TITLE: Calculate integrals involving gamma function
QUESTION [15 upvotes]: What are the usual ways to follow in order to solve the integrals given below? 
$$\begin{align*}
I&=\int_0^1 \ln\Gamma(x)\,dx\\
J&=\int_0^1 x\ln\Gamma(x)\,dx
\end{align*}$$

REPLY [6 votes]: The integral $I$ was mentioned on chat recently, and my solution is different than those given before.
Since $x\Gamma(x)=\Gamma(x+1)$, we have
$$
\int_0^n\log(\Gamma(x))\,\mathrm{d}x+\int_0^n\log(x)\,\mathrm{d}x
=\int_1^{n+1}\log(\Gamma(x))\,\mathrm{d}x\tag{1}
$$
Subtracting $\int_1^n\log(\Gamma(x))\,\mathrm{d}x$ from $(1)$ gives
$$
\int_0^1\log(\Gamma(x))\,\mathrm{d}x+\int_0^n\log(x)\,\mathrm{d}x
=\int_n^{n+1}\log(\Gamma(x))\,\mathrm{d}x\tag{2}
$$
Stirling's approximation says
$$
\log(\Gamma(x))=x\log(x)-x-\frac12\log(x)+\frac12\log(2\pi)+o(1)\tag{3}
$$
Integrating $(3)$ between $n$ and $n+1$ yields
$$
\begin{align}
&\int_n^{n+1}\log(\Gamma(x))\,\mathrm{d}x\\
&=\left[\frac12x^2\log(x)-\frac14x^2-\frac12x^2-\frac12x\log(x)+\frac12x\right]_n^{n+1}+\frac12\log(2\pi)+o(1)\\
&=n\log(n)-n+\frac12\log(2\pi)+o(1)\tag{4}
\end{align}
$$
Furthermore,
$$
\int_0^n\log(x)\,\mathrm{d}x=n\log(n)-n\tag{5}
$$
In light of $(2)$, subtracting $(5)$ from $(4)$ gives
$$
\begin{align}
\int_0^1\log(\Gamma(x))\,\mathrm{d}x
&=\frac12\log(2\pi)+o(1)\\
&=\frac12\log(2\pi)\tag{6}
\end{align}
$$