Theorem: Gauss's Integral Form of Digamma Function

:$\ds \map \psi z = \int_0^\infty \paren {\frac {e^{-t} } t - \frac {e^{-z t} } {1 - e^{-t} } } \rd t$

Proof:
From Extension of Harmonic Number to Non-Integer Argument, we have:
:$\map H x = \gamma + \dfrac {\map {\Gamma'} {x + 1} } {\map \Gamma {x + 1} }$
which is equivalent to:
:$\ds \map \psi z = -\gamma + H_{z - 1}$
We aim to demonstrate:
:$\ds \int_0^\infty \paren {\frac {e^{-t} } t - \frac {e^{-z t} } {1 - e^{-t} } } \rd t = -\gamma + H_{z - 1}$
We have:
{{begin-eqn}}
{{eqn | l = \map \psi z 
      | r = \int_0^\infty \paren {\frac {e^{-t} } t - \frac {e^{-z t} } {1 - e^{-t} } } \rd t
      | c = 
}}
{{eqn | r = \int_0^\infty \frac {e^{-t} } t  \rd t - \int_0^\infty \frac {e^{-z t} } {1 - e^{-t} } \rd t
      | c = Linear Combination of Integrals
}}
{{eqn | r = \int_0^\infty \frac {e^{-t} } t  \rd t - \int_0^\infty \frac {e^{-t} } {1 - e^{-t} } \rd t  + \int_0^\infty \frac {e^{-t} } {1 - e^{-t} } \rd t  - \int_0^\infty \frac {e^{-z t} } {1 - e^{-t} } \rd t
      | c = adding $0$
}}
{{eqn | r = \int_0^\infty \frac {e^{-t} } t  \rd t - \int_0^\infty \frac {e^{-t} } {1 - e^{-t} } \rd t + \int_0^\infty \frac {e^{-t} - e^{-t z } } {1 - e^{-t} } \rd t
      | c = Linear Combination of Integrals
}}
{{eqn | r = \int_0^\infty \frac {e^{-t} } t  \rd t - \int_0^\infty \frac {e^{-t} } {1 - e^{-t} } \times \frac {e^t} {e^t} \rd t + \int_0^\infty \frac {e^{-t} - e^{-t z } } {1 - e^{-t} } \rd t
      | c = multiplying by $1$
}}
{{eqn | r = \int_0^\infty \frac {e^{-t} } t  \rd t - \int_0^\infty \frac 1 {e^t - 1 } \rd t + \int_0^\infty \frac {e^{-t} - e^{-t z } } {1 - e^{-t} } \rd t
      | c = 
}}
{{eqn | r = \int_0^\infty \paren {\frac {e^{-t} } t - \frac 1 {e^t - 1 } } \rd t + \int_0^\infty \frac {e^{-t} - e^{-t z } } {1 - e^{-t} } \rd t
      | c = Linear Combination of Integrals
}}
{{eqn | r = -\gamma + \int_0^\infty \paren {\frac {1 - e^{-t\paren {z - 1} } } {1 - e^{-t} } } e^{-t} \rd t
      | c = Definite Integral to Infinity of Reciprocal of Exponential of x minus One minus Exponential of -x over x
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = e^{-t}
      | c = Integration by Substitution
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d t}
      | r = -e^{-t}
      | c = Derivative of Exponential Function: Corollary 1
}}
{{end-eqn}}
We also have:
{{begin-eqn}}
{{eqn | l = \lim_{t \mathop \to 0} e^{-t}
      | r = 1
      | c = Exponential of Zero
}}
{{eqn | l = \lim_{t \mathop \to \infty} e^{-t}
      | r = 0
      | c = Complex Exponential Tends to Zero
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \map \psi z
      | r = -\gamma + \int_{\to 1}^{\to 0} \paren {\dfrac {1 - u^{z - 1} } {1 - u} } \paren {-\rd u}
      | c = 
}}
{{eqn | r = -\gamma + \int_{\to 0}^{\to 1} \paren {\dfrac {1 - u^{z - 1} } {1 - u} } \rd u
      | c = reversing limits of integration
}}
{{eqn | r = -\gamma + \int_{\to 0}^{\to 1} \sum_{k \mathop = 0}^{z - 2} u^k
      | c = Sum of Geometric Sequence
}}
{{eqn | r = -\gamma + \sum_{k \mathop = 0}^{z - 2} \int_{\to 0}^{\to 1} u^k
      | c = Tonelli's Theorem
}}
{{eqn | r = -\gamma + \sum_{k \mathop = 0}^{z - 2} \intlimits {\dfrac {u^{k + 1} } {k + 1} } 0 1
      | c = Integral of Power
}}
{{eqn | r = -\gamma + \sum_{k \mathop = 0}^{z - 2} \frac 1 {k + 1}
      | c = 
}}
{{eqn | r = -\gamma + \sum_{k \mathop = 1}^{z - 1} \frac 1 k
      | c = reindexing the sum
}}
{{eqn | r = -\gamma + H_{z - 1}
      | c = {{Defof|Harmonic Numbers}}
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Carl Friedrich Gauss|cat = Gauss}}