Hermite's Formula for Hurwitz Zeta Function

Theorem

 * $\ds \map \zeta {s, q} = \frac 1 {2 q^s} + \frac { q^{1 - s} } {s - 1} + 2 \int_0^\infty \frac {\map \sin {s \arctan \frac x q} } {\paren {q^2 + x^2}^{\frac 1 2 s} \paren {e^{2 \pi x} - 1} } \rd x$

where:
 * $\zeta$ is the Hurwitz zeta function
 * $\map \Re s > 1$
 * $\map \Re q > 0$.

Proof
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:


 * $\ds \DD_q^n \Ln \map \Gamma q = \DD_q^n \paren {q - \frac 1 2} \Ln q - \DD_q^n q + \DD_q^n \frac 1 2 \ln 2 \pi + \DD_q^n 2 \int_0^\infty \frac {\map \arctan {x / q} } {e^{2 \pi x} - 1} \rd x$

Assuming $\map \Re n \ge 2$, we can turn the expression into:


 * $\ds \bspsi^{\paren {n - 1} } q = \DD_q^n \paren {q \map \ln q} - \frac 1 2 \DD_q^{n - 1} {\paren {\frac 1 q} } + \DD_q^{n - 2} \paren 0 + \DD_q^{n - 1} \paren 0 + 2 \int_0^\infty \frac {\DD_q^n \paren {\map \arctan {x / q} } } {e^{2 \pi x} - 1} \rd x$

Then, by solving the various fractional derivatives, we can say:


 * $\ds \bspsi^{\paren {n - 1} } q = \frac { q^{1 - s} } {s - 1} \map \Gamma n \paren {-1}^n - \frac 1 2 \frac {\paren {-1}^{n + 1} \map \Gamma n} {q^n} + 2 \map \Gamma n \paren {-1}^n \int_0^\infty \frac {\map \sin {n \arctan \frac x q} } {\paren {q^2 + x^2}^{\frac 1 2 n} \paren {e^{2 \pi x} - 1} } \rd x$


 * $\ds \paren {-1}^n \map \Gamma n \map \bszeta {n, q} = \frac { q^{1 - s} } {s - 1} \map \Gamma n \paren {-1}^n - \frac 1 2 \frac {\paren {-1}^{n + 1} \map \Gamma n} {q^n} + 2 \map \Gamma n \paren {-1}^n \int_0^\infty \frac {\map \sin {n \arctan \frac x q} } {\paren {q^2 + x^2}^{\frac 1 2 n} \paren {e^{2 \pi x} - 1} } \rd x$

Dividing both sides by the common factor:


 * $\ds \map \bszeta {n, q} = \frac {q^{1 - n} } {n - 1} + \frac 1 {2 q^n} + 2 \int_0^\infty \frac {\map \sin {n \arctan \frac x q} } {\paren {q^2 + x^2}^{\frac 1 2 n} \paren {e^{2 \pi x} - 1} } \rd x$

the proof is complete.

Warning: the proof I created assumes $\map \Re n \ge 2$, which is not exacly the statement found in the theorem ( $\map \Re n \ge 1$ ).