integration - Integral $int_0^1 log left(Gammaleft(x+alpharight)right),{rm d}x=frac{logleft( 2 piright)}{2}+alpha logleft(alpharight) -alpha$

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.