FUNCTIONAL EQUATION FOR THE HURWITZ ZETA FUNCTION

In today´s post we will derive a functional equation for the Hurwitz Zeta Function, namely:

$\zeta\left(1-s,\frac{h}{k}\right)=\frac{2\Gamma(s)}{(2 \pi k)^s}\sum_{r=1}^k\cos\left(\frac{ \pi s}{2}}-\frac{2 \pi r h}{k}\right)\zeta\left(s,\frac{r}{k}\right)$

To this end we introduce the periodic zeta function defined by the following expression:

$F(x,s)=\sum_{n=1}^\infty\frac{e^{2 \pi i n x}}{n^s}$ (1)

Theorem 1:

$\zeta(1-s,a)=\frac{\Gamma(s)}{(2 \pi)^s}\left(e^{-\frac{i \pi s}{2}}\operatorname{F}(a,s)+e^{\frac{i \pi s}{2}}\operatorname{F}(-a,s)\right)$ (2)

for $\sigma>1$ and $0<a\leq1$

Proof:

First recall the previously proved fourier expansion (see here)

$\zeta(1-s,a)=\frac{2\Gamma(s) }{(2 \pi)^{s}} \left(\cos\left(\frac{\pi s}{2}\right)\sum_{n=1}^\infty \frac{\cos\left(2a \pi n \right)}{n^{s}}+\sin\left(\frac{\pi s}{2}\right)\sum_{n=1}^\infty \frac{\sin\left(2a \pi n \right)}{n^{s}}\right)$ (3)

for $\sigma>1$ and $0<a\leq1$

Now lets expand (2) by means of Euler´s formula

$\begin{aligned} \zeta(1-s,a)&=\frac{\Gamma(s)}{(2 \pi)^s}\left(e^{-\frac{i \pi s}{2}}\operatorname{F}(a,s)+e^{\frac{i \pi s}{2}}\operatorname{F}(-a,s)\right)\\ &=\frac{\Gamma(s)}{(2 \pi)^s}\left(e^{-\frac{i \pi s}{2}}\sum_{n=1}^\infty\frac{e^{2 \pi i n a}}{n^s}+e^{\frac{i \pi s}{2}}\sum_{n=1}^\infty\frac{e^{-2 \pi i n a}}{n^s}\right)\\ &=\frac{\Gamma(s)}{(2 \pi)^s}\left(e^{-\frac{i \pi s}{2}}\sum_{n=1}^\infty\frac{\cos(2 \pi n a)+i\sin(2 \pi n a)}{n^s}+e^{\frac{i \pi s}{2}}\sum_{n=1}^\infty\frac{\cos(2 \pi n a)-i\sin(2 \pi n a)}{n^s}\right)\\ &=\frac{\Gamma(s)}{(2 \pi)^s}\left(\left(e^{\frac{i \pi s}{2}}+e^{-\frac{i \pi s}{2}}\right)\sum_{n=1}^\infty\frac{\cos(2 \pi n a)}{n^s}-i\left(e^{\frac{i \pi s}{2}}-e^{-\frac{i \pi s}{2}}\right)\sum_{n=1}^\infty\frac{i\sin(2 \pi n a)}{n^s}\right)\\ &=\frac{2\Gamma(s) }{(2 \pi)^{s}} \left(\cos\left(\frac{\pi s}{2}\right)\sum_{n=1}^\infty \frac{\cos\left(2a \pi n \right)}{n^{s}}+\sin\left(\frac{\pi s}{2}\right)\sum_{n=1}^\infty \frac{\sin\left(2a \pi n \right)}{n^{s}}\right) \qquad \blacksquare \end{aligned}$

Lemmma 1:

Let h and k be two integers, $1\leq h \leq k$ , then for $\sigma>1$

$F\left(\frac{h}{k},s\right)=\sum_{n=1}^\infty\frac{e^{2 \pi i n \frac{h}{k}}}{n^s}=\frac{1}{k^s}\sum_{r=1}^k e^{2 \pi i r \frac{h}{k}}\zeta\left(s,\frac{r}{k}\right)$ (4)

Proof:

