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where $\ \displaystyle\text{B}(n,m)=\int_0^1 x^{n-1}(1-x)^{m-1}\ dx=\frac{\Gamma(n)\Gamma(m)}{\Gamma(n+m)}\ $is the beta function, defined over positive $\ n,m>0$.

The point of this post is to provide a proof for $\ \displaystyle\frac{\partial}{\partial m}\text{B}(n,m)=-\text{B}(n,m)\sum_{k=0}^{n-1}\frac{1}{k+m}$ for, $n$ is a positive integer, so that we can use its' applications as a reference in our solutions and here is some of the applications:

$$\int_0^1x^{n-1}\ln(1-x)\ dx=-\frac{H_n}{n}$$ $$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$ $$\int_0^1x^{n-1}\ln^3(1-x)\ dx=-\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$$ $$\int_0^1x^{n-1}\ln^4(1-x)\ dx=\frac{H_n^4+6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2+6H_n^{(4)}}{n}$$

Its worth to mention that the first two identities were obtained by Cornel and can be found in his paper here using simple integration by parts and clever manipulations, but these three identities and more can be found also in Cornel's book, (Almost) Impossible Integral, Sum, and Series page $59-63$ using only series manipulations.

Also, Ramya showed in his paper here the first three identities using the derivative of $\frac{\partial}{\partial m}\text{B}(n,m)=-\text{B}(n,m)\sum_{k=0}^{n-1}\frac{1}{k+m}$ which we intend here to prove.

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\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}

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(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.

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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.

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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.

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  • $\begingroup$ Nice work Omegadot. I'll provide my solution in couple days. $\endgroup$ – Ali Shather Jun 19 at 6:39
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Here's how to prove the three identities in the post using the definition of beta function and its' derivatives. $$\frac{\partial}{\partial m} \mathrm{B}(n, m)=\int_0^1x^{n-1}(1-x)^{m-1}\ln(1-x)\ dx = \mathrm{B}(m, n) \big(\psi(m) - \psi(m + n)\big)\tag{1}$$ let $m$ approach $1$, we get

$$\int_0^1x^{n-1}\ln(1-x)\ dx = \mathrm{B}(n, 1) \big(\psi(1) - \psi(1+n)\big)=-\frac{H_n}{n}$$

Differentiate $(1)$, we get $$\frac{\partial^2}{\partial m^2} \mathrm{B}(n, m)=\int_0^1x^{n-1}(1-x)^{m-1}\ln^2(1-x)\ dx = \mathrm{B}(m, n)\left(\left(\psi(m)-\psi(n+m)\right)^2-\psi^{(1)}(m+n)+\psi^{(1)}(m)\right)\tag{2}$$ and by letting $m$ approach $1$, we get

$$\int_0^1x^{n-1}\ln^2(1-x)\ dx = \mathrm{B}(n,1)\left(\left(\psi(1)-\psi(1+n)\right)^2-\psi^{(1)}(1+n)+\psi^{(1)}(1)\right)\\=\frac1n\left(H_n^2+H_n^{(2)}\right)$$

Differentiate $(2)$, we get $$\frac{\partial^3}{\partial m^3} \mathrm{B}(n, m)=\int_0^1x^{n-1}(1-x)^{m-1}\ln^3(1-x)\ dx =\small{ \mathrm{B}(m, n)\left(\left(\psi(m)-\psi(m+n)\right)^3+3\left(\psi^{(1)}(m)-\psi^{(1)}(m+n)\right)\left(\psi(m)-\psi(m+n) \right)-\psi^{(2)}(m+n)+\psi^{(2)}(m)\right)}$$

and by letting $m$ approach $1$, we get

$$\int_0^1x^{n-1}\ln^3(1-x)\ dx = \small{\mathrm{B}(1, n)\left(\left(\psi(1)-\psi(1+n)\right)^3+3\left(\psi^{(1)}(1)-\psi^{(1)}(1+n)\right)\left(\psi(m)-\psi(1+n) \right)-\psi^{(2)}(1+n)+\psi^{(2)}(1)\right)}\\=-\frac1n\left(H_n^3+3H_n^{(2)}H_n+2H_n^{(3)}\right)$$

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