Taylor Series of Logarithm of Gamma Function

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Theorem

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:

\(\ds \map \Log {\map \Gamma z}\) \(=\) \(\ds -\map \gamma {z - 1} + \sum_{k \mathop = 2}^\infty \frac {\paren {-1}^k \map \zeta k} k \paren {z - 1}^k\)

which is valid for all $z \in \C$ such that $\cmod {z - 1} < 1$.


Proof

\(\ds \map \Gamma {x + 1}\) \(=\) \(\ds x \map \Gamma x\) Gamma Difference Equation
\(\ds \) \(=\) \(\ds \paren {x } \paren {x - 1 } \map \Gamma {x - 1 }\)
\(\ds \) \(=\) \(\ds \paren {x } \paren {x - 1 } \paren {x - 2 } \map \Gamma {x - 2 }\)
\(\ds \) \(=\) \(\ds \paren {x } \paren {x - 1 } \paren {x - 2 } \cdots \paren {x - \floor x } \map \Gamma {x -\floor x }\) For $\floor x$, see the Definition of Floor Function


Hence:

\(\ds \map \Log {\map \Gamma {x + 1} }\) \(=\) \(\ds \map \Log { \paren {x } \paren {x - 1 } \paren {x - 2 } \cdots \paren {x - \floor x } \map \Gamma {x - \floor x } }\)
\(\ds \) \(=\) \(\ds \map \Log x + \map \Log {x - 1 } + \map \Log {x - 2 } + \cdots + \map \Log {x - \floor x } + \map \Log {\map \Gamma {x - \floor x } }\) Sum of Logarithms


Therefore:

\(\ds \frac \d {\d x} \map \Log {\map \Gamma {x + 1} }\) \(=\) \(\ds \frac \d {\d x} \paren {\map \Log x + \map \Log {x - 1} + \map \Log {x - 2 } + \cdots + \map \Log {x - \floor x } + \map \Log {\map \Gamma {x - \floor x} } }\)
\(\ds \) \(=\) \(\ds \frac \d {\d x} \map \Log x + \frac \d {\d x} \map \Log {x - 1} + \frac \d {\d x} \map \Log {x - 2} + \cdots + \frac \d {\d x} \map \Log {x - \floor x} + \frac \d {\d x} \map \Log {\map \Gamma {x - \floor x} }\) Sum Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac 1 x + \frac 1 {x - 1} + \frac 1 {x - 2} + \cdots + \frac 1 {x - \floor x} + \frac \d {\d x} \map \Log {\map \Gamma {x - \floor x} }\) Derivative of Natural Logarithm Function
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{\floor x} \frac 1 {x - k} + \frac \d {\d x} \map \Log {\map \Gamma {x - \floor x} }\)

Setting:

$z = \paren {x - \floor x} \leadsto \d z = \d x$
$M = \floor x + 1$:

we have:

$z + M = x + 1$

Hence:

\(\ds \frac \d {\d x} \map \Log {\map \Gamma {x - \floor x} }\) \(=\) \(\ds \frac \d {\d x} \map \Log {\map \Gamma {x + 1} } - \sum_{k \mathop = 0}^{\floor x} \frac 1 {x - k}\) rearranging $(1)$
\(\ds \frac \d {\d z} \map \Log {\map \Gamma z}\) \(=\) \(\ds \frac \d {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1 } \frac 1 {\paren {z + M - 1} - k}\) substituting with the values above
\(\text {(2)}: \quad\) \(\ds \frac \d {\d z} \map \Log {\map \Gamma z}\) \(=\) \(\ds \frac \d {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1 } \frac 1 {z + k}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d^n} {\d z^n} \map \Log {\map \Gamma z}\) \(=\) \(\ds \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {z + M} } + \sum_{k \mathop = 0}^{M - 1} \frac {\paren {-1}^n \paren {n - 1}!} {\paren {z + k}^n}\) Sum Rule for Derivatives and Nth Derivative of Reciprocal of Mth Power


Recall Stirling's Formula for Gamma Function:

