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Evaluate \[\displaystyle\int_{0}^{1}\log({\Gamma(x+1)})\,dx\,,\] where \(\Gamma(x)\) is the gamma function.

We will use the identity \(\displaystyle \Gamma(z+1)=z\cdot \Gamma(z)\), so

\[ \int_{0}^{1}\log \left(\Gamma(x+1)\right)\;\mathbb{d}x = \int_{0}^{1}\ln\left(x\cdot \Gamma(x)\right)\;\mathbb{d}x=\int_{0}^{1}\log(x)\;\mathbb{d}x+\int_{0}^{1}\log\left(\Gamma(x)\right)\;\mathbb{d}x=-1+\underbrace{\int_{0}^{1}\log\left(\Gamma(x)\right)\;\mathbb{d}x}_{\rm I}\;\;\left(\color{red}\dagger\right)\] and we will have to compute the integral denoted by I. Using the formula \(\displaystyle \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}\), we will make the "trick" presented in the computations below
\[\begin{eqnarray*} \textrm{I} &=& \int_{0}^{1}\log\left(\Gamma(x)\right)\;\mathbb{d}x\\ &=& \frac{1}{2}\left(\int_{0}^{1}\log\left(\Gamma(x)\right)\;\mathbb{d}x+\int_{0}^{1}\log\left(\Gamma(1-x)\right)\;\mathbb{d}x\right)\\&=&\frac{1}{2}\int_{0}^{1}\log\left(\Gamma(x)\Gamma(1-x)\right)\;\mathbb{d}x\\&=& \frac{1}{2}\int_{0}^{1}\log\left(\frac{\pi}{\sin(\pi x)}\right)\;\mathbb{d}x\\ &=& \frac{1}{2}\int_{0}^{1}\log(\pi)-\log(\sin(\pi x ))\;\mathbb{d}x\\&=& \frac{1}{2}\log(\pi)-\frac{1}{2}\int_{0}^{1}\log(\sin(\pi x ))\;\mathbb{d}x\\&=&\frac{1}{2}\log(\pi)-\frac{1}{2\pi } \underbrace{\int_{0}^{\pi}\log(\sin( x ))\;\mathbb{d}x}_{\textrm{J}}\;\;\left(\color{red}\ddagger\right)\end{eqnarray*}\]
and we are left with the computation of the integral J.
\[\begin{eqnarray*} \textrm{J} &=& \int_{0}^{\pi}\log(\sin( x ))\;\mathbb{d}x\\ &=&2 \int_{0}^{\frac{\pi}{2}}\log(\sin(x))\;\mathbb{d}x\;\;\color{yellow}{(*)}\\ &=&2 \int_{0}^{\frac{\pi}{2}}\log\left(\sin\left(\frac{\pi}{2}-x\right)\right)\;\mathbb{d}x\\ &=&2 \int_{0}^{\frac{\pi}{2}}\log\left(\cos(x)\right)\;\mathbb{d}x\;\;\color{yellow}{(**)}\end{eqnarray*}\]
so \(\displaystyle 2\textrm{J}=2\int_{0}^{\frac{\pi}{2}}\log(\sin(x))+\log(\cos(x))\;\mathbb{d}x\Rightarrow \textrm{J}=\int_{0}^{\frac{\pi}{2}}\log\left(\sin(x)\cos(x)\right)\;\mathbb{d}x=\int_{0}^{\frac{\pi}{2}}\log\left(\frac{\sin(2x)}{2}\right)\;\mathbb{d}x\).
\[\begin{eqnarray*} \textrm{J} &=& \int_{0}^{\frac{\pi}{2}}\log\left(\frac{\sin(2x)}{2}\right)\;\mathbb{d}x\\ &=& \int_{0}^{\frac{\pi}{2}}\log(\sin(2x))\;\;\mathbb{d}x-2\int_{0}^{\frac{\pi}{2}}\log(2)\;\mathbb{d}x\\&=&\frac{1}{2}\underbrace{\int_{0}^{\pi}\log(\sin(x))\;\mathbb{d}x}_{\textrm{J}}-\frac{\pi}{2}\log(2)\end{eqnarray*}\]
and \(\textrm{J}=\frac{1}{2}\textrm{J}-\frac{\pi}{2}\log(2)\Rightarrow \textrm{J}=-\pi\log(2)\). And finally going back to \(\displaystyle \left(\ddagger\right)\), we get \(\displaystyle \textrm{I}=\frac{1}{2}\log(\pi)-\frac{1}{2\pi }\left(-\pi \log(2)\right)=\frac{1}{2}\log(2\pi)\) and plugging this in the equation \(\displaystyle \left(\dagger\right)\) we get
\[\boxed{\displaystyle\int_{0}^{1}\log\left(\Gamma(x+1)\right)\;\mathbb{d}x=\frac{1}{2}\log(2\pi)-1}\]

An alternative calculation based on some "quite heavy" equalities:
\begin{align*}
\displaystyle\int_{0}^{1}\log({\Gamma(x+1)})\,dx&=\int_{0}^{1}-\gamma x +\sum_{n=2}^\infty \frac{\zeta(n)}{n} \, (-x)^{n}\,dx\\
& =-\frac{\gamma}{2}+\sum_{n=2}^\infty \frac{\zeta(n)}{n}\int_{0}^{1}(-x)^{n}\,dx\\
& =-\frac{\gamma}{2}+\sum_{n=2}^\infty \frac{\zeta(n)}{n}\,\frac{(-1)^n}{n+1}\\
& =-\frac{\gamma}{2}+\sum_{n=2}^\infty \frac{(-1)^n\,\zeta(n)}{n\,(n+1)}\\
& =-\frac{\gamma}{2}+\frac{1}{2}\,(\log2+\log\pi+\gamma-2)\\
& =-\frac{1}{2}\,\log(2\pi)-1\,,
\end{align*} where \(\zeta\) is Riemann zeta function and \(\gamma\) is Euler–Mascheroni constant.

Is there a "direct" way computing of $\displaystyle\sum_{n=2}^\infty \frac{(-1)^n\,\zeta(n)}{n\,(n+1)}$ ?

akotronis wrote:Is there a "direct" way computing of $\displaystyle\sum_{n=2}^\infty \frac{(-1)^n\,\zeta(n)}{n\,(n+1)}$ ?

See Series involving Riemann zeta function.