Theorem: Taylor Series of Logarithm of Gamma Function

Let $\gamma$ denote the Euler-Mascheroni constant.
Let $\map \zeta s$ denote the Riemann zeta function.
Let $\map \Gamma z$ denote the gamma function.
Let $\Log$ denote the natural logarithm.
Then $\map \Log {\map \Gamma z}$ has the power series expansion:
{{begin-eqn}}
{{eqn | l = \map \Log {\map \Gamma z}
      | r = -\map \gamma {z - 1} + \sum_{k \mathop = 2}^\infty \frac {\paren {-1}^k \map \zeta k} k \paren {z - 1}^k 
}}
{{end-eqn}}
which is valid for all $z \in \C$ such that $\cmod {z - 1} < 1$.

Proof:
{{begin-eqn}}
{{eqn | l = \map \Gamma {x + 1}
      | r = x \map \Gamma x
      | c = Gamma Difference Equation
}}
{{eqn | r = \paren {x } \paren {x - 1 } \map \Gamma {x - 1 }
      | c = 
}}
{{eqn | r = \paren {x } \paren {x - 1 } \paren {x - 2 } \map \Gamma {x - 2 }
      | c = 
}}
{{eqn | r = \paren {x } \paren {x - 1 } \paren {x - 2 } \cdots \paren {x - \floor x } \map \Gamma {x -\floor x }
      | c = For $\floor x$, see the {{Defof|Floor Function}}
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \map \Log {\map \Gamma {x + 1} }
      | r = \map \Log { \paren {x } \paren {x - 1 } \paren {x - 2 } \cdots \paren {x - \floor x } \map \Gamma {x - \floor x } }
      | c = 
}}
{{eqn | r = \map \Log x + \map \Log {x - 1 } + \map \Log {x - 2 } + \cdots + \map \Log {x - \floor x } + \map \Log {\map \Gamma {x - \floor x } }
      | c = Sum of Logarithms
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} \map \Log {\map \Gamma {x + 1} }
      | r = \frac \d {\d x} \paren {\map \Log x + \map \Log {x - 1} + \map \Log {x - 2 } + \cdots + \map \Log {x - \floor x } + \map \Log {\map \Gamma {x - \floor x} } }
      | c = 
}}
{{eqn | r = \frac \d {\d x} \map \Log x + \frac \d {\d x} \map \Log {x - 1} + \frac \d {\d x} \map \Log {x - 2} + \cdots + \frac \d {\d x} \map \Log {x - \floor x} + \frac \d {\d x} \map \Log {\map \Gamma {x - \floor x} }
      | c = Sum Rule for Derivatives
}}
{{eqn | r = \frac 1 x + \frac 1 {x - 1} + \frac 1 {x - 2} + \cdots + \frac 1 {x - \floor x}  + \frac \d {\d x} \map \Log {\map \Gamma {x - \floor x} }
      | c = Derivative of Natural Logarithm Function
}}
{{eqn | n = 1
      | r = \sum_{k \mathop = 0}^{\floor x} \frac 1 {x - k}  + \frac \d {\d x} \map \Log {\map \Gamma {x - \floor x} }
      | c = 
}}
{{end-eqn}}
Setting:
:$z = \paren {x - \floor x} \leadsto \d z = \d x$
:$M = \floor x + 1$:
we have:
:$z + M = x + 1$
Hence:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} \map \Log {\map \Gamma {x - \floor x} }
      | r = \frac \d {\d x} \map \Log {\map \Gamma {x + 1} } - \sum_{k \mathop = 0}^{\floor x} \frac 1 {x - k}
      | c = rearranging $(1)$
}}
{{eqn | l = \frac \d {\d z} \map \Log {\map \Gamma z}
      | r = \frac \d {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1 } \frac 1 {\paren {z + M - 1} - k}
      | c = substituting with the values above
}}
{{eqn | n = 2
      | l = \frac \d {\d z} \map \Log {\map \Gamma z}
      | r = \frac \d {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1 } \frac 1 {z + k}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma z}
      | r = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {z + M} } + \sum_{k \mathop = 0}^{M - 1} \frac {\paren {-1}^n \paren {n - 1}!} {\paren {z + k}^n}
      | c = Sum Rule for Derivatives and Nth Derivative of Reciprocal of Mth Power
}}
{{end-eqn}}
Recall Stirling's Formula for Gamma Function:
{{begin-eqn}}
{{eqn | l = \map \Gamma {z + 1}
      | r = \sqrt {2 \pi z} \, z^z e^{-z} \paren {1 + \dfrac 1 {12 z} + \dfrac 1 {288 z^2} - \dfrac {139} {51 \, 480 z^3} + \cdots}
      | c = 
}}
{{eqn | l = \map \Log {\map \Gamma {z + 1} }
      | r = \frac 1 2 \map \Log {2 \pi } + \frac 1 2 \map \Log z + z \map \Log z - z + \map \log {1 + \dfrac 1 {12 z} + \dfrac 1 {288 z^2} - \dfrac {139} {51 \, 480 z^3} + \cdots}
      | c = Sum of Logarithms and Logarithm of Power/Natural Logarithm
}}
{{eqn | l = \frac \d {\d z} \map \Log {\map \Gamma {z + 1} }
      | r = 0 + \dfrac 1 {2 z} + \paren {\map \Log z + \dfrac z z } - 1 + \map \OO {\frac 1 z}
      | c = Product Rule for Derivatives and Derivative of Natural Logarithm Function
}}
{{eqn | r = \map \Log z + \map \OO {\frac 1 z}
      | c = For $\map \OO {\frac 1 z}$, see {{Defof|Big-O Notation}}
}}
{{eqn | ll = \leadsto