\(\ds \map \Gamma {z + 1}\) \(=\) \(\ds \sqrt {2 \pi z} \, z^z e^{-z} \paren {1 + \dfrac 1 {12 z} + \dfrac 1 {288 z^2} - \dfrac {139} {51 \, 480 z^3} + \cdots}\)
\(\ds \map \Log {\map \Gamma {z + 1} }\) \(=\) \(\ds \frac 1 2 \map \Log {2 \pi } + \frac 1 2 \map \Log z + z \map \Log z - z + \map \log {1 + \dfrac 1 {12 z} + \dfrac 1 {288 z^2} - \dfrac {139} {51 \, 480 z^3} + \cdots}\) Sum of Logarithms and Logarithm of Power/Natural Logarithm
\(\ds \frac \d {\d z} \map \Log {\map \Gamma {z + 1} }\) \(=\) \(\ds 0 + \dfrac 1 {2 z} + \paren {\map \Log z + \dfrac z z } - 1 + \map \OO {\frac 1 z}\) Product Rule for Derivatives and Derivative of Natural Logarithm Function
\(\ds \) \(=\) \(\ds \map \Log z + \map \OO {\frac 1 z}\) For $\map \OO {\frac 1 z}$, see Definition of Big-O Notation
\(\ds \leadsto \ \ \) \(\ds \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {z + 1} }\) \(=\) \(\ds \map \OO {\frac 1 z}\) For $n > 1$


We now have:

$\ds \frac \d {\d z} \map \Log {\map \Gamma {z + 1} } = \map \Log z + \map \OO {\frac 1 z}$


Therefore:

\(\ds \frac \d {\d z} \map \Log {\map \Gamma {z + M } }\) \(=\) \(\ds \map \Log {z + M - 1} + \map \OO {\frac 1 {z + M - 1} }\)
\(\ds \frac \d {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}\) \(=\) \(\ds \map \Log {z + M - 1} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} + \map \OO {\frac 1 {z + M - 1} }\) subtracting $\ds \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}$ from both sides of the equation
\(\ds \lim_{M \mathop \to \infty} \paren {\frac \d {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} }\) \(=\) \(\ds \lim_{M \mathop \to \infty} \paren {\map \Log {z + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} } + \lim_{M \mathop \to \infty} \map \OO {\frac 1 {z + M} }\) Definition of Limit of Real Function, taking the limit as M goes to $\infty$ on both sides of the equation
\(\ds \) \(=\) \(\ds \lim_{M \mathop \to \infty} \paren {\map \Log {z + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} }\) $\ds \lim_{M \mathop \to \infty} \map \OO {\frac 1 {z + M} } = 0$


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 $(2)$:

\(\ds \frac \d {\d z} \map \Log {\map \Gamma 1}\) \(=\) \(\ds \lim_{M \mathop \to \infty} \frac \d {\d z} \map \Log {\map \Gamma {1 + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {1 + k}\)
\(\ds \) \(=\) \(\ds \lim_{M \mathop \to \infty} \map \Log {1 + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {1 + k}\)
\(\ds \) \(=\) \(\ds \lim_{M \mathop \to \infty} \map \Log {1 + M} - \sum_{j \mathop = 1}^{M} \frac 1 j\) setting $j = 1 + k$
\(\ds \) \(=\) \(\ds -\gamma\) Definition of Euler-Mascheroni Constant


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 $n > 1$:


\(\ds \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}\) \(=\) \(\ds \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {1 + M} } + \sum_{k \mathop = 1}^M \frac {\paren {-1}^n \paren {n - 1}!} {k^n}\)
\(\ds \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}\) \(=\) \(\ds \lim_{M \mathop \to \infty} \paren {\frac {\d^n} {\d z^n} \map \Log {\map \Gamma {1 + M} } + \sum_{k \mathop = 1}^M \frac {\paren {-1}^n \paren {n - 1}!} {k^n} }\)
\(\ds \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}\) \(=\) \(\ds \paren {-1}^n \paren {n - 1}! \sum_{k \mathop = 1}^{\infty} \frac 1 {k^n}\)
\(\ds \) \(=\) \(\ds \paren {-1}^n \paren {n - 1}! \map \zeta n\) Definition of Riemann Zeta Function

Thus by definition of Taylor series:

\(\ds \map f z\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \map {f^{\paren n} } a \frac {\paren {z - a}^n} {n!}\)
\(\ds \) \(=\) \(\ds \map f a + \map {f^{\paren 1} } a \frac {\paren {z - a}^1} {1!} + \map {f^{\paren 2} } a \frac {\paren {z - a}^2} {2!} + \cdots\)
\(\ds \map \Log {\map \Gamma z}\) \(=\) \(\ds \map \Log {\map \Gamma 1} - \gamma \paren {z - 1} + \sum_{k \mathop = 2}^\infty \frac {\paren {-1}^k \map \zeta k \paren {k - 1}!} {k!} \paren {z - 1}^k\)
\(\ds \) \(=\) \(\ds -\gamma \paren {z - 1} + \sum_{k \mathop = 2}^\infty \frac{\paren {-1}^k \map \zeta k} k \paren {z - 1}^k\) Gamma of One is One and Natural Logarithm of 1 is 0

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 $\cmod {1 - 0} = 1$.

$\blacksquare$