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

Let $z \in \C$ with a strictly positive real part and $\size {\arg z} < \dfrac \pi 2$.
Then: 
:$\ds \map \Ln {\map \Gamma z} = \paren {z - \dfrac 1 2} \map \Ln z - z + \dfrac {\ln 2 \pi} 2 + \sum_{n \mathop = 1}^d \frac {B_{2 n} } {2 n \paren {2 n - 1} z^{2 n - 1} } + \OO \paren {\dfrac 1 {z^{2 d + 1} } }$
where: 
:$\Gamma$ is the gamma function
:$\Ln$ is the principal branch of the complex logarithm
:$B_{2 n}$ is the $2 n$th Bernoulli number
:$\OO$ is big-$\OO$ notation.

Proof:
From Binet's Formula for Logarithm of Gamma Function Formulation 2, we have:
:$\ds \Ln \map \Gamma z = \paren {z - \dfrac 1 2} \Ln z - z + \dfrac 1 2 \ln 2 \pi + 2 \int_0^\infty \dfrac {\map \arctan {t / z} } {e^{2 \pi t} - 1} \rd t$
:Let $A = \paren {z - \dfrac 1 2} \Ln z - z + \dfrac 1 2 \ln 2 \pi$
Then:
{{begin-eqn}}
{{eqn | l = \Ln \map \Gamma z
      | r = A + 2 \int_0^\infty \frac {\map \arctan {t / z} } {e^{2 \pi t} - 1} \rd t
      | c = Binet's Formula for Logarithm of Gamma Function Formulation 2
}}
{{eqn | r = A + \int_0^\infty \paren {\frac 2 {e^{2 \pi t} - 1} } \paren {\frac t z - \frac 1 3 \frac {t^3} {z^3} + \frac 1 5 \frac {t^5} {z^5} - \cdots + \frac {\paren {-1}^{d - 1} } {\paren {2 d - 1} } \frac {t^{2 d - 1} } {z^{2 d - 1} } + \sum_{n \mathop = d + 1}^\infty \frac {\paren {-1}^{n - 1} t^{2 n - 1} } {\paren {2 n - 1} z^{2 n - 1} } } \rd t
      | c = Power Series Expansion for Real Arctangent Function
}}
{{eqn | r = A + \int_0^\infty \paren {\frac 2 {e^{2 \pi t} - 1} } \paren {\frac t z - \frac 1 3 \frac {t^3} {z^3} + \frac 1 5 \frac {t^5} {z^5} - \cdots + \frac {\paren {-1}^{d - 1} } {\paren {2 d - 1} } \frac {t^{2 d - 1} } {z^{2 d - 1} } + \sum_{n \mathop = d}^\infty \frac {\paren {-1}^{\paren {n + 1} - 1 } t^{2 \paren {n + 1} - 1} } {\paren {2 \paren {n + 1} - 1} z^{2 \paren {n + 1} - 1} } } \rd t
      | c = reindexing: $n \to n + 1$
}}
{{eqn | r = A + \int_0^\infty \paren {\frac 2 {e^{2 \pi t} - 1} } \paren {\frac t z - \frac 1 3 \frac {t^3} {z^3} + \frac 1 5 \frac {t^5} {z^5} - \cdots + \frac {\paren {-1}^{d - 1} } {\paren {2 d - 1} } \frac {t^{2 d - 1} } {z^{2 d - 1} } + \sum_{n \mathop = d}^\infty \frac {\paren {-1}^n t^{2 n + 1} } {\paren {2 n + 1} z^{2 n + 1} } } \rd t
      | c = 
}}
{{eqn | r = A + \int_0^\infty \paren {\frac 2 {e^{2 \pi t} - 1} } \paren {\frac t z - \frac 1 3 \frac {t^3} {z^3} + \frac 1 5 \frac {t^5} {z^5} - \cdots + \frac {\paren {-1}^{d - 1} } {\paren {2 d - 1} } \frac {t^{2 d - 1} } {z^{2 d - 1} } + \sum_{n \mathop = d}^\infty \frac {\paren {-1}^n } { z^{2 n + 1} } \int_0^t u^{2 n} \rd u } \rd t
      | c = Primitive of Power
}}
