Hi I am trying to prove$$
I:=\int_0^1 \log\left(\,\Gamma\left(x+\alpha\right)\,\right)\,{\rm d}x
=\frac{\log\left(2\pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha\,,\qquad \alpha \geq 0.
$$
I am not sure whether to use an integral representation or to somehow use the Euler reflection formula
$$
\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z}
$$
since a previous post used that to solve this kind of integral. Other than this method, we can use the integral representation
$$
\Gamma(t)=\int_0^\infty x^{t-1} e^{-x}\, dx.
$$
Also note $\Gamma(n)=(n-1)!$.
Answer
This one is deceptively simple. Differentiate with respect to $\alpha$ and note that your integrand becomes $\dfrac{\Gamma'(x+\alpha)}{\Gamma(x+\alpha)} $. You can view this also as $(\log\Gamma(x+\alpha))'$ (where the derivative is taken with respect to $x$ now). At this point you have
$$\begin{align}\int_0^1(\log\Gamma(x+\alpha))'dx &= \log\Gamma(x+\alpha)\bigg|_0^1 \\ &= \log\Gamma(1+\alpha)-\log\Gamma(0+\alpha) \\ &= \log(\alpha\Gamma(\alpha))-\log\Gamma(\alpha) \\ &= \log\alpha+\log\Gamma(\alpha)-\log\Gamma(\alpha) \\ &=\log\alpha \end{align}$$
So $I'(\alpha) = \log(\alpha)$ which gives that $I(\alpha) = \alpha\log\alpha-\alpha+C$. To determine the constant of integration, take $\alpha = 0$. This gives
$$I(0) = C = \int_0^1\log\Gamma(x)dx.$$
From here, refer to achille's answer on a different question to evaluate this integral.
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