More terms in asymptotic expansion of $\sum_{n=-\infty}^\infty \arctan \frac{D}{2n+1} \log\left(\frac{D}{|2n+1|}\right) \frac{1}{n+3/4}$

Question

Problem: Find the first two terms in the asymptotic expansion of (as $D \to \infty$) $$S = \sum_{n=-\infty}^\infty \arctan \frac{D}{2n+1} \log\left(\frac{D}{|2n+1|}\right) \frac{1}{n+3/4}.$$

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

My attempt: We have \begin{align} S &= \sum_{n=0}^\infty \arctan \frac{D}{2n+1} \left(\log \frac{D}{2n+1}\right) \left(\frac{1}{n+3/4} + \frac{1}{n+1/4} \right)\\ &= \sum_{n=0}^\infty \arctan \frac{D}{2n+1} \left(\log \frac{D}{2n+1}\right) \frac{4}{2n+1}\\ &\quad + \sum_{n=0}^\infty \arctan \frac{D}{2n+1} \left(\log \frac{D}{2n+1}\right) \frac{4}{(4n+3)(4n+1)(2n+1)}\\ &= \sum_{n=0}^\infty \arctan \frac{D}{2n+1} (\log D) \frac{4}{2n+1}\\ &\quad - \sum_{n=0}^\infty \arctan \frac{D}{2n+1} (\log (2n+1)) \frac{4}{2n+1}\\ &\quad + \sum_{n=0}^\infty \frac{\pi}{2} (\log D) \frac{4}{(4n+3)(4n+1)(2n+1)}\\ &\quad - \sum_{n=0}^\infty \arctan \frac{2n+1}{D} (\log D) \frac{4}{(4n+3)(4n+1)(2n+1)}\\ &\quad - \sum_{n=0}^\infty \arctan \frac{D}{2n+1} (\log (2n+1)) \frac{4}{(4n+3)(4n+1)(2n+1)}\\ &= I_1 - I_2 + I_3 - I_4 - I_5 \end{align} where we have used $\arctan \frac{D}{2n+1} + \arctan \frac{2n+1}{D} = \frac{\pi}{2}$ and $\frac{1}{n+3/4} + \frac{1}{n+1/4} = \frac{4}{2n+1} + \frac{4}{(4n+3)(4n+1)(2n+1)}$.

Clearly, $I_4 = O(1)$ and $I_5 = O(1)$. Also, $I_3 = \pi \ln 2 \log D$.

Since $\arctan \frac{D}{2n+1} = \int_0^D \frac{2n+1}{(2n+1)^2 + t^2} \mathrm{d} t$ and $\sum_{n=0}^\infty \frac{4}{(2n+1)^2 + t^2} = \frac{\pi}{t}\tanh \frac{\pi t}{2}$, we have \begin{align} I_1 &= \log D \sum_{n=0}^\infty \frac{4}{2n+1}\int_0^D \frac{2n+1}{(2n+1)^2 + t^2} \mathrm{d} t \\ &= \log D \int_0^D \sum_{n=0}^\infty \frac{4}{(2n+1)^2 + t^2} \mathrm{d} t \\ &= \log D \int_0^D \frac{\pi}{t}\tanh \frac{\pi t}{2} \mathrm{d} t\\ &= \log D \left[\int_0^1 \frac{\pi}{t}\tanh \frac{\pi t}{2} \mathrm{d} t + \int_1^D \frac{\pi}{t} \mathrm{d} t - \int_1^\infty \frac{\pi}{t}\left(1 - \tanh \frac{\pi t}{2}\right) \mathrm{d} t \right.\\ &\qquad\qquad + \left. \int_D^\infty \frac{\pi}{t}\left(1 - \tanh \frac{\pi t}{2}\right) \mathrm{d} t \right]\\ &= \pi \log^2 D + \alpha \pi \log D + o(1) \end{align} where $\alpha = \int_0^1 \frac{1}{t}\tanh \frac{\pi t}{2} \mathrm{d} t - \int_1^\infty \frac{1}{t}\left(1 - \tanh \frac{\pi t}{2}\right) \mathrm{d} t \approx 1.27$, and we have used $\log D \int_D^\infty \frac{\pi}{t}\left(1 - \tanh \frac{\pi t}{2}\right) \mathrm{d} t = o(1)$.

It remains to estimate $$I_2 = \sum_{n=0}^\infty \arctan \frac{D}{2n+1} (\log (2n+1)) \frac{4}{2n+1}.$$ Consider $$I_2' = \int_0^\infty \arctan \frac{D}{2x+1} (\log (2x+1)) \frac{4}{2x+1} \mathrm{d} x.$$ We have \begin{align} I_2' &= \int_0^D \arctan u \left(\log \frac{D}{u}\right)\frac{2}{u} \mathrm{d}u\\ &= \int_0^D \arctan u (\log D)\frac{2}{u} \mathrm{d}u - \int_0^D \arctan u (\log u)\frac{2}{u} \mathrm{d}u\\ &= 2\log D \left(\arctan D (\log D) - \int_0^D \frac{\log u}{1+u^2} \mathrm{d}u\right)\\ &\qquad - \left(\arctan D (\log D)^2 - \int_0^D \frac{\log^2 u}{1+u^2} \mathrm{d}u \right)\\ &= \arctan D (\log^2 D) + O(1) \end{align} where we have used $2\log D \int_0^D \frac{\ln u}{1+u^2} \mathrm{d}u = o(1)$ and $\int_0^D \frac{\log^2 u}{1+u^2} \mathrm{d}u = O(1)$.

I $\color{blue}{\textrm{GUESS}}$ that $I_2 = I_2' + O(1)$. Is it true? I try to find a rigorous argument.

Any comments and solutions are welcome.

Answer 1

Some thoughts:

Fact 1: Let $f$ is continuously differentiable on $[0, \infty)$. Then $$\sum_{m=1}^k f(m) = \int_0^k f(x) \mathrm{d} x + \int_0^k (x - \lfloor x \rfloor - \tfrac{1}{2})f'(x) \mathrm{d} x + \frac{1}{2}(f(k)-f(0)).$$ See: https://www.math.ucdavis.edu/~tracy/courses/math205A/EulerMaclaurinSummation.pdf

Let $$f(x) = \arctan \frac{D}{2x+1} (\log (2x+1)) \frac{4}{2x+1}.$$ We have $$f'(x) = -\frac{8D\log (2x+1)}{(2x+1)^3 + (2x+1)D^2} + \frac{8\arctan \frac{D}{2x+1}}{(2x+1)^2} - \frac{8\arctan \frac{D}{2x+1} \ln (2x+1)}{(2x+1)^2}.$$ By Fact 1, roughly, to prove $I_2 = I_2' + O(1)$, it suffices to prove that $\int_0^\infty |f'(x)| \mathrm{d}x = O(1)$.

Clearly, $\int_0^\infty \frac{8\arctan \frac{D}{2x+1}}{(2x+1)^2} \mathrm{d} x = O(1)$, and $\int_0^\infty \frac{8\arctan \frac{D}{2x+1} \ln (2x+1)}{(2x+1)^2} \mathrm{d} x = O(1)$. Also, since $$\frac{8D\log (2x+1)}{(2x+1)^3 + (2x+1)D^2} \le \frac{8D\log (2x+1)}{2(2x+1)^2D} = \frac{8\log (2x+1)}{2(2x+1)^2},$$ we have $\int_0^\infty \frac{8D\log (2x+1)}{(2x+1)^3 + (2x+1)D^2} \mathrm{d} x = O(1)$.