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For any $\delta>0$ I understand the proof of $$ \log \Gamma (s) =(s-1/2)\log s -s +\log \sqrt{2 \pi}+O(|s|^{-1})$$ in $\{s||\arg (s)| < \pi -\delta\}$.

But I cannot prove following corollary.

for any fixed real number $\sigma_1<\sigma _2$ we get $$\log \Gamma (\sigma +it) =(\sigma +it-1/2)\log (it) +\log \sqrt{2\pi}+O(|t|^{-1})$$ in $\{s=\sigma + it | \sigma_1 < \sigma <\sigma_2, |t|\ge 1\}$.

Please teach me this proof.

yuu
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1 Answers1

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Note that $|t|$ is large and $\sigma$ is fixed. We can write, using Stirling's formula, \begin{align*} &\log \Gamma (\sigma + \mathrm{i}t) = \left( {\sigma + \mathrm{i}t - \tfrac{1}{2}} \right)\log (\sigma + \mathrm{i}t) - \sigma - \mathrm{i}t + \log \sqrt {2\pi } + \mathcal{O}(\left| {\sigma + \mathrm{i}t} \right|^{ - 1} ) \\ & = \left( {\sigma + \mathrm{i}t - \tfrac{1}{2}} \right)\log \left( {\mathrm{i}t\left( {1 + \frac{\sigma }{{\mathrm{i}t}}} \right)} \right) - \sigma - \mathrm{i}t + \log \sqrt {2\pi } +\mathcal{O}(\left| {\sigma + \mathrm{i}t} \right|^{ - 1} ) \\ & = \left( {\sigma + \mathrm{i}t - \tfrac{1}{2}} \right)\log (\mathrm{i}t) + \left( {\sigma + \mathrm{i}t - \tfrac{1}{2}} \right)\log \left( {1 + \frac{\sigma }{{\mathrm{i}t}}} \right) - \sigma - \mathrm{i}t + \log \sqrt {2\pi } + \mathcal{O}(\left| {\sigma + \mathrm{i}t} \right|^{ - 1} ). \end{align*} For small $|z|$, we have $$ \log (1 + z) = z + \mathcal{O}(\left| z \right|^2 ). $$ Thus, $$ \left( {\sigma + \mathrm{i}t - \tfrac{1}{2}} \right)\log \left( {1 + \frac{\sigma }{{\mathrm{i}t}}} \right) = \sigma + \mathcal{O}(\left| t \right|^{ - 1} ). $$ Therefore, $$ \log \Gamma (\sigma + \mathrm{i}t) = \left( {\sigma + \mathrm{i}t - \tfrac{1}{2}} \right)\log (\mathrm{i}t) - \mathrm{i}t + \log \sqrt {2\pi } + \mathcal{O}(\left| {\sigma + \mathrm{i}t} \right|^{ - 1} ). $$ Since $|t|$ is large and $\sigma$ is fixed, we also have $$ \mathcal{O}(\left| {\sigma + \mathrm{i}t} \right|^{ - 1} ) = \mathcal{O}(\left| t \right|^{ - 1} ). $$ Consequently, $$ \log \Gamma (\sigma + \mathrm{i}t) = \left( {\sigma + \mathrm{i}t - \tfrac{1}{2}} \right)\log (\mathrm{i}t) - \mathrm{i}t + \log \sqrt {2\pi } + \mathcal{O}(\left| t \right|^{ - 1} ). $$ This is the correct result. In your statement, you are missing the term $-\mathrm{i}t$.

Gary
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