Problem
Evaluate
\[I=\int_0^1\ln(\Gamma(x))\,\mathrm dx.\]Solution
Let $x\mapsto1-x$ so
\[I=\int_0^1\ln(\Gamma(1-x))\,\mathrm dx.\]If we take the average of these two integrals and apply the reflection formula, $\Gamma(x)\Gamma(1-x)=\pi/\sin(\pi x)$, we see that
\[\begin{align*}I &=\frac{1}{2}\int_0^1\ln(\Gamma(x)\Gamma(1-x))\,\mathrm dx \\ &=\frac{1}{2}\int_0^1\ln\left(\frac{\pi}{\sin(\pi x)} \right )\,\mathrm dx \\ &=\frac{1}{2}\ln(\pi)-\frac{1}{2}\underset{J}{\underbrace{\int_0^1\ln(\sin(\pi x))\,\mathrm dx}} \\ &=\frac{1}{2}\ln(\pi)-\frac{1}{2}\int_0^1\ln\left(2\sin\left(\frac{\pi x}{2}\right)\cos\left(\frac{\pi x}{2}\right) \right )\,\mathrm dx \\ &=\frac{1}{2}\ln(\pi)-\frac{1}{2}\ln(2)-\int_0^{1/2}\ln(\sin(\pi x)\cos(\pi x))\,\mathrm dx,\qquad x\mapsto 2x \\ &= \frac{1}{2}\ln(\pi)-\frac{1}{2}\ln(2)-\int_0^{1/2}\ln(\sin(\pi x))\,\mathrm dx-\int_0^{1/2}\ln(\cos(\pi x))\,\mathrm dx \\ &= \frac{1}{2}\ln(\pi)-\frac{1}{2}\ln(2)-\frac{1}{2}J-\frac{1}{2}J \\ &= \frac{1}{2}\ln(\pi)-\frac{1}{2}\ln(2)-J.\end{align*}\]We can solve
\[\frac{1}{2}\ln(\pi)-\frac{1}{2}J=\frac{1}{2}\ln(\pi)-\frac{1}{2}\ln(2)-J\]to get
\[\int_0^1\ln(\sin(\pi x))\,\mathrm dx=-\ln(2),\]which tells us
\[\int_0^1\ln(\Gamma(x))\,\mathrm dx=\frac{1}{2}\ln(2\pi).\]Some bonus series
Taking the logarithm of the product
\[\frac{\sin(\pi x)}{\pi x}=\pi x\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2}\right)\]yields the infinite series
\[\ln(\sin(\pi x))=-\ln(\pi)-\ln(x)-\sum_{n=1}^\infty\frac{\zeta(2n)}{n}x^{2n},\qquad |x|<1.\]Integrating this from $0$ to $1$ shows
\[\sum_{n=1}^\infty\frac{\zeta(2n)}{n(2n+1)}=\ln(2\pi)-1.\]Similarly, integrating the series for the log-gamma function,
\[\ln(\Gamma(x+1))=-\gamma x+\sum_{n=2}^\infty\frac{(-1)^n\zeta(n)}{n}x^n,\]results in the identity
\[\sum_{n=2}^\infty\frac{(-1)^n\zeta(n)}{n(n+1)}=\frac{1}{2}\ln(2\pi)+\frac{\gamma}{2}-1.\]