What's the easiest way to derive an approximation of $ \frac{d}{dz} \log \Gamma(z) $ from the literature?

Question

My main problem is with the error term. I tried to start with Stirling's formula in the form $$ \log\Gamma(z) = ... + \mu(z) $$ with the holomorphic function $$ \mu(z) := - \int_0^\infty \frac{P(t)}{z+t} dt, \quad P(t) := t - \lfloor t \rfloor - \frac{1}{2}. $$ From the source of this approximation I have $ \mu(z) = O(|z|^{-1}) $. It's easy to calculate the derivatives of the dominant terms in the approximation of $ \Gamma(z) $, but I would also need to show that $ \mu'(z) = O(|z|^{-2}) $ holds, which seems unnecessarily complicated. What would be an easier way to obtain such an approximation from the literature? Or is there some textbook where this was already worked out and I can quote it directly?

Answer 1

I am not sure about the literature itself, but here is a fairly simple argument for dealing with your situation:

Let $f : \mathbb{R} \to \mathbb{R}$ be a locally integrable, $1$-periodic function such that $\int_{0}^{1}f(t)\,\mathrm{d}t=0$, and let $\mu$ be defined by

$$ \mu(z) = \int_{0}^{\infty} \frac{f(t)}{z+t} \, \mathrm{d}t. $$

Then by letting $F(x)=\int_{0}^{x}f(t)\,\mathrm{d}t$, we see that $F$ is continuous and $1$-periodic. Also, by integration by parts,

\begin{align*} \mu(z) = \lim_{R\to\infty} \biggl( \left[ \frac{F(t)}{z+t} \right]_{0}^{R} + \int_{0}^{R} \frac{F(t)}{(z+t)^2} \, \mathrm{d}t \biggr) = \int_{0}^{\infty} \frac{F(t)}{(z+t)^2} \, \mathrm{d}t. \end{align*}

This allows to prove the desired claims. For example, this shows that

$$ \mu^{(k)}(z) = (-1)^k (k+1)! \int_{0}^{\infty} \frac{F(t)}{(z+t)^{k+2}} \, \mathrm{d}t, $$

and then for any $z \in \mathbb{C}\setminus(-\infty,0]$, substituting $t = |z|s$ and $z/|z|=e^{i\theta}$ tells that

$$ \left| \mu^{(k)}(z) \right| \leq \frac{(k+1)!(\sup|F|)}{|z|^{k+1}} \int_{0}^{\infty} \frac{1}{\left|e^{i\theta}+s\right|^{k+2}} \, \mathrm{d}s. $$

Then, for each $\delta \in (0, \pi)$ and with $\left|\theta\right|\leq \pi-\delta$, adopting the idea in @Gary's comment shows that

\begin{align*} |e^{i\theta}+s| &= \sqrt{(s+1)^2 - 4s\sin^2(\theta/2)} \\ &\geq \sqrt{(s+1)^2 - (s+1)^2\sin^2(\theta/2)} = (s+1)\cos(\theta/2) \geq (s+1)\sin(\delta/2). \end{align*}

Here, we utilized the fact that $\cos\theta=1-2\sin^2(\theta/2)$ and $4s\leq(s+1)^2$ for $s \geq 0$. Now plugging this to the above bound, we have

$$ \left| \mu^{(k)}(z) \right| \leq \frac{C_{k,f,\delta}}{|z|^{k+1}} \quad\text{whenever}\quad \left|\arg(z)\right|<\pi-\delta $$

with the constant $C_{k,f,\delta}$ given by

$$ C_{k,f,\delta} =\frac{(k+1)!(\sup|F|)}{\sin^{k+2}(\delta/2)} \int_{0}^{\infty} \frac{\mathrm{d}s}{(s+1)^{k+2}} = \frac{k!(\sup|F|)}{\sin^{k+2}(\delta/2)}.$$