Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In my professor's notes he has written this:

$$\int_1^N \frac{\{t\} - \frac{1}{2}}{t}dt = \int_1^N\frac{1}{t}d \left(\int_1^t B(y)dy \right) = \int_1^t B(y)dy|_1^N + \int_1^N \frac{\int_1^t B(y)dy}{t^2}dt$$

Where $\{x\}$ indicates the fractional part of $x$ and $B(x) = \{x\} - \frac{1}{2}$.

This is Abel summation with the Riemann-Stieltjes integral.

What I don't understand is I feel first that the lower bound of the $\displaystyle \int_1^t$ integrals should all be $\dfrac{1}{2}$ and not $1$, and second since this is integration by parts there should be a $\dfrac{1}{t}$ in front of the $\displaystyle \left. \int_1^t B(y)dy \right \vert_1^N$ term.

Am I missing something? Thanks.

share|cite|improve this question
up vote 2 down vote accepted

There should be a $\dfrac1t$ infront of the integral $\displaystyle \left( \int_1^{t} B(y) dy \right)$. $$\int_1^{N} \dfrac{\{t\} - \dfrac12}t dt = \int_1^{N} \dfrac1t d \left( \int_1^t B(y) dy\right)\\ = \left. \left(\dfrac1t \left( \int_1^{t} B(y) dy\right) \right) \right \rvert_{1}^{N} + \int_1^{N} \dfrac1{t^2} \left( \int_1^{t} B(y) dy\right) dt$$ The lower bound for the inner integral $\displaystyle \int_1^t (\cdot) dy$ can in principle be anything since $$d \left( \int_a^t B(y) dy\right) = B(t) dt = \left(\{t\} - \dfrac12 \right) dt$$ where $a$ is any fixed real number.

share|cite|improve this answer
    
I made an edit 14min ago it's the 1 to t integrals that are the problem, not the 1 to N.. not sure if you saw that or not. The reason is that B(1) = 1/2 when you want the lower bound to give zero since B(t) = {t} - 1/2 by itself – esproff Jul 1 '12 at 4:47
1  
@NollieTré The lower bound doesn't matter for the inner integral since it is within $d \left(\cdot \right)$. Note that $$d \left( \int_1^t B(y) dy\right) = B(t) dt$$ – user17762 Jul 1 '12 at 4:49
1  
Ohhhhh that's right! Ok cool thanks. – esproff Jul 1 '12 at 4:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.