How to prove $\displaystyle \int_0^1 \ln \left (\Gamma(x) \right) \,dx = -{\zeta'}(0)$?

Question

How do you prove $$\displaystyle \int_0^1 \ln \left (\Gamma(x) \right) \,dx = -{\zeta'}(0)$$ where $\zeta$ is the Riemann Zeta function.

Answer 1

Hint: $$\begin{align*} 2\int_{0}^{1} \ln \Gamma(x) \mathrm{d} x&=\int_{0}^{1} \ln \Gamma(x) \mathrm{d} x+ \int_{0}^{1} \ln \Gamma(1-x) \mathrm{d} x\\[1ex] &=\int_{0}^{1} \ln (\Gamma(x) \Gamma(1-x)) \mathrm{d} x \\[1ex] &=\int_{0}^{1} \ln (2 \pi) \mathrm{d} x-\int_{0}^{1} \ln (2 \sin (\pi x)) \mathrm{d} x \\[1ex] &=\ln (2 \pi). \end{align*}$$ The last step follows from equality $$\begin{equation*} -\ln (2 \sin (\pi x))=\sum_{n=1}^{\infty} \frac{1}{n} \cos (2 n \pi x) \end{equation*}$$ end the fact you can interchange order of sum with integral. At the end it suffices to use Particular values of the Riemann zeta function or Computing the value of $\zeta'(0)$ if you wish to calculate zerta part also.