Approximation to Reciprocal times Derivative of Gamma Function

Theorem

Let $\Gamma$ denote the gamma function.

For all $z \in \C$ such that $\cmod {\map \arg z} < \pi - \epsilon, \cmod z > 1$:

$\dfrac {\map {\Gamma'} z} {\map \Gamma z} = \ln z + \map {\OO_\epsilon} {z^{-1} }$

where:

$\map \OO {z^{-1} }$ denotes big-$\OO$ notation

the implied constant depends on $\epsilon$.

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Proof

From Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function:

$\ln \map \Gamma z = \paren {z - \dfrac 1 2} \ln z - z + \dfrac {\ln 2 \pi} 2 + \map \OO {z^{-1} }$

Taking the derivative with respect to $z$:

$(1): \quad \dfrac {\map {\Gamma'} z} {\map \Gamma z} = \ln z - \dfrac 1 {2 z} + \dfrac \d {\d z} \map \OO {z^{-1} }$

Since there exists $\map c \epsilon > 0$ such that:

$\forall \size z > 1: -\dfrac c {\size {z^{-1} } } < \size {\map \OO {z^{-1} } } < \dfrac c {\size {z^{-1} } }$

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it follows directly that the third term in $(1)$ is $\map \OO {z^{-1} }$.

The validity of the material on this page is questionable. In particular: Why? $\dfrac \d {\d z} \map \OO {z^{-1} }$ is not estimated at all You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

$\blacksquare$