Theorem: Hermite's Formula for Hurwitz Zeta Function

:$\ds \map \zeta {s, q} = \frac 1 {2 q^s} + \frac { q^{1 - s} } {s - 1} + 2 \int_0^\infty \frac {\map \sin {s \arctan \frac x q} } {\paren {q^2 + x^2}^{\frac 1 2 s} \paren {e^{2 \pi x} - 1} } \rd x$
where: 
:$\zeta$ is the Hurwitz zeta function
:$\map \Re s > 1$
:$\map \Re q > 0$.

Proof:
{{MissingLinks|Some of these are to pages which doing exist yet}}
To prove this theorem, we can make use of Binet's Second Formula for Log Gamma:
Let $q$ be a complex number with a positive real part.
Then: 
:$\ds \Ln \map \Gamma q = \paren {q - \frac 1 2} \Ln q - q + \frac 1 2 \ln 2 \pi + 2 \int_0^\infty \frac {\map \arctan {x / q} } {e^{2 \pi x} - 1} \rd x$
Applying the $n$th fractional derivative with respect to $q$ on both sides, we get:
{{begin-eqn}}
{{eqn | l = \DD_q^n \Ln \map \Gamma q 
      | r =  \DD_q^n \paren {q - \frac 1 2} \Ln q - \DD_q^n q + \DD_q^n \frac 1 2 \ln 2 \pi + \DD_q^n 2 \int_0^\infty \frac {\map \arctan {x / q} } {e^{2 \pi x} - 1} \rd x
}}
{{eqn | l = \ds \bspsi^{\paren {n - 1} } q 
      | r = \DD_q^n \paren {q \map \ln q} - \frac 1 2 \DD_q^{n - 1} {\paren {\frac 1 q} } + \DD_q^{n - 2} \paren 0 + \DD_q^{n - 1} \paren 0 + 2 \int_0^\infty \frac {\DD_q^n \paren {\map \arctan {x / q} } } {e^{2 \pi x} - 1} \rd x
      | c = setting $\map \Re n \ge 2$
}}
{{eqn | r = \DD_q^n \paren {q \map \ln q} - \frac 1 2 \DD_q^{n - 1} {\paren {\frac 1 q} } + 2 \int_0^\infty \frac {\DD_q^n \paren {\map \arctan {x / q} } } {e^{2 \pi x} - 1} \rd x
      | c = simplifying
}}
{{end-eqn}}
Now we can solve the fractional derivatives:
{{begin-eqn}}
{{eqn | l = \DD_q^{n - 1} {\paren {\frac 1 q} }
      | r = \DD_q^{n - 1} {\paren {q^{-1} } }
}}
{{eqn | r = \frac {\paren {-1}^{n + 1} \map \Gamma n} {q^n}
      | c = fractional derivative of a power of $x$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \DD_q^n \paren {q \map \ln q}
      | r = \DD_q^n \paren {\lim_{h \mathop \to 0} q \paren {\frac {q^h - 1} h} }
      | c = limit definition of the logarithm
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac 1 h \DD_q^n \paren {q^{1 + h} - q}
      | c = bringing limits and constants outside
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac 1 h \paren {\DD_q^n \paren {q^{1 + h} }-\DD_q^{n - 2} \paren 0}
      | c = $\map \Re n \ge 2$
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac 1 h \paren {1 + h} \paren h \DD_q^{n - 2} \paren {q^{h - 1} }
      | c = $\map \Re n \ge 2$
}}
{{eqn | r = \DD_q^{n - 2} \paren {q^{-1} }
      | c = the limit can be taken without problems
}}
{{eqn | r = \frac {\paren {-1}^n \map \Gamma {n - 1} } {q^{n - 1} }
      | c = fractional derivative of a power of $x$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \DD_q^n \paren { \map \arctan {x / q} }
      | r = \DD_q^n \int_0^\infty e^{-t q} \frac {\map \sin {t x} } t \rd t
      | c = integral representation for the arctangent
}}
{{eqn | r = \paren {-1}^n \int_0^\infty e^{-tq} \map \sin {tx} t^{n-1} \rd t
      | c = fractional derivative of an exponential
}}
{{eqn | r = \paren {-1}^n \map \Gamma n \paren {q^2 + x^2}^{-\frac 1 2 n} \map \sin {n \map \arctan {\frac x q} }
      | c = integration process (not too difficult)
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \bspsi^{\paren {n - 1} } q 
      | r = \frac {q^{1 - n} } {n - 1} \map \Gamma n \paren {-1}^n - \frac 1 2 \frac {\paren {-1}^{n + 1} \map \Gamma n} {q^n} + 2 \map \Gamma n \paren {-1}^n \int_0^\infty \frac {\map \sin {n \arctan \frac x q} } {\paren {q^2 + x^2}^{\frac 1 2 n} \paren {e^{2 \pi x} - 1} } \rd x
}}
{{eqn | l = \paren {-1}^n \map \Gamma n \map \bszeta {n, q}
      | r = \frac {q^{1 - n} } {n - 1} \map \Gamma n \paren {-1}^n - \frac 1 2 \frac {\paren {-1}^{n + 1} \map \Gamma n} {q^n} + 2 \map \Gamma n \paren {-1}^n \int_0^\infty \frac {\map \sin {n \arctan \frac x q} } {\paren {q^2 + x^2}^{\frac 1 2 n} \paren {e^{2 \pi x} - 1} } \rd x
}}
{{end-eqn}}
Dividing both sides by the common factor, we get the initial formula:
:$\ds \map \bszeta {n, q} = \frac {q^{1 - n} } {n - 1} + \frac 1 {2 q^n} + 2 \int_0^\infty \frac {\map \sin {n \arctan \frac x q} } {\paren {q^2 + x^2}^{\frac 1 2 n} \paren {e^{2 \pi x} - 1} } \rd x$
{{qed}}
{{improve|Warning: the proof above assumes $\map \Re n \ge 2$, which is not exactly the statement found in the theorem ( $\map \Re n \ge 1$ ). It needs to be made so as to be consistent with the problem statement.}}
{{proofread|As stated above, proof isn't exactly correct, it needs some kind of fixing at the moment}}
{{Namedfor|Charles Hermite|cat = Hermite}}