Integral $\int_0^1 \log \left(\Gamma\left(x+\alpha\right)\right)\,{\rm d}x=\frac{\log\left( 2 \pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha$ 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 $\Gamma(n)=(n-1)!$.
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\begin{align}&\totald{}{\alpha}\int_{0}^{1}
\ln\pars{\Gamma\pars{x + \alpha}}\,\dd x
=\int_{0}^{1}\partiald{\ln\pars{\Gamma\pars{x + \alpha}}}{\alpha}\,\dd x
=\int_{0}^{1}\partiald{\ln\pars{\Gamma\pars{x + \alpha}}}{x}\,\dd x
\\[3mm]&=\ln\pars{\Gamma\pars{1 + \alpha} \over \Gamma\pars{\alpha}}
=\ln\pars{\alpha}
\end{align}

\begin{align}&\int_{0}^{1}\ln\pars{\Gamma\pars{x + \alpha}}\,\dd x
=\alpha\ln\pars{\alpha} - \alpha + \int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x
\end{align}

\begin{align}&\int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x
=\int_{0}^{1}\ln\pars{\Gamma\pars{1 - x}}\,\dd x
=\int_{0}^{1}\ln\pars{\pi \over \Gamma\pars{x}\sin\pars{\pi x}}\,\dd x
\\[3mm]&=\ln\pars{\pi} - \int_{0}^{1}\ln\pars{\sin\pars{\pi x}}\,\dd x
-\int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x
\end{align}

\begin{align}&\int_{0}^{1}\ln\pars{\Gamma\pars{x}}\,\dd x
=\half\,\ln\pars{\pi}
-{1 \over 2\pi}\
\underbrace{\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\,\dd x}_{\ds{-\pi\ln\pars{2}}}
=\half\,\ln\pars{2\pi}
\end{align}
  The above $\ds{\ul{\ln\pars{\sin\pars{\cdots}}\!\mbox{-integral}}}$ is a well known result and it appears frequently in M.SE.

$$\color{#66f}{\large%
\int_{0}^{1}\ln\pars{\Gamma\pars{x + \alpha}}\,\dd x
={\ln\pars{2\pi} \over 2} + \alpha\ln\pars{\alpha} - \alpha}
$$
A: Another way to show $$\int_{0}^{1} \log \Gamma(x+ \alpha) \, dx = \int_{0}^{1} \log \Gamma(x) \, dx + \alpha \log \alpha - \alpha $$
is to rewrite the integral as
$$ \begin{align} \int_{0}^{1} \log \Gamma (x+\alpha) \, dx &= \int_{\alpha}^{\alpha+1} \log \Gamma(u) \, du \\ &= \int_{0}^{1} \log \Gamma (u) \, du + \int_{1}^{\alpha+1} \log \Gamma (u) \, du - \int_{0}^{\alpha} \log \Gamma (u)  \, du \\ &= \int_{0}^{1} \log \Gamma (u) \, du + \int_{0}^{\alpha} \log \Gamma (w+1) \, dw - \int_{0}^{\alpha} \log \Gamma (u) \, du \end{align}$$
and then combine the 2nd and 3rd integrals and use the functional equation $\frac{\Gamma(x+1)}{\Gamma (x)} = x.$
A: 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.
A: Here's a general method you could use to calculate $I(\alpha)$ if you already know $I(0),I(1)$.
After you've differentiated w.r.t. $\alpha$ under the integral, you could always use that
$$(\log\Gamma(x+\alpha))'=-\gamma+\sum_{k \ge1}\frac{1}{k}-\frac{1}{(k+x+\alpha-1)}$$
and differentiate again to give
$$(\log\Gamma(x+\alpha))''=\sum_{k \ge1}\frac{1}{(k+x+\alpha-1)^2}.$$
Thus by Tornelli we swap integral and summation order, giving
$$I''(\alpha)=\sum_{k \ge 1}\int_0^1\frac{dx}{(k+x+\alpha-1)^2}=\sum_{k \ge 1}\frac{1}{(k+\alpha-1)}-\frac{1}{(k+\alpha)}=\frac{1}{\alpha}$$
$$I'(\alpha)=\log(\alpha)+k$$
$$I(\alpha)=\alpha\log(\alpha)+k\alpha+c$$
$$I(\alpha)=\alpha\log(\alpha)+(I(1)-I(0))\alpha+I(0).$$
