Prove that $ \int_0^1 x^2 \psi(x) \, dx = \ln\left(\frac{A^2}{\sqrt{2\pi}} \right) $ Basically what the title says:

Prove that  $ \displaystyle \int_0^1 x^2 \psi(x) \, dx = \ln\left(\dfrac{A^2}{\sqrt{2\pi}} \right). $

where $A\approx 1.2824$ denotes the Glaisher–Kinkelin constant and $\psi(x) $ denote the digamma function.
I've tried reflection formula and some Feynmann method on the integrals in the wikipedia but nothing works.
(I've copied this from AoPS because there's no response there)

A little late, but here is yet another solution not shown previously.
I will first show that
$$\int_0^1 x \ln\left(\Gamma(x)\right)\,dx=\frac{1}{4} \ln \frac{2 \pi}{A^4} \tag{1}$$
And then from $(1)$ derive the desired result.

Recall Kummer´s fourier expansion for LogGamma $0<x<1$
$$\ln\left(\Gamma(x)\right)=\frac{\ln\left(2 \pi\right)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos\left(2 \pi k x\right) + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\sin\left(2 \pi k x \right) \tag{2}$$
Then, if we plug $(2)$ in $(1)$
$$
\begin{aligned}
\int_0^1 x \ln\left(\Gamma(x)\right)\,dx&=\frac{\ln 2 \pi}{2}\int_0^1 x\,dx+ \sum_{k=1}^{\infty}\frac{1}{2k}\int_0^1 x \cos\left(2 \pi k x\right)\,dx + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\int_0^1 x \sin\left(2 \pi k x \right)\,dx\\
&=\frac{\ln 2 \pi}{4}+ \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\left(-\frac{1}{2 \pi k}  \right)\\
&=\frac{\ln 2 \pi}{4}-\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{1}{2\pi^2}\sum_{k=1}^{\infty}\frac{\ln\left(2 \pi k\right)}{k^2}\\
&=\frac{\ln 2 \pi}{4}-\frac{\gamma}{12}-\frac{\ln 2 \pi}{2\pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{1}{2\pi^2}\sum_{k=1}^{\infty}\frac{\ln  k}{k^2}\\
&=\frac{\ln 2 \pi}{4}-\frac{\gamma}{12}-\frac{\ln 2 \pi}{12}-\frac{1}{2\pi^2}\sum_{k=1}^{\infty}\frac{\ln k}{k^2}\\
&=\frac{\ln 2 \pi}{6}-\frac{\gamma}{12}+\frac{1}{2\pi^2}\zeta^\prime(2)\\
&=\frac{\ln 2 \pi}{6}-\frac{\gamma}{12}+\frac{\zeta(2)}{2\pi^2} \left(-12 \ln A + \gamma + \ln2 \pi \right)\\
&=\frac{\ln 2 \pi}{6}- \ln A +\frac{\ln2 \pi}{12} \\
&=\frac{\ln 2 \pi}{4}- \ln A  \\
&=\frac{1}{4} \ln \frac{2 \pi}{A^4}  \qquad \blacksquare\\
\end{aligned}
$$
We used that $\zeta^\prime(2) =\zeta(2)\left(-12 \ln A + \gamma + \ln2 \pi \right)$

On the other hand, if we integrate by parts the L.H.S. of $(1)$ we have that ($du=\psi(x)$ and $v=\frac{x^2}{2}$):
$$\int_0^1 x^2 \psi(x)\,dx =-2\int_0^1 x \ln\left(\Gamma(x)\right)\,dx\tag{3}$$
Which imiediatly gives the desired result
$$\boxed{\int_0^1 x^2 \psi(x) \, dx = \ln\left(\dfrac{A^2}{\sqrt{2\pi}} \right)}$$

Appendix
$$
\begin{aligned}
\int_0^1 x \cos(2 \pi kx)\,dx&=\frac{1}{(2 \pi k)^2}\int_0^{2 \pi k} x \cos(x)\,dx \qquad (2 \pi kx \to x)\\
&=\frac{1}{(2 \pi k)^2}\left(x\sin(x)\Big|_0^{2 \pi k} -\int_0^{2 \pi k}\sin(x)\,dx\right)\\
&=\frac{1}{(2 \pi k)^2}\left(-\cos(x)\Big|_0^{2 \pi k}\right)\\
&=\frac{1}{(2 \pi k)^2}\left(-1+1\right)\\
&=0 \qquad \blacksquare
\end{aligned}
$$

$$
\begin{aligned}
\int_0^1 x \sin(2 \pi kx)\,dx&=\frac{1}{(2 \pi k)^2}\int_0^{2 \pi k} x \sin(x)\,dx \qquad (2 \pi kx \to x)\\
&=\frac{1}{(2 \pi k)^2}\left(-x\cos(x)\Big|_0^{2 \pi k} +\int_0^{2 \pi k}\cos(x)\,dx\right)\\
&=\frac{1}{(2 \pi k)^2}\left(-2 \pi k\right)\\
&=-\frac{1}{2 \pi k}\qquad \blacksquare
\end{aligned}
$$
