Taylor Series of Logarithm of Gamma Function

From ProofWiki
Jump to navigation Jump to search

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:

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

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


Proof

From Gamma Difference Equation:

$\map \Gamma {z + 1} = z \map \Gamma z$

Hence:

\(\displaystyle \frac {\d} {\d z} \map \Log {\map \Gamma {z + 1} }\) \(=\) \(\displaystyle \frac {\d} {\d z} \map \Log {z \map \Gamma z}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\d} {\d z} \map \Log {\map \Gamma z} + \map \Log z\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\d} {\d z} \map \Log {\map \Gamma z} + \frac {\d} {\d z} \map \Log z\) Sum Rule for Derivatives
\(\displaystyle \) \(=\) \(\displaystyle \frac {\d} {\d z} \map \Log {\map \Gamma z} + \frac 1 z\) Derivative of Natural Logarithm Function


Successive use of this identity gives us

\(\displaystyle \frac {\d} {\d z} \map \Log {\map \Gamma z}\) \(=\) \(\displaystyle \frac {\d} {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}\)

and thus from the Sum Rule for Derivatives:

\(\displaystyle \frac {\d^n} {\d z^n} \map \Log {\map \Gamma z}\) \(=\) \(\displaystyle \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {z + M} } + \sum_{k \mathop = 0}^{M - 1} \frac {\paren {-1}^n \paren {n - 1}!} {\paren {z + k}^n}\)

From Stirling's Formula for Gamma Function:

\(\displaystyle \map \Log {\map \Gamma {z + 1} }\) \(=\) \(\displaystyle z \map \Log {\map \Gamma z} - z + \frac 1 2 \map \Log z + \map \OO {\frac 1 z}\)
\(\displaystyle \frac {\d} {\d z} \map \Log {\map \Gamma {z + 1} }\) \(=\) \(\displaystyle \map \Log z + \map \OO {\frac 1 z}\)
\(\displaystyle \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {z + 1} }\) \(=\) \(\displaystyle \map \OO {\frac 1 z}\)
\(\displaystyle \frac {\d} {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}\) \(=\) \(\displaystyle \map \Log {z + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} + \map \OO {\frac 1 {z + M} }\)
\(\displaystyle \lim_{M \to \infty} \frac {\d} {\d z} \map \Log {\map \Gamma {z + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}\) \(=\) \(\displaystyle \lim_{M \to \infty} \map \Log {z + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k} + \lim_{M \to \infty} \map \OO {\frac 1 {z + M} }\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{M \to \infty} \map \Log {z + M} - \sum_{k \mathop = 0}^{M - 1} \frac 1 {z + k}\)

which by definition of the Euler-Mascheroni constant:

$\displaystyle \frac {\d} {\d z} \map \Log {\map \Gamma 1} = \lim_{M \to \infty} \frac {\d} {\d z} \map \Log {\map \Gamma {1 + M} } - \sum_{k \mathop = 0}^{M - 1} \frac 1 {1 + k} = -\gamma$


Also:

$\dfrac {\d^{1 + k} } {\d z^{1 + k}} \map \Log {\map \Gamma {z + 1} } = \map \OO {\frac 1 z}$

shows that:

$\displaystyle \lim_{M \to \infty} \frac {\d^{1 + k} } {\d z^{1 + k} } \map \Log {\map \Gamma {M + 1} } = 0$

thus for $n > 1$:

\(\displaystyle \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}\) \(=\) \(\displaystyle \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {1 + M} } + \sum_{k \mathop = 1}^M \frac {\paren {-1}^n \paren {n - 1}!} {k^n}\)
\(\displaystyle \frac {\d^n} {\d z^n} \map \Log {\map \Gamma 1}\) \(=\) \(\displaystyle \lim_{M \to \infty} \frac {\d^n} {\d z^n} \map \Log {\map \Gamma {1 + M} } + \sum_{k \mathop = 1}^M \frac {\paren {-1}^n \paren {n - 1}!} {k^n}\)
\(\displaystyle \) \(=\) \(\displaystyle \map \zeta n \paren {n - 1}! \paren {-1}^n\)

Thus by definition of Taylor series:

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

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

$\blacksquare$