# 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:

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

Now we can solve the fractional derivatives:

 $\ds \DD_q^{n - 1} {\paren {\frac 1 q} }$ $=$ $\ds \DD_q^{n - 1} {\paren {q^{-1} } }$ $\ds$ $=$ $\ds \frac {\paren {-1}^{n + 1} \map \Gamma n} {q^n}$ fractional derivative of a power of $x$
 $\ds \DD_q^n \paren {q \map \ln q}$ $=$ $\ds \DD_q^n \paren {\lim_{h \mathop \to 0} q \paren {\frac {q^h - 1} h} }$ limit definition of the logarithm $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac 1 h \DD_q^n \paren {q^{1 + h} - q}$ bringing limits and constants outside $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac 1 h \paren {\DD_q^n \paren {q^{1 + h} }-\DD_q^{n - 2} \paren 0}$ $\map \Re n \ge 2$ $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac 1 h \paren {1 + h} \paren h \DD_q^{n - 2} \paren {q^{h - 1} }$ $\map \Re n \ge 2$ $\ds$ $=$ $\ds \DD_q^{n - 2} \paren {q^{-1} }$ the limit can be taken without problems $\ds$ $=$ $\ds \frac {\paren {-1}^n \map \Gamma {n - 1} } {q^{n - 1} }$ fractional derivative of a power of $x$
 $\ds \DD_q^n \paren { \map \arctan {x / q} }$ $=$ $\ds \DD_q^n \int_0^\infty e^{-t q} \frac {\map \sin {t x} } t \rd t$ integral representation for the arctangent $\ds$ $=$ $\ds \paren {-1}^n \int_0^\infty e^{-tq} \map \sin {tx} t^{n-1} \rd t$ fractional derivative of an exponential $\ds$ $=$ $\ds \paren {-1}^n \map \Gamma n \paren {q^2 + x^2}^{-\frac 1 2 n} \map \sin {n \map \arctan {\frac x q} }$ integration process (not too difficult)

So:

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

Dividing both sides by the common factor, we get the initial formula:

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

$\blacksquare$

## Source of Name

This entry was named for Charles Hermite.