Asymptotic analysis of $\sum_{n=-\infty}^\infty \tan^{-1} \left(\frac{D}{2n+1}\right) \log\left(\frac{D}{|2n+1|}\right) \frac{1}{n+3/4}$ 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.
 A: 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 $D\to\infty$
$$S(D)=\frac{\pi}{2}\log^2(D)+O(\log(D))$$
A: 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
$$ \sum_{k=1}^{n}\frac{1}{k}=\log n+O(1),\qquad \sum_{k=1}^{n}\frac{\log n}{n}=\frac{1}{2}\log^2(n)+O(\log n), $$
$$ \sum_{k=1}^{n}\log(k) = n\log n+O(\log n),\qquad \sum_{k\geq n}\frac{1}{k^2}=\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$
$$ \sum_{k\geq n}\frac{\log k}{k^2}=\frac{\log n}{n}+O\left(\frac{1}{n}\right).$$
