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I was messing around with the digamma function the other day, and I discovered this identity: $$\psi\left(\frac ab\right)=\sum_{\substack{\large\rho^b=1\\\large\rho\ne1}}(\rho^a-1)\ln(1-\bar\rho)-\gamma$$ when $0<\dfrac ab\le1$. It's unusual in that it sums over the $b$-eth roots of unity (which I don't see very often). (Note that $\bar\rho=\rho^{-1}$.) It also gives explicit values of the digamma function for all rational arguments, but no irrational ones.

Another thing that's interesting is that it's well-defined. For example, I know that $\psi\left(\dfrac12\right)=\psi\left(\dfrac24\right)$, but that's not immediately obvious if I plug those values into the RHS. It's also not immediately obvious—looking at the RHS—that the function is continuous.

Example: $\psi\left(\dfrac14\right)=(i-1)\ln(1+i)+(-1-1)\ln(1+1)+(-i-1)\ln(1-i)-\gamma=\\-3\ln2-\dfrac\pi2-\gamma$.

Since $H_n=\dfrac1n+\gamma+\psi(n)$, we can get a similar identity for the Harmonic numbers.

Is this identity well-known? Does it have a name?

(And, if so, is there a way to integrate it, so that I get a similar formula for the Gamma function?)

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  • $\begingroup$ This might also be used to derive the reflection formula for digamma… $\endgroup$ Commented Oct 1, 2014 at 14:46
  • $\begingroup$ See Gauss's digamma theorem. $\endgroup$
    – Lucian
    Commented Oct 1, 2014 at 15:52
  • $\begingroup$ @Lucian Oh, wow, that's interesting. It's probably equivalent to my thing. I do feel like my version is "more elegant," though. It's also shorter. $\endgroup$ Commented Oct 1, 2014 at 16:19
  • $\begingroup$ Very nice! Another approach: math.stackexchange.com/questions/1357091/… $\endgroup$ Commented Jan 12, 2016 at 13:43
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    $\begingroup$ So I wrote this way back in 2014. In the five years since, I have completely forgotten how I found this equation, or why it's true. So, um, if anyone could rediscover my proof, I'd be pretty grateful. $\endgroup$ Commented Mar 29, 2019 at 12:13

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Your formula can in fact be simplified a bit more by splitting off the term $$-\sum_{\substack{\large\rho^b=1\\\large\rho\ne1}}\ln(1-\bar\rho)=-\ln \prod_{\substack{\large\rho^b=1\\\large\rho\ne1}}(1-\bar\rho) \tag{1}$$ from the summation. Since $\bar\rho$ is also a root of unity if $\rho$ is, we may replace $\rho\to \bar\rho$ in each factor without changing the product. But the resulting product is the value of the polynomial $\dfrac{z^b-1}{z-1}$ evaluated at $z=1$, since this polynomial is monic with roots at all $b$th roots of unity save $z=1$. Hence the value of Eq. $(1)$ is $$-\ln\left(\frac{z^b-1}{z-1}\right)_{z\to 1}=-\ln(1+z+z^2+\cdots+z^{b-1})_{z\to 1}=-\ln b.$$ Thus the desired formula may be written as $$\psi\left(\frac ab\right)=\sum_{\substack{\large\rho^b=1\\\large\rho\ne1}}\rho^{-a}\ln(1-\rho)-\ln b-\gamma \tag{2}$$ where I have replaced $\rho\to \bar\rho=1/\rho$ as before. We now note that every root of unity $\rho$ can be written in the form $\rho=\omega^n$ for some $n=0,1,\cdots,b-1$ where $\omega=e^{2\pi i/b}$. Hence the sum over roots (other than unity) in Eq. $(2)$ may be written as

$$\psi\left(\frac ab\right)=\sum_{k=1}^{b-1}\omega^{-na}\ln(1-\omega)-\ln b-\gamma.$$

But this identity has already appeared on this site, in an answer showing how to derive Gauss's digamma theorem. (See the equation after "Let $t\to 1^{-}$", with $p/q$ instead of $a/b$). This validates the desired identity.

As such I'm inclined to consider your formula as being an equivalent form of the Gauss digamma function. The only substantial difference (aside from the $-\ln b$ term) is that the Gauss digamma function is explicitly real and has been written in the form of trigonometric functions.

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