{{eqn | r = A + \int_0^\infty \paren {\frac 2 {e^{2 \pi t} - 1} } \paren {\frac t z - \frac 1 3 \frac {t^3} {z^3} + \frac 1 5 \frac {t^5} {z^5} - \cdots + \frac {\paren {-1}^{d - 1} } {\paren {2 d - 1} } \frac {t^{2 d - 1} } {z^{2 d - 1} } + \int_0^t \sum_{n \mathop = d}^\infty \frac {\paren {-1}^n } { z^{2 n + 1} } u^{2 n} \rd u } \rd t
      | c = Tonelli's Theorem
}}
{{eqn | r = A + \int_0^\infty \paren {\frac 2 {e^{2 \pi t} - 1} } \paren {\frac t z - \frac 1 3 \frac {t^3} {z^3} + \frac 1 5 \frac {t^5} {z^5} - \cdots + \frac {\paren {-1}^{d - 1} } {\paren {2 d - 1} } \frac {t^{2 d - 1} } {z^{2 d - 1} } + \int_0^t \dfrac 1 z \sum_{n \mathop = d}^\infty \paren {\frac {-u^2} {z^2} }^n \rd u } \rd t
      | c = Product of Powers
}}
{{eqn | r = A + \int_0^\infty \paren {\frac 2 {e^{2 \pi t} - 1} } \paren {\frac t z - \frac 1 3 \frac {t^3} {z^3} + \frac 1 5 \frac {t^5} {z^5} - \cdots + \frac {\paren {-1}^{d - 1} } {\paren {2 d - 1} } \frac {t^{2 d - 1} } {z^{2 d - 1} } + \int_0^t \dfrac 1 z \dfrac {\paren {\dfrac {-u^2} {z^2} }^d } {1 + \dfrac {u^2} {z^2} } \rd u } \rd t
      | c = {{Corollary|Sum of Infinite Geometric Sequence|3}}: $\ds \sum_{n \mathop = N}^\infty z^n = \frac {z^N } {1 - z}$
}}
{{eqn | r = A + \int_0^\infty \paren {\frac 2 {e^{2 \pi t} - 1} } \paren {\frac t z - \frac 1 3 \frac {t^3} {z^3} + \frac 1 5 \frac {t^5} {z^5} - \cdots + \frac {\paren {-1}^{d - 1} } {\paren {2 d - 1} } \frac {t^{2 d - 1} } {z^{2 d - 1} } + \int_0^t \dfrac z z \times \dfrac 1 z \dfrac {\paren {\dfrac {-u^2} {z^2} }^d } {1 + \dfrac {u^2} {z^2} } \rd u } \rd t
      | c = multiplying top and bottom by z 
}}
{{eqn | r = A + \int_0^\infty \paren {\frac 2 {e^{2 \pi t} - 1} } \paren {\frac t z - \frac 1 3 \frac {t^3} {z^3} + \frac 1 5 \frac {t^5} {z^5} - \cdots + \frac {\paren {-1}^{d - 1} } {\paren {2 d - 1} } \frac {t^{2 d - 1} } {z^{2 d - 1} } + \frac {\paren {-1}^d } {z^{2 d - 1} } \int_0^t \dfrac {u^{2 d} } {z^2 + u^2 } \rd u } \rd t
      | c =
}}
{{eqn | r = A + \frac 1 z \int_0^\infty \frac {2 t^{2 \paren {1} - 1} } {e^{2 \pi t} - 1} \rd t  -  \frac 1 {3 z^3} \int_0^\infty \frac {2 t^{2 \paren {2} - 1} } {e^{2 \pi t} - 1} \rd t + \frac 1 {5 z^5} \int_0^\infty \frac {2 t^{2 \paren {3} - 1} } {e^{2 \pi t} - 1} \rd t - \cdots + \frac {\paren {-1}^{d - 1} } {\paren {2 d - 1} z^{2 d - 1} } \int_0^\infty \frac {2 t^{2 d - 1} } {e^{2 \pi t} - 1} \rd t + \frac {\paren {-1}^d } {z^{2 d - 1} } \int_0^\infty \paren {\int_0^t \frac {u^{2 d} \rd u} {u^2 + z^2}  } \paren {\frac 2 {e^{2 \pi t} - 1} } \rd t
      | c = 
}}
{{eqn | r = A + \frac {2 \size {B_{2 \paren {1 } } } } {4 \paren {1 } z}  - \frac {2 \size {B_{2 \paren {2 } } } } {4 \paren {2 } 3 z^3 }  + \frac {2 \size {B_{2 \paren {3 } } } } {4 \paren {3 } 5 z^5 } - \cdots + \frac {\paren {-1}^{d - 1} 2 \size {B_{2 d } } } {4 d \paren {2 d - 1} z^{2 d - 1} } + \frac {\paren {-1}^d } {z^{2 d - 1} } \int_0^\infty \paren {\int_0^t \frac {u^{2 d} \rd u} {u^2 + z^2}  } \paren {\frac 2 {e^{2 \pi t} - 1} } \rd t
