I was looking a bit around on the site and disovered (to my surprise) that
$$ \int_0^1 \log \Gamma(x+t)\,\mathrm{d}x = \log \Bigl(\sqrt{2\pi}\Bigr) - t + t \log t $$
and using this and the fact that $B(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)\\$ I think that
$$ \iint_{\Omega} \log \big(B(x,y)\big) \,d\Omega = \log \Bigl(\sqrt{2\pi}\Bigr) + \frac{3}{4} $$
Where $\Omega = [0,1] \times [0,1]$ is the unit cube. Is this correct? Or is there a different way to show this?
I have a cheeseball argument for the first integral. First note that we may write the integrand as follows via series expansion at $t=0$:
$$ \log \Gamma(x+t) = \log\Gamma(x)+\sum_{n=0}^{\infty}\frac{\partial_{t}^{(n+1)} \log \Gamma(x+t)|_{t=0}}{(n+1)!}t^{n+1} $$
Therefore,
$$ \int_{0}^{1}\log \Gamma(x+t)\,dx= \int_{0}^{1}\log \Gamma(x)\,dx + \sum_{n=0}^{\infty}\frac{t^{n+1}}{(n+1)!}\int_{0}^{1} \partial_{t}^{(n+1)} \log \Gamma(x+t)|_{t=0}\,dx $$
Note that $\partial_{t}f(x+t)=\partial_{x}f(x+t)$, and therefore we can replace the differential in the integrand by $\partial_{x}^{(n+1)}$ so that integration simply gives $\partial_{x}^{n} = \partial_{t}^{n}$. Thus:
$$ \int_{0}^{1} \partial_{t}^{(n+1)} \log \Gamma(x+t)|_{t=0}\,dx= \partial_{t}^{(n)} \log \Gamma(x+t)|_{t=0}\bigg|_{x=0}^{x=1} \\ =\partial_{t}^{(n)} \log \Gamma(t+1)|_{t=0} - \partial_{t}^{(n)} \log \Gamma(t)|_{t=0} $$
But then $\Gamma(t+1)=t\Gamma(t)$, so:
$$ \partial_{t}^{(n)} \log \Gamma(t+1)|_{t=0}=\partial_{t}^{(n)} [\log t+\log\Gamma(t)|_{t=0} $$
The terms with $\Gamma(t)$ cancel, leaving $\partial_{t}^{(n)}\log t |_{t=0}$, so the sum is now:
$$ \sum_{n=0}^{\infty}\frac{\partial_{t}^{(n)}\log t |_{t=0}}{(n+1)!}t^{n+1}=\int_{0}^{t}\sum_{n=0}^{\infty}\frac{\partial_{t}^{(n)}\log t |_{t=0}}{n!}t^{n}\,dt=\int_{0}^{t}\log t\,dt=t\log t - t $$
The only remaining task is to find $\int_{0}^{1}\log\Gamma(x)\,dx$, see Theorem 1 here for a short proof that its value is $\log(\sqrt{2\pi})$. Therefore: $$ \int_{0}^{1}\int_{0}^{1}\log(B(x,y))\,dxdy=\int_{0}^{1}\int_{0}^{1}[\log\Gamma(x) + \log\Gamma(y) - \log\Gamma(x+y)]\,dxdy \\ = \int_{0}^{1}[\log\Gamma(y) +y - y\log y]\,dy=\log(\sqrt{2\pi})+\frac{3}{4} $$ Mathematica finds the same result, so I would say it's correct. Also, apparently there are much faster ways of obtaining the first integral, mostly related to the fact that $$ \frac{\partial}{\partial x}\zeta_{H}(x,y)\bigg|_{x=0} = \log\left(\frac{\Gamma(y)}{\sqrt{2\pi}} \right) $$ Where $\zeta_{H}(x, y)$ is the Hurwitz zeta function. I guess maybe if you know a lot about zeta functions (I don't) the first result is obvious from this formula.