# Complex logarithm for Stirling’s formula

analytic-number-theorycomplex-analysis

In complex analysis, The Stirling formula,

\begin{align*} \Gamma(s) = \left(\frac{2\pi}{s}\right)^{1/2}\left(\frac{s}{e}\right)^{s}\left(1+O\left(\frac{1}{|s|}\right)\right) \end{align*}
is valid in the angle $$|\arg(s)|\leq \pi-\varepsilon$$ with the implied constant depening on $$\varepsilon$$.

Hence for $$s=\sigma+it$$, $$t\ne 0$$, $$\sigma$$ fixed
\begin{align*} \Gamma(\sigma+it) = \sqrt{2\pi} (it)^{\sigma-\frac{1}{2}}e^{-\frac{\pi}{2}|t|} \left(\frac{|t|}{e}\right)^{it}\left(1+O\left(\frac{1}{|t|}\right)\right). \end{align*}

If I consider complex logarithm then I can write

\begin{align*} \log\Gamma(\sigma+it) = \log|\Gamma(\sigma+it)|+i\arg(\Gamma(\sigma+it)) \end{align*}

In absolute value we have
\begin{align*} |\Gamma(\sigma+it)| = \sqrt{2\pi}|t|^{\sigma-\frac{1}{2}}e^{-\frac{\pi}{2}|t|} \left(1+O\left(\frac{1}{|t|}\right)\right). \end{align*}

How to calculate $$\arg(\Gamma(\sigma+it))$$?

When $$\sigma$$ lies in a fixed interval and $$t\to+\infty$$, a convenient version of Stirling's formula (See equation 4.12.1 of Titchmarh's The theory of the Riemann zeta-function) is stated as follows:
$$\log\Gamma(\sigma+it)=\left(\sigma+it-\frac12\right)\log(it)-it+\frac12\log2\pi+\mathcal O\left(1\over t\right)$$
$$\arg\Gamma(\sigma+it)=t\log t-t+\left(\sigma-\frac12\right)\frac\pi2+\mathcal O\left(1\over t\right)$$