Evaluation of the integral $\int_0^1 \log{\Gamma(x+1)}\mathrm dx$

Question

As it says in the title, I'd like to know how to solve the definite integral $\int_0^1 \log{\Gamma(x+1)}\mathrm dx$. Mathematica gives the answer $\frac{1}{2}\log (2\pi)-1$ but I have no idea how one would obtain that (also I feel like Stirling's Approximation is somehow involved since it seems to like the number $\log (2\pi)$ and concerns itself with the relationship between exponentials and factorials).

Motivation: This gives a pretty cool relation, even if it probably can't be used for anything. $$ \frac{\sqrt{2\pi}}{e}=e^{\int_o^1 \log(\Gamma(x+1))\mathrm dx}=\lim_{n\to\infty} e^{\frac{1}{n}\sum_{k=0}^n \log(\Gamma(\frac{k}{n}+1))}=\lim_{n\to\infty} \left(\prod_{k=0}^n\Gamma\left(\frac{k}{n}+1\right)\right)^{1/n} $$

TL;DR: How can we solve the integral, is the value related to Stirling, and if so, how?

Answer 1

Note that

$$ \begin{align*}I &= \int_0^1 \log \Gamma(x+1) \; dx = \int_0^1 \log \left(x \Gamma(x) \right)\; dx = \int_0^1 \log(x)dx+\int_0^1 \log\Gamma(x) \; dx \\ &=-1+\int_0^1 \log\Gamma(x) \; dx \tag{1} \end{align*}$$

On the substitution $x\mapsto 1-x$, this becomes

$$ I=-1+\int_0^1 \log\Gamma(1-x) \; dx \tag{2} $$

Averaging $(1)$ and $(2)$,

$$ \begin{align*} I &= -1+\frac{1}{2}\int_0^1 \log\left( \Gamma(x)\Gamma(1-x) \right)\; dx \\ &= -1+\frac{1}{2}\int_0^1 \log \frac{\pi}{\sin(\pi x)}dx \tag{3}\\ &= -1+\frac{1}{2}\log(\pi)-\frac{1}{2}\int_0^1 \log(\sin \pi x) dx \\ &= -1+\frac{1}{2}\log(\pi)-\frac{1}{2\pi}\int_0^\pi \log(\sin x)dx \\ &= -1+\frac{1}{2}\log(\pi)+\frac{1}{2}\log(2) \tag{4}\\ &= \log\sqrt{2\pi}-1 \end{align*} $$

Equation $(3)$ follows from the reflection formula of the Gamma Function. To get equation $(4)$, I used the well known result:

$$\int_0^\pi \log(\sin x)dx=-\pi \log(2)$$