How to solve $\frac{\partial{\rm B}}{\partial b}\left(0^+,1\right)=-\frac{\pi^2}{6}$ Could you help me to prove

$$\frac{\partial{\rm B}}{\partial b}\left(0^+,1\right)=-\frac{\pi^2}{6}$$

where ${\rm B}(a,b)$ is Beta function.
 A: Knowing that
$$\text{B}\,(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
and
$$\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}\quad\Longrightarrow\quad\Gamma'(x)=\psi(x)\Gamma(x)$$
where $\Gamma(x)$ is gamma function and $\psi(x)$ is digamma function, then
$$\begin{align}\require\cancel\frac{\partial\text{B}}{\partial b}&=\frac{[\Gamma(a)\Gamma(b)]'\cdot\Gamma(a+b)-\Gamma'(a+b)\cdot\Gamma(a)\Gamma(b)}{\Gamma^2(a+b)}\\&=\frac{[\Gamma'(a)\Gamma(b)+\Gamma(a)\Gamma'(b)]\Gamma(a+b)-\Gamma'(a+b)\cdot\Gamma(a)\Gamma(b)}{\Gamma(a+b)\Gamma(a+b)}\\&=\frac{[0+\Gamma(a)\psi(b)\Gamma(b)]\cancel{\Gamma(a+b)}-\psi(a+b)\cancel{\Gamma(a+b)}\cdot\Gamma(a)\Gamma(b)}{\Gamma(a+b)\cancel{\Gamma(a+b)}}\\&=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\big[\psi(b)-\psi(a+b)\big]\\&=\text{B}\,(a,b)\big[\psi(b)-\psi(a+b)\big]\end{align}$$
Hence
$$\begin{align}\frac{\partial\text{B}}{\partial b}\left(0^+,1\right)&=\lim_{a\to0^+}\text{B}\,\left(a,1\right)\big[\psi(1)-\psi(a+1)\big]\\&=\lim_{a\to0^+}\frac{1}{a}\left[-\gamma+\gamma -\sum_{k=1}^\infty (-1)^{k+1}\zeta (k+1)\;a^k\right]\tag{1}\\&=-\lim_{a\to0^+}\sum_{k=1}^\infty (-1)^{k+1}k\,\,\zeta (k+1)\;a^{k-1}\tag{2}\\&=-\lim_{a\to0^+}\left(\zeta(2)-2\zeta(3)a+3\zeta(4)a^2-4\zeta(5)a^3+\cdots\right)\tag{3}\\&=-\frac{\pi^2}{6}\tag{4}\end{align}$$

Explanation :
$(1)$ Use series representations of $\displaystyle\text{B}\,\left(x,1\right)=\frac{1}{x}$ and $\displaystyle\psi(x+1)=-\gamma +\sum_{k=1}^\infty (-1)^{k+1}\zeta (k+1)\;x^k$ for $|x|<1$
$(2)$ Use L'Hôpital's rule because of indeterminate form $\dfrac{0}{0}$
$(3)$ Expanding the series form
$(4)$ Use $\zeta(2)=\dfrac{\pi^2}{6}$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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With $\ds{a\ >\ 0}$ and $\ds{b\ >\ 0}$:

$$
{\rm B}\pars{a,b}\equiv\int_{0}^{1}t^{a - 1}\pars{1 - t}^{b - 1}\,\dd t
$$

$$
\partiald{{\rm B}\pars{a,b}}{b}
=\int_{0}^{1}t^{a - 1}\pars{1 - t}^{b - 1}\ln\pars{1 - t}\,\dd t
$$

\begin{align}
\left.\partiald{{\rm B}\pars{a,b}}{b}\right\vert_{\, b\ =\ 1}
&=\int_{0}^{1}t^{a - 1}\ln\pars{1 - t}\,\dd t
=\int_{0}^{1}t^{a - 1}\bracks{-\sum_{n\ =\ 1}^{\infty}{t^{n} \over n}}\,\dd t
\\[5mm]&=-\sum_{n\ =\ 1}^{\infty}{1 \over n}\int_{0}^{1}t^{n + a - 1}\,\dd t
=-\sum_{n\ =\ 1}^{\infty}{1 \over n\pars{n + a}}
\end{align}

$$\color{#66f}{\large%
\lim_{a\ \to\ 0^{+}}\left.\partiald{{\rm B}\pars{a,b}}{b}\right\vert_{\, b\ =\ 1}}
=-\sum_{n\ =\ 1}^{\infty}{1 \over n^{2}}
=\color{#66f}{\large -\,{\pi^{2} \over 6}}
$$
A: Here is another slightly different approach.
As Venus showed in her answer, the partial derivative of the beta function $\text{B} (a,b)$ with respect to $b$ is 
$$\partial_b \text{B}(a,b) = \text{B} (a,b) \left [\psi (b) - \psi (a + b) \right ],$$
where $\psi (x)$ is the digamma function. 
When evaluated at $b = 1$ we have
$$\partial_b \text{B} (a,b) \Big{|}_{b = 1} = \text{B} (a,1) \left [\psi (1) - \psi (1 + a) \right ].$$
Now observe that
$$\text{B} (a,1) = \frac{\Gamma (a) \Gamma (1)}{\Gamma (a + 1)} = \frac{\Gamma (a)}{a \Gamma (a)} = \frac{1}{a}.$$
Thus
$$\lim_{a \to 0^+} \partial_b \text{B} (a,1) \Big{|}_{b = 1} = \lim_{a \to 0^+} \frac{\psi (1) - \psi(1 + a)}{a}.$$
The limit is indeterminate and of the form $\frac{0}{0}$. So for its evaluation we may apply l'Hôpital's rule. Doing so yields
$$\lim_{a \to 0^+} \partial_b \text{B} (a,1) \Big{|}_{b = 1} = \lim_{a \to 0^+} -\psi^{(1)}(1 + a) = -\psi^{(1)}(1),$$
where we note the derivative of the digamma function $\psi (a)$ gives the first order polygamma gamma (or trigamma) function $\psi^{(1)} (a)$.
We now find the value for $\psi^{(1)} (1)$. The polygamma function of order $m$ is defined by
$$\psi^{(m)} (z) = (-1)^{m + 1} m! \sum_{n = 0}^\infty \frac{1}{(z + n)^{m + 1}}.$$
To find the value of $\psi^{(1)}(1)$, setting $m = 1$ and $z = 1$ gives
$$\psi^{(1)} (1) = \sum^\infty_{n = 0} \frac{1}{(n + 1)^2}.$$
Shifting the index in the sum by $n \mapsto n - 1$ gives
$$\psi^{(1)}(1) = \sum^\infty_{n = 1} \frac{1}{n^2} = \frac{\pi}{6},$$
the sum corresponding to the very famous Basel problem.
So finally for our limit we have
$$\lim_{a \to 0^+} \partial_b \text{B} (a,1) \Big{|}_{b = 1} = - \frac{\pi^2}{6},$$
as required.
