Asymptotic analysis of $\sum_{n=-\infty}^\infty \tan^{-1} \left(\frac{D}{2n+1}\right) \log\left(\frac{D}{|2n+1|}\right) \frac{1}{n+3/4}$

Question

For large positive constant $D$, I want an asymptotic evaluation of the sum $$\sum_{n=-\infty}^\infty \tan^{-1} \left(\frac{D}{2n+1}\right) \log\left(\frac{D}{|2n+1|}\right) \frac{1}{n+3/4}.$$

Note that the sum is convergent since for large $n$ since $\tan^{-1}(D/(2n+1)) \approx D/(2n+1)$ and therefore the term decays fastly. This question is motivated from a calculation of Feynman diagram in quantum field theory.

Answer 1

We follow the analogous development in this answer. Let $S(D)$ be given by

$$\begin{align} S(D)&=\sum_{n=-\infty}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{|2n+1|}\right)}{n+3/4}\\\\&=\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}+\sum_{n=-\infty}^{-1} \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{|2n+1|}\right)}{n+3/4}\\\\ &=\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}+\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+1/4}\tag1 \end{align}$$

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We analyze the first series on the right-hand side of $(1)$. We begin by writing

$$\begin{align} \sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}&=\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}\\\\ &+\sum_{2n+1> D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4} \end{align}\tag2$$

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For the first series on the right-hand side of $(2)$ we have

$$\begin{align} \sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}&=\log(D)\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)}{n+3/4}\\\\ &-\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(2n+1\right)}{n+3/4}\tag3 \end{align}$$

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For the first series on the right-hand side of $(3)$ we find using the Euler-McLaurin Summation Formula that

$$\begin{align} \sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)}{n+3/4}&=\frac\pi2 \sum_{2n+1\le D}\frac1{n+3/4}-\sum_{2n+1\le D}\frac{\arctan\left(\frac {2n+1}{D}\right)}{n+3/4}\\\\ &=\frac\pi2\left(\log(D)+O(1)\right)-O(1)\\\\ &=\frac\pi2 \log(D)+O(1)\tag4 \end{align}$$

For the second series on the right-hand side of $(3)$ we find using the Euler-McLaurin Summation Formula that

$$\begin{align} \sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log(2n+1)}{n+3/4}&=\frac\pi2 \sum_{2n+1\le D}\frac{\log(2n+1)}{n+3/4}\\\\ &-\sum_{2n+1\le D}\frac{\arctan\left(\frac{2n+1}{D}\right)\log(2n+1)}{n+3/4}\\\\ &=\frac\pi4 \log^2(D)+O\left(\frac{\log(D)}{D}\right)-O(1)\tag5 \end{align}$$

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Using $(4)$ and $(5)$ in $(3)$ reveals

$$\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}=\frac\pi4 \log^2(D)+O(\log(D))$$

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Next, we anlayze the second series on the right-hand side of $(2)$. It is evident that

$$\left|\sum_{2n+1> D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}\right|\le D^2\sum_{2n+1>D}\frac{1}{(2n+1)^2(n+3/4)}=O(1)$$

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Putting it all together, we find that for $D\to\infty$

$$S(D)=\frac{\pi}{2}\log^2(D)+O(\log(D))$$

Answer 2

Let us break the series in a few pieces:

$$ \sum_{n=0}^{D}\arctan\left(\frac{D}{2n+1}\right)\frac{\log D-\log(2n+1)}{n+3/4} =\sum_{n=0}^{D}\left[\frac{\pi}{2}-\arctan\left(\frac{2n+1}{D}\right)\right]\frac{\log D-\log(2n+1)}{n+3/4} $$ behaves like $$ \frac{\pi}{4}\log^2(D)+O(\log D)+\frac{2}{D}\sum_{n=0}^{D}\log(2n+1)=\frac{\pi}{4}\log^2(D)+O(\log D) $$ while $$ \sum_{n>D}\arctan\left(\frac{D}{2n+1}\right)\frac{\log D-\log(2n+1)}{n+3/4} $$ behaves like $$ D\log D\sum_{n>D}\frac{1}{2n^2}-D\sum_{n>D}\frac{\log(2n)}{2n^2}=\frac{\log D}{2}-\frac{1+\log(2D)}{2}+o(\log D)=o(\log D) $$ so $$\sum_{n\geq 0}\arctan\left(\frac{D}{2n+1}\right)\frac{\log D-\log(2n+1)}{n+3/4} =\frac{\pi}{4}\log^2(D)+O(\log D).$$ The series on $n<0$ can be managed in a similar fashion. We exploited

$$ \sum_{k=1}^{n}\frac{1}{k}=\log n+O(1),\qquad \sum_{k=1}^{n}\frac{\log n}{n}=\frac{1}{2}\log^2(n)+O(\log n), $$ $$ \sum_{k=1}^{n}\log(k) = n\log n+O(\log n),\qquad \sum_{k\geq n}\frac{1}{k^2}=\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$ $$ \sum_{k\geq n}\frac{\log k}{k^2}=\frac{\log n}{n}+O\left(\frac{1}{n}\right).$$