The (real) digamma function is new to me and I notice that under some circumstances a sum of $\log\Gamma$ functions will have a derivative consisting of a sum of digamma functions that converge to a constant as x goes to infinity.
It appears true, for example, that $s_n= \sum\limits_{k=1}^{n}a_k \log\Gamma(b_k x) $ will have a derivative of the sort above if $\sum\limits_{k=1}^{n}a_kb_k = 0$. Then $\lim\limits_{x\to\infty}^{}\frac{ds_n}{dx} = $ constant.
An example:
$s_n = \log\Gamma(x)- \frac{3}{2}\log\Gamma(\frac{x}{2})-\frac{1}{3}\log\Gamma(\frac{x}{12})-\frac{8}{9}\log\Gamma(\frac{x}{4}). $
The sum
$\sum a_kb_k = 1 - \frac{3}{4}-\frac{1}{36}-\frac{8}{36} = 0 $
And, writing $\psi(x)$ for the digamma function,
$\frac{ds_n}{dx} = \psi(x) - \frac{3}{4}\psi(\frac{x}{2})-\frac{1}{36}\psi(\frac{x}{12})-\frac{8}{36}\psi(\frac{x}{4}) $.
Here is my question. If I ask Mathematica to find $\lim\limits_{x\to \infty}\frac{ds_n}{dx}$, it offers:
$\frac{1}{36}(45\log(2)+\log(3))$. Can someone explain how this number was calculated? Thanks.