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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.

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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.

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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 –  Ron Jeremy Jul 1 '12 at 4:47
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@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
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Ohhhhh that's right! Ok cool thanks. –  Ron Jeremy Jul 1 '12 at 4:54
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