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.
Best Answer
We follow the analogous development in this answer. Let $S(D)$ be given by
$$\begin{align} S(D)&=\sum_{n=-\infty}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{|2n+1|}\right)}{n+3/4}\\\\&=\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}+\sum_{n=-\infty}^{-1} \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{|2n+1|}\right)}{n+3/4}\\\\ &=\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}+\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+1/4}\tag1 \end{align}$$
We analyze the first series on the right-hand side of $(1)$. We begin by writing
$$\begin{align} \sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}&=\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}\\\\ &+\sum_{2n+1> D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4} \end{align}\tag2$$
For the first series on the right-hand side of $(2)$ we have
$$\begin{align} \sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}&=\log(D)\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)}{n+3/4}\\\\ &-\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(2n+1\right)}{n+3/4}\tag3 \end{align}$$
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
$$\begin{align} \sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log(2n+1)}{n+3/4}&=\frac\pi2 \sum_{2n+1\le D}\frac{\log(2n+1)}{n+3/4}\\\\ &-\sum_{2n+1\le D}\frac{\arctan\left(\frac{2n+1}{D}\right)\log(2n+1)}{n+3/4}\\\\ &=\frac\pi4 \log^2(D)+O\left(\frac{\log(D)}{D}\right)-O(1)\tag5 \end{align}$$
Using $(4)$ and $(5)$ in $(3)$ reveals
$$\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}=\frac\pi4 \log^2(D)+O(\log(D))$$
Next, we anlayze the second series on the right-hand side of $(2)$. It is evident that
$$\left|\sum_{2n+1> D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}\right|\le D^2\sum_{2n+1>D}\frac{1}{(2n+1)^2(n+3/4)}=O(1)$$
Putting it all together, we find that for $D\to\infty$
$$S(D)=\frac{\pi}{2}\log^2(D)+O(\log(D))$$