Prove $\frac{\partial}{\partial m}\text{B}(n,m)=-\text{B}(n,m)\sum_{k=0}^{n-1}\frac{1}{k+m}$

Solution 1:

\begin{align} \text{B}(n,m)=\frac{\Gamma(n)\Gamma(m)}{\Gamma(n+m)}=\frac{(n-1)!}{m(m+1)...(m+n-1)}=(n-1)!\prod_{k=0}^{n-1}\frac{1}{m+k} \end{align} take the log to both sides, we get \begin{align} \ln\text{B}(n,m)=\ln(n-1)!+\sum_{k=0}^{n-1}\ln\left(\frac{1}{m+k}\right) \end{align} differentiate both sides with respect to $\ m$, we get \begin{align} \frac{\frac{\partial}{\partial m}\text{B}(n,m)}{\text{B}(n,m)}=-\sum_{k=0}^{n-1}\frac1{m+k}\quad \Longrightarrow \frac{\partial}{\partial m}\text{B}(n,m)=-\text{B}(n,m)\sum_{k=0}^{n-1}\frac1{m+k} \end{align}

Solution 2:

(A partial hint/answer but too long for a comment)

I'm not an expert on the subject, but by this one has $$\frac{\partial}{\partial x} \mathrm{B}(x, y) = \mathrm{B}(x, y) \big(\psi(x) - \psi(x + y)\big)$$ and by this one has $$\psi(w + 1) - \psi(z + 1) = H_w - H_z$$ so combining these two seems promising.

Solution 3:

If $n$ is an integer you can succesively integrate by parts i.e. \begin{align} &\quad \, \, \int_0^1 x^{n-1} (1-x)^{m-1} \log(1-x) \, {\rm d}x \\ &=\frac{1}{m} \int_0^1 (1-x)^{m} \left\{ -\frac{x^{n-1}}{1-x} + (n-1)x^{n-2}\log(1-x) \right\} {\rm d}x \\ &= -\frac{B(n,m)}{m} + \frac{(n-1)}{m(m+1)} \int_0^1 (1-x)^{m+1} \left\{ -\frac{x^{n-2}}{1-x} + (n-2)x^{n-3}\log(1-x) \right\} {\rm d}x \\ &= -\frac{B(n,m)}{m} - \frac{(n-1) B(n-1,m+1)}{m(m+1)} \\ &\quad + \frac{(n-1)(n-2)}{m(m+1)(m+2)} \int_0^1 (1-x)^{m+2} \left\{ -\frac{x^{n-3}}{1-x} + (n-3)x^{n-4}\log(1-x) \right\} {\rm d}x \\ &= \dots \\ &=-\frac{B(n,m)}{m} - \sum_{i=1}^k\frac{(n-1)\cdots(n-i)B(n-i,m+i)}{m(m+1)\cdots(m+i)} \\ &\quad + \frac{(n-1) \cdots (n-1-k)}{m(m+1)\cdots(m+k)} \int_0^1 (1-x)^{m+k} x^{n-2-k}\log(1-x) \, {\rm d}x \, . \end{align} For $k=n-1$ the last term vanishes, since $\log(1)=0$ in the last boundary term. It is then just a simple matter of fact to use $$\frac{(n-1)\cdots(n-i)B(n-i,m+i)}{m(m+1)\cdots(m+i-1)} = B(n,m)$$ as can be seen from the Gamma representation.

Solution 4:

Following through on @b00n heT's idea, since $$\frac{\partial}{\partial m} \operatorname{B}(n,m) = \operatorname{B} (n,m) \big{(} \psi (m) - \psi (m + n) \big{)},$$ where $\psi (x)$ denotes the digamma function, making use of the fact that $\psi (a) = H_{a - 1} - \gamma$, we see that \begin{align} \psi (m) - \psi (m + n) &= H_{m - 1} - H_{m + n - 1}\\ &= \left (1 + \frac{1}{2} + \cdots + \frac{1}{m - 1} \right ) - \left (1 + \frac{1}{2} + \cdots + \frac{1}{m - 1} + \frac{1}{m} + \cdots + \frac{1}{m + n - 1} \right )\\ &= -\left (\frac{1}{m} + \frac{1}{m + 1} + \cdots + \frac{1}{m + n - 1} \right )\\ &= -\sum_{k = 0}^{n - 1} \frac{1}{k + m}, \end{align} allowing us to arrive at $$\frac{\partial}{\partial m} \operatorname{B} (n,m) = - \operatorname{B} (n,m) \sum_{k = 0}^{n - 1} \frac{1}{k + m},$$ as desired.