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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,

  1. $D_{\beta,\gamma}(x)=(\beta+\gamma)x\sum_{n\geq 0}\frac{1}{(n+x\beta)(n-\gamma x)}.$
  2. $\beta\not= \gamma$ since if $\beta=\gamma$ the function is periodic.
  3. [checked experimentaly] It seems that $D_{\beta,\gamma}(x+T)\approx D_{\beta,\gamma}(x)$ for large x and $T=\beta+\gamma.$
111
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  • It follows directly from $\Gamma(s+1) = s \Gamma(s)$ that $\psi(s)-\psi(1-s)$ is periodic. What do you get for your function ? – reuns Jun 05 '19 at 10:18
  • This proves $\psi(x)-\psi(-x)=-\pi \cot(\pi x) -1/x.$ Also proves that $D_{b,b}(x)$ is of the same form. Can't say anything for $D_{\beta,\gamma}(x).$ – 111 Jun 05 '19 at 10:50
  • $\psi(x)-\psi(1-x)=-\pi \cot(\pi x)$ is more difficult than just saying $\psi(x)-\psi(1-x)$ is periodic and it doesn't follow only from $\Gamma(x+1) = x\Gamma(x), \psi(x) = \Gamma'(x)/\Gamma(x)$ – reuns Jun 05 '19 at 10:53
  • true, you need also cosecant formula of gamma function – 111 Jun 05 '19 at 11:13
  • I am not sure that $D_{\beta,\gamma}(x)$ is periodic, do you have a proof based on gamma properties? – 111 Jun 05 '19 at 14:06
  • It is not ${}{}{}{}$ – reuns Jun 05 '19 at 14:18
  • sorry I meant $\psi(\beta x)-\psi(1-\gamma x)$ (and not $D_{\beta,\gamma}(x).$) Because in your first comment I understood that, based on gamma properties you can prove that $\psi(\beta x)-\psi(1-\gamma x)$ is periodic, as in the case of $\psi(x)-\psi(1-x).$ – 111 Jun 05 '19 at 14:27

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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 $O(1/x).$

111
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