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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 something has changed. –  Mhenni Benghorbal Oct 4 '12 at 0:53
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1 Answer

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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Tks Mhenni Benghorbal for response –  Joao Pereira Oct 4 '12 at 8:03
    
Applying integration by parts as you suggest we get for the last integral –  Joao Pereira Oct 4 '12 at 8:04
    
Applying integration by parts as you suggest we get for the last integral K*[b-w] - K*[a-w] - INT(a to b ) ([x-w]d(k)= K*(b-a) - 0 =K*(b-a) assuming K,w integers. But this result is not 0 as it should be as I claimmed above. So it means that there is somewhere a mistake in my doing,don,t you agree?Can you make things clear?Thks –  Joao Pereira Oct 4 '12 at 8:17
    
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, –  Andrew Aug 7 '13 at 16:04
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