Digamma equation identification 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?)
 A: 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.
