Is well known that $$\psi(x)-\psi(-x)=-\pi \cot(\pi x) - \frac{1}{x}.$$ I am wondering if a similar property holds for the following function, $$D_{\beta,\gamma}(x) = \psi(\beta x)-\psi(-\gamma x),\ \beta,\gamma \in {\mathbb{Z}_{>0}},$$ i.e. if $D_{\beta,\gamma}(x) =\text{a periodic function} + O\Big(\frac{1}{x}\Big)?$ Any ideas?
Edit.
It may be useful the following,
- $D_{\beta,\gamma}(x)=(\beta+\gamma)x\sum_{n\geq 0}\frac{1}{(n+x\beta)(n-\gamma x)}.$
- $\beta\not= \gamma$ since if $\beta=\gamma$ the function is periodic.
- [checked experimentaly] It seems that $D_{\beta,\gamma}(x+T)\approx D_{\beta,\gamma}(x)$ for large x and $T=\beta+\gamma.$
By setting $z=-\gamma x$ in the reflection formula $\psi(z) - \psi(1-z) = -\pi \cot(\pi z)$ we get $$\psi(-\gamma x) - \psi(1+\gamma x) = \pi \cot(\pi \gamma x).$$ So $$D_{\beta,\gamma}(x) = \psi(\beta x) - \psi(1+\gamma x) - \pi \cot(\pi \gamma x). $$ From the well know relation $$\psi(z) - \ln{z} = -\frac{1}{2z}+O(\frac{1}{z^2})$$ we get $$D_{\beta,\gamma}(x) = [\psi(\beta x) - \ln(\beta x)] - [\psi((1+\gamma x) - \ln(1+\gamma x)] +$$ $$ \ln{(\frac{\beta x}{1+\gamma x})} - \pi \cot(\pi \gamma x)=$$ $$\ln(\frac{\beta}{\gamma})-\pi\cot(\pi\gamma x)+O(\frac{1}{x}).$$ So, indeed $D_{\beta,\gamma}(x)$ for large $x$ is a sum of a periodic function with period 1 and a term $O(1/x).$