<s>[INST] How to calculate the value of the special integral I get
$${\left. \frac{\partial ^2}{\partial n^2} \left( \frac{\partial ^2}{\partial m^2} B(m,n) \right) \right|_{m = \frac{1}{2},n = 0}} = \int_0^1 \frac{\ln^2 x \ln^2 (1 - x)}{\sqrt x (1 - x)} \, dx  =\text{ ?}$$
but how to calculate the value of beta function derivative.
where $B(m,n)$ is beta function.
 [/INST] For the integral:
$${\left. \frac{\partial ^2}{\partial n^2} \left( \frac{\partial ^2}{\partial m^2} B(m,n) \right) \right|_{m = \frac{1}{2},n = 0}} = \int_0^1 \frac{\ln^2 x \ln^2 (1 - x)}{\sqrt x (1 - x)} \, dx $$

Since 
\begin{align}
\partial_{m} B(m,n) = B(m,n) \left[ \psi(m) - \psi(m+n) \right]
\end{align}
then
\begin{align}
\partial_{m}^{2} B(m,n) = B(m,n) \left[ \psi'(m) - \psi'(m+n) + \left( \psi(m) - \psi(m+n) \right)^{2} \right]
\end{align}
and 
\begin{align}
\partial_{m}^{2} \left. B(m,n) \right|_{m=\frac{1}{2}} = B(1/2,n) \left[ \psi'(1/2) - \psi'(n+1/2) + \left( \psi(1/2) - \psi(n+1/2) \right)^{2} \right].
\end{align}
Now
\begin{align}
\partial_{n}\partial_{m}^{2} \left. B(m,n) \right|_{m=\frac{1}{2}} &= B(1/2,n) \left[ - \psi''(n+1/2) -2 \psi'(n+1/2) \left( \psi(1/2) - \psi(n+1/2) \right) \right. \\
& \left. + (\psi(n) - \psi(n+1/2))(\psi'(1/2) - \psi'(n+1/2) + \left( \psi(1/2) - \psi(n+1/2) \right)^{2} )\right].
\end{align}
Also 
\begin{align}
\partial_{n}^{2} \partial_{m}^{2} \left. B(m,n) \right|_{m=\frac{1}{2}} &= B(1/2,n) \left[ - \psi'''(n+1/2) -2 \psi''(n+1/2) \left( \psi(1/2) - \psi(n+1/2) \right) \right. \\
& \left. + 2 \left( \psi'(n+1/2)\right)^{2} 
+ (\psi(n) - \psi(n+1/2))( - \psi''(n+1/2) -2 \psi'(n+1/2) \cdot \right. \\
& \left. \left( \psi(1/2) - \psi(n+1/2) \right) )\right]  \\
& + B(1/2,n) (\psi(n) - \psi(n+1/2)) \left[ - \psi''(n+1/2) -2 \psi'(n+1/2) \left( \psi(1/2) - \psi(n+1/2) \right) \right. \\
& \left. + (\psi(n) - \psi(n+1/2))(\psi'(1/2) - \psi'(n+1/2) + \left( \psi(1/2) - \psi(n+1/2) \right)^{2} )\right]
\end{align}
When $n=0$ this becomes
\begin{align}
\partial_{n}^{2} \partial_{m}^{2} \left. B(m,n) \right|_{m=\frac{1}{2}}^{n=0} &= B(1/2,0) \left[ - \psi'''(1/2) + 2 \left( \psi'(1/2)\right)^{2} 
+ 2(\psi(0) - \psi(1/2))( - \psi''(1/2)) \right].  
\end{align}
This is value trying to be sought. Notice that $\Gamma(0) = \infty$ and thus the result is invalid.

Correct Process

Consider the integral
\begin{align}
I =  \int_{0}^{1} \frac{\ln^2 x \ln^2 (1 - x)}{(1-x)\sqrt{x}} \, dx
\end{align}
for which integration by parts, in the form,
\begin{align}
I &=  \int_{0}^{1} \frac{\ln^2 x}{\sqrt{x}} \cdot \frac{\ln^2 (1 - x)}{(1-x)} \, dx \\
&= \left[ - \frac{1}{3} \ln^{3}(1-x) \cdot \frac{\ln^2 x}{\sqrt{x}} \right]_{0}^{1} + \frac{1}{3} \int_{0}^{1} \ln^{3}(-x) \, D\left( x^{-1/2} \, \ln^{2}(x) \right) \, dx \\
&= \frac{1}{3} \int_{0}^{1} \ln^{3}(-x) \, D\left( x^{-1/2} \, \ln^{2}(x) \right) \, dx \\
&= \frac{2}{3} \int_{0}^{1} x^{-3/2} \ln(x) \, \ln^{3}(1-x) \, dx - \frac{1}{6} \int_{0}^{1} x^{-3/2} \ln^{2}(x) \, \ln^{3}(1-x) \, dx \\
&= \frac{2}{3} \partial_{x} \partial_{y}^{3} \left. B(x,y) \right|_{x=-1/2}^{y=1} - \frac{1}{6} \partial_{x}^{2} \partial_{y}^{3} \left. B(x,y) \right|_{x=-1/2}^{y=1}.  
\end{align}
</s>