      | l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {z + 1} }
      | r = \map \OO {\frac 1 z}
      | c = For $n > 1$
}}
{{end-eqn}}
We now have:
:$\ds \frac \d {\d z} \map \Log {\map \Gamma {z + 1} } = \map \Log z + \map \OO {\frac 1 z}$
Therefore:
{{begin-eqn}}
{{eqn | l = \frac \d {\d z} \map \Log {\map \Gamma {z + M } }
      | r = \map \Log {z + M - 1} + \map \OO {\frac 1 {z + M - 1} }
      | c = 
}}
{{eqn | l = \frac \d {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}
      | r = \map \Log {z + M - 1} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} + \map \OO {\frac 1 {z + M - 1} }
      | c = subtracting $\ds \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}$ from both sides of the equation
}}
{{eqn | l = \lim_{M \mathop \to \infty} \paren {\frac \d {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} }
      | r = \lim_{M \mathop \to \infty} \paren {\map \Log {z + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} } + \lim_{M \mathop \to \infty} \map \OO {\frac 1 {z + M} }
      | c = {{Defof|Limit of Real Function}}, taking the limit as M goes to $\infty$ on both sides of the equation 
}}
{{eqn | r = \lim_{M \mathop \to \infty} \paren {\map \Log {z + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} }
      | c = $\ds \lim_{M \mathop \to \infty} \map \OO {\frac 1 {z + M} } = 0$
}}
{{end-eqn}}
We now have:
:$\ds \lim_{M \mathop \to \infty} \frac \d {\d z} \map \Log {\map \Gamma {z + M} } = \lim_{M \mathop \to \infty} \map \Log {z + M} $
Setting $z = 1$ into $(2)$:
{{begin-eqn}}
{{eqn | l = \frac \d {\d z} \map \Log {\map \Gamma 1}
      | r = \lim_{M \mathop \to \infty} \frac \d {\d z} \map \Log {\map \Gamma {1 + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {1 + k}
      | c = 
}}
{{eqn | r = \lim_{M \mathop \to \infty} \map \Log {1 + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {1 + k}
      | c = 
}}
{{eqn | r = \lim_{M \mathop \to \infty} \map \Log {1 + M} - \sum_{j \mathop = 1}^{M} \frac 1 j
      | c = setting $j = 1 + k$
}}
{{eqn | r = -\gamma
      | c = {{Defof|Euler-Mascheroni Constant}}
}}
{{end-eqn}}
Also:
:$\ds \frac {\d^{1 + k} } {\d z^{1 + k}} \map \Log {\map \Gamma {z + 1} } = \map \OO {\frac 1 z}$
shows that:
:$\ds \lim_{M \mathop \to \infty} \frac {\d^{1 + k} } {\d z^{1 + k} } \map \Log {\map \Gamma {M + 1} } = 0$
thus for $n > 1$:
{{begin-eqn}}
{{eqn | l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}
      | r = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {1 + M} } + \sum_{k \mathop = 1}^M \frac {\paren {-1}^n \paren {n - 1}!} {k^n}
      | c = 
}}
{{eqn | l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}
      | r = \lim_{M \mathop \to \infty} \paren {\frac {\d^n} {\d z^n} \map \Log {\map \Gamma {1 + M} } + \sum_{k \mathop = 1}^M \frac {\paren {-1}^n \paren {n - 1}!} {k^n} }
}}
{{eqn | l = \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}
      | r = \paren {-1}^n \paren {n - 1}! \sum_{k \mathop = 1}^{\infty} \frac 1 {k^n}
}}
{{eqn | r = \paren {-1}^n \paren {n - 1}! \map \zeta n
      | c = {{Defof|Riemann Zeta Function}}
}}
{{end-eqn}}
Thus by definition of Taylor series:
{{begin-eqn}}
{{eqn | l = \map f z
      | r = \sum_{n \mathop = 0}^\infty \map {f^{\paren n} } a \frac {\paren {z - a}^n} {n!}
}}
{{eqn | r = \map f a + \map {f^{\paren 1} } a \frac {\paren {z - a}^1} {1!} + \map {f^{\paren 2} } a \frac {\paren {z - a}^2} {2!} + \cdots
}}
{{eqn | l = \map \Log {\map \Gamma z}
      | r = \map \Log {\map \Gamma 1} - \gamma \paren {z - 1} + \sum_{k \mathop = 2}^\infty \frac {\paren {-1}^k \map \zeta k \paren {k - 1}!} {k!} \paren {z - 1}^k 
}}
{{eqn | r = -\gamma \paren {z - 1} + \sum_{k \mathop = 2}^\infty \frac{\paren {-1}^k \map \zeta k} k \paren {z - 1}^k
      | c = Gamma of One is One and Natural Logarithm of 1 is 0
}}
{{end-eqn}}
From Zeroes of Gamma Function, we see that $\map \Gamma z$ is non-zero everywhere.
Thus $\map \Log {\map \Gamma z}$ has poles only where $\Gamma$ does, that is, the negative integers.
Since the radius of convergence of a power series is equal to the distance of its center to the closest point where the function is not analytic:
The radius of convergence of $\map \Log {\map \Gamma z}$ is $\cmod {1 - 0} = 1$.
{{qed}}
Category:Gamma Function
Category:Natural Logarithms
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