TITLE: Taylor series of $\sum_{k=0}^\infty \frac{x^k}{k!} \log(\Gamma(k+1))$
QUESTION [3 upvotes]: Does a the following Taylor series has a closed form expression 
\begin{align}
 \sum_{k=0}^\infty  \frac{x^k}{k!} \log(k!).
\end{align}
Note that $\Gamma(k+1)=k!$.
Also, note that this power series converges (see for example here).

REPLY [1 votes]: This is not answer, but it is a little long for a comment.
Let 
$$I(a)=\sum_{k=0}^{+\infty} \frac{x^k}{k!}\log \Gamma(k+a)$$
Then
$$I'(a)=\sum_{k=0}^{+\infty} \frac{x^k}{k!}\psi^{(0)}(k+a)$$
And because 
$$\psi^{(0)}(k+a)=H_{k+a-1}-\gamma$$
We can find that
\begin{align}
I'(a)&=\sum_{k=0}^{+\infty} \frac{x^k}{k!}H_{k+a-1}-\sum_{k=0}^{+\infty}\gamma \frac{x^k}{k!} \\
&=\psi^{(0)}(1)e^x+\sum_{k=0}^{+\infty} \frac{x^k}{k!}H_{k+a-1}
\end{align}
However, I am not sure how to evaluate the last sum.
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