int_{3}^{6}[x]dx = \int_{3}^{4} 3 dx + \int_{4}^{5} 4 dx + \int_{5}^{6} 5 dx \,. $

Now, what do you think the value of the following integral is?

$ \int_{3}^{6}d[x] = ?$

Just apply the above theorem and see what you get.

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    Again, thanks for this link. Maybe I could ask a question. I think the relation following "Note that" refers to the integral on the RHS. If this is correct, then I would think $f(x) = [x]$ and $g(x) = x$. If this is correct, I think this would lead to the integrand on the LHS of x. If this is all correct, I was wondering why you posed the question $\int d[x]$ rather than $\int x d[x]$? Thanks again. Regards,2013-08-07