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Let $\psi(x)$ denote the digamma function $$ \psi(x)=\Gamma(x)\frac{\partial}{\partial x} \Gamma(x). $$ Consider $x=x_1 +x_2+\dots +x_m$, where $x_j>0$, for $j=1, \ldots,m$. Is there any formula to decompose $\psi(x)$ in terms of $\psi(x_1),\ldots,\psi(x_m)$?

I know that in the very special case of $m=2$ and $x_1=x_2=x/2$, with $x>0$, Legendre duplication formula allows to claim $$ \psi(x_1+x_2)=\log 2 +\frac{1}{2}\left( \psi(x_1) + \psi(x_2+1/2) \right) $$ and I was wondering whether something more general than that is known in the literature.

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I'm pretty sure the answer is no. In the case that $x_1=...=x_m=z$ however there is the nice formula $$\psi^{[n]}(mz)=\delta_{n,0}\ln m +\frac{1}{m^{n+1}}\sum_{k=0}^{m-1}\psi^{[n]}\left(z+\frac{k}{m}\right)$$

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  • $\begingroup$ Ok, many thanks. Could you point me to a reference? $\endgroup$ Oct 12, 2020 at 9:32
  • $\begingroup$ It's on Wolfram Mathworld although I'm unsure of the original source, as Mathworld does not state which reference it comes from. The list of references is at the bottom of the article, but you'll have to do your own digging to find which one it came from. I unfortunately could not find a proof for the general polygamma, but Flammable Maths has a proof for the Gamma function. Perhaps some of the methods used there will prove useful. $\endgroup$
    – K.defaoite
    Oct 12, 2020 at 11:54
  • $\begingroup$ I suspect that this identity comes from Abramowitz and Stegun's Handbook of Mathematical Functions although there are several other references cited. $\endgroup$
    – K.defaoite
    Oct 12, 2020 at 11:58
  • $\begingroup$ Many thanks, I'll do the search. $\endgroup$ Oct 12, 2020 at 12:03
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    $\begingroup$ Sure, thanks. Meanwhile, I've found the formula in Abramowitz and Stegun's book, under the name "Multiplication Formula" on page 260. $\endgroup$ Oct 12, 2020 at 13:42

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