In the paper by Kraskov et al (2004) there is a rather large jump in calculations. I am wondering if someone could fill out the gap for the equation below (equation 17 in the paper):
$$ k\binom{N-1}{k}\int_0^1 p^{k-1} (1-p)^{N-k-1} \ln(p) dp = \psi(k) - \psi(N) $$
where $\psi(x)$ is the dimgamma function.
Reference: Kraskov, A., Stögbauer, H., & Grassberger, P. (2004). Estimating mutual information. Physical review E, 69(6), 066138.
\begin{align*} \int_0^1 p^{k-1} (1-p)^{N-k-1} \ln(p) dp =& \int_0^1 p^{k-1} (1-p)^{N-k-1} \left(\lim_{t \to 0+} \frac{d}{dt} p^t\right)dp \\ =& \lim_{t \to 0+} \frac{d}{dt}\int_0^1 p^{k+t-1} (1-p)^{N-k-1} dp \\ =& \lim_{t \to 0+} \frac{d}{dt} B(k+t,N-k)\\ =& \lim_{t \to 0+} \left[\psi(k+t)-\psi(n+t)\right]B(k+t,N-k)\\ =& \left[\psi(k)-\psi(N)\right]B(k,N-k)\\ =& \left[\psi(k)-\psi(N)\right]\frac{\Gamma(k)\Gamma(N-k)}{\Gamma(N)}\\ =& \left[\psi(k)-\psi(N)\right]\frac{(k-1)!(N-k-1)!}{(N-1)!}\\ =& \left[\psi(k)-\psi(N)\right]\frac{k!(n-k-1)!}{k(N-1)!}\\ =& \left[\psi(k)-\psi(N)\right]\frac{1}{k\binom{N-1}{k}} \end{align*}
$$ \boxed{ k\binom{N-1}{k}\int_0^1 p^{k-1} (1-p)^{N-k-1} \ln(p) dp = \psi(k)-\psi(N)} $$