Prove $\lim_{n\mapsto 0}[(\psi(n)+\gamma)\psi^{(1)}(n)-\frac12\psi^{(2)}(n)]=2\zeta(3)$ How to prove that
$$\lim_{n\mapsto 0}[(\psi(n)+\gamma)\psi^{(1)}(n)-\frac12\psi^{(2)}(n)]=2\zeta(3)\ ?$$
I encountered this limit while I was trying to solve $\int_0^1\frac{\ln x\ln(1-x)}{x(1-x)}dx$ using the derivative of beta function but I have no idea how to tackle this limit.
We know that this integral is very simple :
$$\int_0^1\frac{\ln x\ln(1-x)}{x(1-x)}dx=\int_0^1\frac{\ln x\ln(1-x)}{x}dx+\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{1-x}dx}_{1-x\to x}$$
$$=2\int_0^1\frac{\ln x\ln(1-x)}{x}dx=2\zeta(3)$$
but using integration does not always work for high-power log integrals and beta function would be the right tool but my problem is only when $n\to 0$. Any help would be appreciated.
Note: No solutions using asymptotic expansion please.
 A: We shall be using the following reflection formulas to evalute the limit.

$$\psi_0(1-x)-\psi(x)=\frac{\pi}{\tan \pi x}\\ \psi_1(1-x)+\psi_1(x)=\frac{\pi^2}{\sin^2 \pi x}\\\psi_2(1-x)-\psi_2(x)=\frac{d^2}{dx^2}(\pi\cot\pi x)=\frac{2\pi^3\cot\pi x}{\sin^2\pi x}$$

These give us $$\lim_{x\to 0}\left[\left(\psi(1-x)+\gamma -\frac{\pi}{\tan \pi x}\right)\left(\frac{\pi^2}{\sin^2\pi x}-\psi_1(1-x)\right)-\frac{1}{2}\left(\psi_2(1-x)-\frac{2\pi^3\cot\pi x}{\sin^2\pi x}\right)\right]$$ Making the  use of $\displaystyle \lim_{x\to 0} x^{-1}\sin x =1=\lim_{x\to 0}x^{-1} \tan x$  the last expression can be reduced to
$$\lim_{x\to 0}\left[\left(\psi(1-x)+\gamma -\frac{1}{ x}\right)\left(\frac{1}{ x^2}-\psi_1(1-x)\right)-\frac{1}{2}\left(\psi_2(1-x)-\frac{2}{x^3}\right)\right] \\=-\frac{1}{2}\psi_2(1-x)+\lim_{x\to 0} \left(\frac{\psi(1-x)+\gamma +x\psi_1(1-x)}{x^2}\right)$$ since the latter limit obtained attains $0/0$ form so we evaluate it by L-hopital's rule $$-\frac{1}{2}\psi_2(1-x)+\lim_{x\to 0}\frac{-\psi_1(1-x)+0+\psi_1(1-x)-x\psi_2(1-x)}{2x}=\lim_{x\to 0}\left(-\frac{1}{2}\psi_2(1-x)-\frac{1}{2}\psi_2(1-x)\right)=-\psi_2(1)= -(-1)^{3} 2!\zeta(3,1)=2\zeta(3)$$
A: Solution due to my friend Khalef Ruhemi ( he is not a MSE user ):
Let $f(x)=(\psi(x)+\gamma)\psi^{(1)}(x)-\frac12\psi^{(2)}(x)$
Since
$$\psi(x)=\psi(x+1)-\frac1x$$
$$\Longrightarrow  \psi^{(1)}(x)=\psi^{(1)}(x+1)+\frac1{x^2}$$
$$\Longrightarrow  \psi^{(2)}(x)=\psi^{(2)}(x+1)-\frac{2}{x^3}$$
we have
$$f(x)=(\gamma+\psi(x+1))\psi^{(1)}(x+1)+\frac{\gamma+\psi(x+1)-\frac12x^2\psi^{(2)}(x+1)-x\psi^{(1)}(x+1)}{x^2}$$
Thus,
$$\lim_{x\to 0}f(x)=$$
$$\underbrace{\lim_{x\to 0}(\gamma+\psi(x+1))\psi^{(1)}(x+1)}_{0\ \text{as $\psi(1)=-\gamma$}}+\underbrace{\lim_{x\to 0}\frac{\gamma+\psi(x+1)-\frac12x^2\psi^{(2)}(x+1)-x\psi^{(1)}(x+1)}{x^2}}_{\text{L'Hopital}}$$
$$=\lim_{x\to 0}\frac{\psi^{(1)}(x+1)-x\psi^{(2)}(x+1)-\frac12 x^2\psi^{(3)}(x+1)-\psi^{(1)}(x+1)-x\psi^{(2)}(x+1)}{2x}$$
$$=\lim_{x\to 0}(-\psi^{(2)}(x+1)-\frac14 x\psi^{(3)}(x+1))=-\psi^{(2)}(1)=2\zeta(3)$$
