# Calculating the limit of the derivative of a sum of digamma functions

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.

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When $x$ is large, it is known that $\psi(x) = \ln x + O(\frac1x)$. Therefore$$\frac{ds_n}{dx} = \ln x - \frac34 (\ln x - \ln 2) - \frac1{36} (\ln x - \ln 12) - \frac8{36} (\ln x-\ln 4) + O(1/x),$$ and the right-hand side is $\frac1{36}(45\ln 2+\ln 3) + O(1/x)$.