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

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?

• @SangchulLee Wow! That should be accepted as answer, not comment! Thanks very much. – eigenvalue Jun 26 '20 at 14:41
• Glad it helped. I will try a bit to see if I can extract further terms and then write an answer :) – Sangchul Lee Jun 26 '20 at 14:44

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 $$|D| \to \infty$$.