How can I show that $$\int_{0}^{\frac{1}{2}} x \log\Gamma(x) \mathrm dx = \frac{1}{96} \Big[\log(16 \pi^6 A^{24}) - \frac{21 \zeta(3)}{\pi^2}\Big]≈0.135973$$ where $\Gamma(x)$ is the Gamma function, $\zeta(x)$ is the Zeta function and $A$ is the Glaisher-Kinkelin Constant.
The integral is simply an arbitrary adaptation of some similar integrals found on this site, evaluated using Wolfram alpha. The most famous "log-gamma" integral is due to Raabe, $\int_{0}^{1} \log \Gamma(z) \mathrm dx = \frac{1}{2} \log 2\pi$. Very similar integrals (to this post as well) are found here. My attempts are adapted from a solution given in the previous post.
Integration by parts and taking the appropriate limits gives
$$\int_0^\frac12 t\log\Gamma(t)\,\mathrm dt=-\frac12\int_0^\frac12 t^2\,\psi(t)\,\mathrm dt$$
Consider the identity due to Victor Adamchik, where $B_n$ and $B_n(z)$ are the Bernoulli numbers and polynomials, $H_n=\sum_{j=1}^n\frac1{j}$ is a harmonic number, and $\zeta^\prime(s,a)=\left.\frac{\mathrm d}{\mathrm dt}\zeta(t,a)\right|_{t=s}$ is the derivative of the Hurwitz zeta function.
$$\int_{0}^{z} x^n \psi(x) \mathrm dx = (-1)^n \Big(\frac{B_{n+1}H_n}{n+1} - \zeta'(-n)\Big) + \sum_{k=0}^{n} (-1)^k \binom{n}{k} z^{n-k} \Big(\zeta'(-k, z) - \frac{B_{k+1}(z) H_k}{k+1}\Big)$$
Thus, we need to evaluate the expression for $z=\frac12$ and $n=2$. However, I am unsure how to get the closed form for the original integral (how do I get the $24$th power of the Glaisher-Kinkelin constant).
Another attempt I tried is below was using the Fourier series for $\ln\Gamma(x)$ discovered by E.E. Kummer in 1847. Although we can evaluate some of the integral terms, I am unsure how they evaluate to the closed form given at the very beginning of the post.
$$\ln\Gamma(x)=\frac{\ln 2\pi}{2}+\sum_{n=1}^{\infty}\frac{\cos 2\pi nx}{2n}+\sum_{n=1}^{\infty}\frac{(\gamma+\ln 2\pi n)\sin 2\pi nx}{n\pi}\,(0<x<1)$$
\begin{aligned} \int_{0}^{\frac{1}{2}}x\ln\Gamma(x) \mathrm dx &=\int_{0}^{\frac12} x \frac{\ln 2\pi}{2} dx\\ &+\sum_{n=1}^{\infty}\int_{0}^{\frac12}x\frac{\cos 2\pi nx}{2n} \mathrm dx \\ &+\sum_{n=1}^{\infty}\int_{0}^{\frac12} x \frac{(\gamma+\ln 2\pi n)\sin 2\pi nx}{n\pi} \mathrm dx \end{aligned}
For the fourier series attempt, the difference between the $\int_{0}^{1}$ case and the $\int_{0}^{\frac12}$ case are the trigonometric integrals that have different evaluations. For example, for natural $n$,
$$\int_{0}^{\frac{1}{2}} x \cos(2 \pi n x) \mathrm dx = \frac{\pi n \sin(\pi n) + \cos(\pi n) - 1}{4 \pi^2 n^2}$$
$$\int_{0}^{1} x \cos(2 \pi n x) \mathrm dx = 0$$
This would further complicate the summations; I haven't been able to prove they are equal to the closed form given at the top of this post.
I find this integral quite striking as its closed form includes the 24th power Glaisher-Kinkelin constant, Apery's constant, and some unusual terms! Any input is appreciated, thanks!
