On a log-gamma definite integral A very famous log-gamma integral due to Raabe is
$$\int_0^1 \log \Gamma (x) \, dx = \frac{1}{2} \log (2\pi).$$
Several proofs of this result can be found here.
I would like to known about the evaluation of the similar looking log-gamma integral

$$I = \int_0^1 \frac{\log \Gamma (x)}{1 - x} \, dx.$$

I very much doubt a closed-form solution for this integral can be found, though I would really like to see one if it could be found (we live in hope you see). Here is what I have managed to come up with so far. Integrating by parts gives
$$I = \int_0^1 \log (1 - x) \psi (x) \, dx.$$
Here $\psi (x)$ is the digamma function. Expanding the log term, after swapping the order of the summation with the integration we have
$$I = -\sum_{n = 1}^\infty \frac{1}{n} \int_0^1 x^n \psi (x) \, dx.$$
Now amazingly, a closed form solution for $\int_0^1 x^n \psi (x) \, dx$ exists. It is
$$\int_0^1 x^n \psi (x) \, dx = \sum_{k = 0}^{n - 1} (-1)^k \binom{n}{k} \left (\zeta'(-k) - \frac{B_{k + 1} H_k}{k + 1} \right ).$$
Thus
$$I = \sum_{n = 1}^\infty \sum_{k = 0}^{n - 1} \frac{(-1)^{k + 1}}{n} \binom{n}{k} \left (\zeta'(-k) - \frac{B_{k + 1} H_k}{k + 1} \right ).$$
Here $H_n$ is the $n$th harmonic number, $B_n$ is the $n$th Bernoulli number, while $\zeta' (x)$ denotes the derivative of the Riemann zeta function. I am guessing not much more can be done with these sums.
Any other alternative approaches one could try?
 A: Developing as series around $x=1$, we have
$$\frac{\log (\Gamma (x))}{1-x}=-\sum_{n=0}^\infty \frac{\psi ^{(n)}(1)}{\Gamma (n+2)} (x-1)^n$$
$$\int_0^1\frac{\log (\Gamma (x))}{1-x}\,dx=\sum_{n=0}^\infty (-1)^{n+1}\frac{ \psi ^{(n)}(1)}{(n+1) \Gamma (n+2)}$$
The problem is that the convergence is quite slow. Summing up to $k$, we have
$$\left(
\begin{array}{cc}
 k & \text{partial sum} \\
 10 & 1.32631 \\
 100 & 1.40004 \\
 1000 & 1.40798 \\
\cdots & \cdots \\
\infty & 1.41321
\end{array}
\right)$$
A: By changing $x=1-t$, the integral reads
\begin{align}
 I &= \int_0^1 \frac{\log \Gamma (x)}{1 - x} \, dx\\
 &=\int_0^1\frac{\log \Gamma (1-t)}t \, dt
\end{align}
Then, using the identity (DLMF)
\begin{equation}
 \sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k}t^{k}=-\gamma t+\ln\Gamma\left(1-t\right)
\end{equation}
one obtains, by swapping summation and integral,
\begin{align}
 I&=\int_0^\infty\left[\gamma t+\sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k}t^{k}\right]\frac{dt}{t}\\
 &=\gamma+\sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k^2}
\end{align}
The series can be made more rapidly convergent by introducing the Basel series:
\begin{equation}
 I=\frac{\pi^2}{6}-1+\gamma+\sum_{k=2}^{\infty}\frac{\zeta\left(k\right)-1}{k^2}
\end{equation}
