On the fractional parts of the roots of the Alternating Harmonic Numbers

Question

We define $$\bar{H}_x=\ln2+\cos(\pi x)\left(\psi(x)-\psi\left(\frac x2\right)-\frac1x-\ln2\right)$$As the $x$th Alternating Harmonic Number (test out a few values to see why). Let $x_n$ be the $n$th largest zero of the Alternating Harmonic Numbers, show that $\lim_{n\rightarrow\infty}\{x_n\}=\eta$ converges (where $\{x\}$ denotes the fractional part of $x$) or equivalently $\lim_{n\rightarrow\infty}(x_n-x_{n+1})=1$.

This definition of the alternating harmonic number originates from my terribly formatted paper where I generalize sigma notation for non-integer upper and lower bounds. Graphically speaking, the conjecture (I called it so in my paper because I couldn't solve it) seems to be true and $\eta\approx-0.431$. Some interesting approximations are $\frac{7\pi}{51}$ and this seemingly random integral $$\int_0^1\frac{\bar{H}_x\bar{H}_{-x}\sin(\pi x)}{\pi x}dx\approx0.43141591$$Not only does it hold for three digits, but there is a decimal representation of pi mixed in, up to 5 accurate decimal digits. Anyway, using the reflection formula derived for $\bar{H}_x$: $$\bar{H}_x-\bar{H}_{2-x}=\pi\cot(\pi x)-\left(\frac1{1-x}+\frac1x-\frac1{2-x}\right)\cos(\pi x)$$ we get that the conjecture is equivalent to finding the limit of the fractional part of the roots of $$\bar{H}_x-\bar{H}_{2-x}+\ln2=\ln2+\pi\cot(\pi x)-\left(\frac1{1-x}+\frac1x-\frac1{2-x}\right)\cos(\pi x)$$I got stuck here.

Answer 1

I finally solved the conjecture. Notice that as $x$ approaches negative infinity, the reflection formula becomes $\ln2+\pi\cot(\pi x)$ and so $\ln2+\pi\cot(\pi x_n)\sim0$. Solving for $x_n$ we get that $x_n\sim-n-\frac1{\pi}\arctan\left(\frac{\pi}{\ln2}\right)$ implying that $\eta=-\frac1{\pi}\arctan\left(\frac{\pi}{\ln2}\right)$, which agrees with numerical evidence ($\eta\approx-0.430876945137$) and has nothing to do with the integral in the OP.