      | c = Integral Representation of Bernoulli Number: $\ds \dfrac {\size {B_{2 n} } } {4 n} = \int_0^\infty \dfrac {t^{2 n - 1} } {e^{2 \pi t} - 1} \rd t$
}}
{{eqn | r = A + \frac {B_{2 \paren {1 } } } {2 \paren {1 } z}  + \frac {B_{2 \paren {2 } } } {2 \paren {2 } 3 z^3 }  + \frac {B_{2 \paren {3 } } } {2 \paren {3 } 5 z^5 } - \cdots + \frac {B_{2 d } } {2 d \paren {2 d - 1} z^{2 d - 1} } + \frac {\paren {-1}^d } {z^{2 d - 1} } \int_0^\infty \paren {\int_0^t \frac {u^{2 d} \rd u} {u^2 + z^2}  } \paren {\frac 2 {e^{2 \pi t} - 1} } \rd t
      | c = cancelling $2$ and {{Defof|Bernoulli Numbers}}
}}
{{eqn | r = \paren {z - \dfrac 1 2} \Ln z - z + \dfrac 1 2 \ln 2 \pi + \sum_{n \mathop = 1}^d \frac {B_{2 n} } {2 n \paren {2 n - 1} z^{2 n - 1} } + \frac {\paren {-1}^d } {z^{2 d - 1} } \int_0^\infty \paren {\int_0^t \frac {u^{2 d} \rd u} {u^2 + z^2}  } \paren {\frac 2 {e^{2 \pi t} - 1} } \rd t
      | c = 
}}
{{end-eqn}}
Turning our attention now to the remainder term:
:$\ds \frac {\paren {-1}^d } {z^{2 d - 1} }  \int_0^\infty \paren {\int_0^t \frac {u^{2 d} \rd u} {u^2 + z^2}  } \paren {\frac 2 {e^{2 \pi t} - 1} } \rd t$
Let $K_z = \sup {\set {\size {\dfrac {z^2} {u^2 + z^2} } : u \in \R} }$.
Then: 
{{begin-eqn}}
{{eqn | l = \int_0^\infty \paren {\int_0^t \frac {u^{2 d} \rd u} {u^2 + z^2}  } \paren {\frac 1 {e^{2 \pi t} - 1} } \rd t
      | o = \le
      | r = K_z \paren {\size z}^{-2} \int_0^\infty \paren {\int_0^t u^{2 d} \rd u  } \paren {\frac 1 {e^{2 \pi t} - 1} } \rd t
      | c = 
}}
{{eqn | r = \frac {K_z} {\size {z}^2 \paren {2 d + 1} } \int_0^\infty \frac {t^{2 \paren {d + 1} - 1} } {e^{2 \pi t} - 1} \rd t 
      | c = Primitive of Power
}}
{{eqn | o = \le
      | r = \frac {K_z B_{2 \paren {d + 1} } } {4 \paren {d + 1} \paren {2 d + 1} \size {z}^2 }
      | c = Integral Representation of Bernoulli Number: $\ds \dfrac {\size {B_{2 n} } } {4 n} = \int_0^\infty \dfrac {t^{2 n - 1} } {e^{2 \pi t} - 1} \rd t$
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \frac {\paren {-1}^d } {z^{2 d - 1} } \int_0^\infty \paren {\int_0^t \frac {u^{2 d} \rd u} {u^2 + z^2}  } \paren {\frac 2 {e^{2 \pi t} - 1} } \rd t
      | o = \lt
      | r = \frac {\paren {-1}^d } {z^{2 d - 1} } \frac {2 K_z B_{2 \paren {d + 1} } } {4 \paren {d + 1} \paren {2 d + 1} \size {z}^2 }
      | c = 
}}
{{eqn | r = \frac {K_z B_{2 \paren {d + 1} } } {2 \paren {d + 1} \paren {2 d + 1} \size{z}^{2 d + 1} }
      | c = Product of Powers
}}
{{end-eqn}}
We now observe that if $K_z$ is bounded above, then:
:$\ds \lim_{\size z \mathop \to +\infty} \dfrac {K_z B_{2 \paren {d + 1} } } {2 \paren {d + 1} \paren {2 d + 1} \size{z}^{2 d + 1} } = 0$
Next, we observe that: