For large real onstant $D$, I want an asymptotic evaluation of the sum $$\sum_{n=-\infty}^\infty \left( \tan^{-1} \frac{D}{2n+1} \right)^2 \frac{1}{n+3/4}.$$
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 physics.
One more clue: From the numerical analysis, I suspect the above integral has $\sim \log D$ behavior. Then, what is the coefficient?
Let $f(D)$ denote the sum, and write
\begin{align*} f(D) &= \sum_{n=0}^{\infty} \Biggl[ \arctan^2\left(\frac{D}{2n+1}\right)\frac{1}{n+\frac{3}{4}} + \arctan^2\left(\frac{D}{2(-n-1)+1}\right)\frac{1}{(-n-1)+\frac{3}{4}} \Biggr] \\ &= \sum_{n=0}^{\infty} \arctan^2\left(\frac{D}{2n+1}\right) \biggl(\frac{1}{n+\frac{3}{4}}-\frac{1}{n+\frac{1}{4}}\biggr) \\ &= -\sum_{n=0}^{\infty} \frac{8\arctan^2\left(\frac{D}{2n+1}\right)}{(4n+1)(4n+3)}. \end{align*}
Then the sum converges uniformly on $\mathbb{R}$, and so,
\begin{align*} \lim_{|D| \to \infty} f(D) = \sum_{n=0}^{\infty} \left(\frac{\pi}{2}\right)^2 \biggl(\frac{1}{n+\frac{3}{4}}-\frac{1}{n+\frac{1}{4}}\biggr) =-\frac{\pi^3}{4}. \end{align*}
We then investigate the next term. To this end, assume $D > 0$ without losing the generality and write
\begin{align*} f(D) - \left(-\frac{\pi^3}{4}\right) &= 8 \sum_{n=0}^{\infty} \frac{\frac{\pi^2}{4} - \arctan^2\left(\frac{D}{2n+1}\right)}{(4n+1)(4n+3)} \\ &= 8 \sum_{2n+1 \leq D} \frac{\frac{\pi^2}{4} - \arctan^2\left(\frac{D}{2n+1}\right)}{(4n+1)(4n+3)} + 8 \sum_{2n+1 > D} \frac{\frac{\pi^2}{4} - \arctan^2\left(\frac{D}{2n+1}\right)}{(4n+1)(4n+3)}. \end{align*}
The second term in the last line is easily bounded as
$$ 8 \sum_{2n+1 > D} \frac{\frac{\pi^2}{4} - \arctan^2\left(\frac{D}{2n+1}\right)}{(4n+1)(4n+3)} = \mathcal{O}\left(\frac{1}{D}\right). $$
So we move on to the first term. Using the asymptotic formula $\arctan(x) = x + \mathcal{O}(x^3)$,
\begin{align*} &8 \sum_{2n+1 \leq D} \frac{\frac{\pi^2}{4} - \arctan^2\left(\frac{D}{2n+1}\right)}{(4n+1)(4n+3)} \\ &= 8 \sum_{2n+1 \leq D} \frac{\arctan\left(\frac{2n+1}{D}\right) \left(\pi - \arctan\left(\frac{2n+1}{D}\right) \right)}{(4n+1)(4n+3)} \\ &= 8\pi \sum_{2n+1 \leq D} \frac{\left(\frac{2n+1}{D}\right)}{(4n+1)(4n+3)} + \mathcal{O}\Biggl( \sum_{2n+1 \leq D} \frac{\left(\frac{2n+1}{D}\right)^2}{(4n+1)(4n+3)} \Biggr) \\ &= \frac{\pi \log D}{D} + \mathcal{O}\left( \frac{1}{D} \right) \end{align*}
Combining altogether, we conclude that
$$ f(D) = -\frac{\pi^3}{4} + \frac{\pi \log |D|}{|D|} + \mathcal{O}\left( \frac{1}{D} \right) $$
as $|D| \to \infty$.
Addendum. Using the identity
$$ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2 + a^2} = \frac{\pi}{4a} \tanh \left(\frac{\pi a}{2}\right), $$
the sum $f(D)$ can be recast as
\begin{align*} f(D) &= -\sum_{n=0}^{\infty} \frac{8(2n+1)^2}{4(2n+1)^2 - 1} \int_{0}^{D} \int_{0}^{D} \frac{\mathrm{d}x \mathrm{d}y}{((2n+1)^2 + x^2)((2n+1)^2 + y^2)} \\ % &= \int_{0}^{D} \int_{0}^{D} \pi \biggl[ % -\frac{4}{(1 + 4x^2)(1 + 4y^2)} \\ % &\hphantom{= \pi \int_{0}^{D} \int_{0}^{D} \biggl[} % +\frac{2x \tanh (\pi x/2)}{(1 + 4x^2)(x^2 - y^2)} % -\frac{2y \tanh (\pi y/2)}{(1 + 4y^2)(x^2 - y^2)} % \biggr] \, \mathrm{d}x\mathrm{d}y \\ &= -\pi \arctan^2 (2D) + \int_{0}^{2D}\int_{0}^{2D} \frac{\varphi(x) - \varphi(y)}{x^2 - y^2} \, \mathrm{d}x\mathrm{d}y, \end{align*}
where
$$ \varphi(x) = \frac{x \tanh(\pi x/4)}{1 + x^2}. $$
I think this integral representation may be used instead to study the asymptotic behavior of $f(D)$ more systematically.