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Suppose I have either of the following expressions:

$\newcommand{\rd}{\mathrm{d}} \int\nolimits_1^n\int\nolimits_1^\frac{n}{x} \,\rd y\, \rd x - \sum_{x=2}^n\sum_{y=2}^\frac{n}{x}1 $

$ \int\nolimits_1^n\int\nolimits_1^\frac{n}{x} \log y \,\rd y\, \rd x - \sum_{x=2}^n\sum_{y=2}^\frac{n}{x}\log y $

Are there any simpler forms I could convert either of these into? Or any other ways of expressing these that might yield insight or let me work with them in other interesting ways?

I already know there are several different ways of rewriting just the integrals on the left in isolation ($ n \log n - n + 1 $ at the simplest for the first one, for example), but I'm really looking for expressions that capture the difference between the continuous and the discrete here, kind of the way Euler-Maclaurin summation does.

1 Answers 1

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The double sum $S_n$ in your first expression is $ S_n=\sum\limits_{i=2}^n\sum\limits_k\left[2\leqslant k\leqslant\left\lfloor n/i\right\rfloor\right]=\sum\limits_{i=2}^n\int\left[1\leqslant y\leqslant\lfloor n/i\rfloor\right]\mathrm dy, $ hence $ S_n=\int_1^n\int_1^{u_n(x)}\mathrm dy\mathrm dx,\qquad u_n(x)=\left\lfloor n/\lceil x\rceil\right\rfloor. $ The difference between the integral $I_n$ and the sum $S_n$ is $ I_n-S_n=\int_1^n\int_{u_n(x)}^{n/x}\mathrm dy\mathrm dx=\int_1^n((n/x)-u_n(x))\mathrm dx. $ Since $u_n(x)\leqslant n/x$ for every $x$ in $(1,n)$, a consequence of this formula is that $I_n-S_n\geqslant0$. Another consequence is that $I_n-S_n=nJ+o(n)$ when $n\to\infty$, with $ J=\int_0^1((1/x)-\lfloor 1/x\rfloor )\mathrm dx=\int_1^{+\infty}(x-\lfloor x\rfloor )\frac{\mathrm dx}{x^2}. $ As noted by @Andrew, writing $J$ as a sum of integrals from $n$ to $n+1$ and then as a series gives $ J=\lim\limits_{n\to\infty}\left(\log(n)-\sum\limits_{k=2}^n\frac1k\right)=1-\gamma, $ where $\gamma$ is Euler's constant hence $\gamma=0.577\ldots$ and $J=0.423\ldots$

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    I think that that $o(1)$ has to be too small (though I wish it weren't). $S_n$ is related to the Dirichlet divisor problem - $S_n = \sum_{i=1}^n d(i) - 2 n + 1$, and I think the error term is known to be no smaller than $o(n^\frac{1}{4})$ . Ultimately, I want to use either expression I wrote as parts of larger sums (consisting of related expressions with deeper and deeper sum / integral nesting), so error terms in isolation matter less to me than having new possible ways to manipulate them, if that makes sense.2011-10-20