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How may one go about proving

\displaystyle\frac{\Gamma'(s)}{\Gamma(s)}=O(\log|s|),

(away from the poles) directly? By a direct proof, I mean not to go through the usual Stirling formula with the exact error term. The use of a rough form of Stirling's formula is welcome.

1 Answers 1

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You can use the product representation of $\Gamma(z)$, take logs and differentiate the resulting series.

$\Gamma(z) = \dfrac{e^{-\gamma z}}{z} \prod_{n=1}^{\infty} \left(1 + \dfrac{z}{n}\right)^{-1} \ e^{\frac{z}{n}}$

You can find more information here: Polygamma Function and Digamma Function.

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    @J. M.: That is exactly what is bugging me because to extract nontrivial information you need two fairly different views on the same thing. I am feeling these two views (digamma and harmonic numbers) are too close to each other. It is similar to trying to derive Stirling's formula (especially the behavior of the Gamma function at points with large imaginary parts) from estimates on the factorial. But I should stop complaining and try harder.2010-12-25