<s>[INST] How do you prove this integral involving the Glaisher–Kinkelin constant According to wikipedia on the page Glaisher–Kinkelin constant
$$\int_0^{1/2} \ln\Gamma(x) dx=\frac32\ln \text{A}+\frac5{24}\ln 2+\frac14\ln\pi$$
I got interested in how you possibly could prove something like that, but couldn't find any citations about it on the wiki page.
 [/INST] In general, $$ \begin{align} \int_{0}^{z} \log \Gamma(x)  \  dx &= \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z) \\ &= \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} + \ln A - \frac{1}{12} + \zeta^{'}(-1,z) \end{align}$$
where $\zeta(a,z)$ is the Hurwitz zeta function and where I used the fact that $\zeta'(-1) = \frac{1}{12} - \log A$.
You can find a derivation here.
Then $$ \int_{0}^{1/2} \log \Gamma(x) \ dx = \frac{\log 2 \pi}{4} + \frac{1}{24}  + \log A + \zeta' \left( -1, \frac{1}{2} \right) .$$
But using the identity $$\zeta \left( z, \frac{1}{2} \right) = (2^z-1) \zeta(z) \ , \tag{1}$$
we have
$$\zeta' \left( z, \frac{1}{2} \right) = \log(2) 2^{z} \zeta(z) +(2^z-1) \zeta'(z) . $$
And replacing $z$ with $-1$,  $$\begin{align} \zeta' \left( -1, \frac{1}{2} \right) &= \frac{\log 2}{2} \zeta(-1) - \frac{1}{2} \zeta'(-1) \\ &= - \frac{\log 2}{24} + \frac{\log A}{2} - \frac{1}{24} \end{align}$$
where I used the fact that $\zeta(-1) = -\frac{1}{12}$.
Therefore,
$$ \begin{align} \int_{0}^{1/2}\log\Gamma (x)  \ dx  &= \frac{\log 2 \pi}{4} - \frac{\log 2}{24} + \frac{3}{2} \log A \\ &= \frac{3}{2} \log A +  \frac{5}{24} \log 2 + \frac{\log \pi}{4} . \end{align}$$
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$(1)$ http://mathworld.wolfram.com/HurwitzZetaFunction.html (11)
EDIT: 
A proof showing that $\zeta'(-1) = \frac{1}{12} - \ln A$ can be found here.
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