$\begin{aligned} \sum_{n=1}^\infty\frac{e^{2 \pi i n \frac{h}{k}}}{n^s}&=\sum_{r=1}^k\sum_{q=0}^\infty \frac{e^{2 \pi i r \frac{h}{k}}}{(kq+r)^s}\\ &=\sum_{r=1}^k e^{2 \pi i r \frac{h}{k}}\sum_{q=0}^\infty \frac{1}{(kq+r)^s}\\ &=\frac{1}{k^s}\sum_{r=1}^k e^{2 \pi i r \frac{h}{k}}\sum_{q=0}^\infty \frac{1}{\left(q+\frac{r}{k}\right)^s}\\ &=\frac{1}{k^s}\sum_{r=1}^k e^{2 \pi i r \frac{h}{k}}\zeta\left(s,\frac{r}{k}\right) \qquad \blacksquare \end{aligned}$

To prove this equality used in the proof above

$\sum_{n=1}^\infty\frac{e^{2 \pi i n \frac{h}{k}}}{n^s}&=\sum_{r=1}^k\sum_{q=0}^\infty \frac{e^{2 \pi i r \frac{h}{k}}}{(kq+r)^s}$

Lets expand first it´s L.H.S.

$\begin{aligned} \sum_{n=1}^\infty\frac{e^{2 \pi i n \frac{h}{k}}}{n^s}&=\frac{e^{2 \pi i \frac{h}{k}}}{1^s}+\frac{e^{4 \pi i \frac{h}{k}}}{2^s}+\frac{e^{6 \pi i \frac{h}{k}}}{3^s}+ \cdots+\frac{e^{2 \pi i h}}{k^s}+\frac{e^{2 \pi i \frac{h}{k}}}{(k+1)^s}+\cdots \end{aligned}$

We used the fact that $e^{2 \pi i h m}=1$ for m an integer

Now, lets expand it´s R.H.S. and show that they are equal to each other

$\begin{aligned} \sum_{r=1}^k\sum_{q=0}^\infty \frac{e^{2 \pi i r \frac{h}{k}}}{(kq+r)^s}&=\sum_{r=1}^k e^{2 \pi i r \frac{h}{k}} \left(\sum_{q=0}^\infty \frac{1}{(kq+r)^s}\right)\\ &=\sum_{r=1}^k e^{2 \pi i r \frac{h}{k}} \left(\frac{1}{r^s}+\frac{1}{(k+r)^s}+\frac{1}{(2k+r)^s}+\frac{1}{(3k+r)^s}+\cdots\right)\\ &=\sum_{r=1}^k \left(\frac{e^{2 \pi i r \frac{h}{k}}}{r^s}+\frac{e^{2 \pi i r \frac{h}{k}}}{(k+r)^s}+\frac{e^{2 \pi i r \frac{h}{k}}}{(2k+r)^s}+\frac{e^{2 \pi i r \frac{h}{k}}}{(3k+r)^s}+\cdots\right)\\ &=\sum_{r=1}^k \frac{e^{2 \pi i r \frac{h}{k}}}{r^s}+\sum_{r=1}^k\frac{e^{2 \pi i r \frac{h}{k}}}{(k+r)^s}+\sum_{r=1}^k\frac{e^{2 \pi i r \frac{h}{k}}}{(2k+r)^s}+\sum_{r=1}^k\frac{e^{2 \pi i r \frac{h}{k}}}{(3k+r)^s}+\cdots\\ &=\left(\frac{e^{2 \pi i \frac{h}{k}}}{1^s}+\frac{e^{4 \pi i \frac{h}{k}}}{2^s}+\frac{e^{6 \pi i \frac{h}{k}}}{3^s}+ \cdots+ \frac{e^{2 \pi i h}}{k^s}\right)+\\ &+\left(\frac{e^{2 \pi i \frac{h}{k}}}{(k+1)^s}+\frac{e^{4 \pi i \frac{h}{k}}}{(k+2)^s}+\frac{e^{6 \pi i \frac{h}{k}}}{(k+3)^s}+ \cdots+ \frac{e^{2 \pi i h}}{(2k)^s}\right)+\cdots\\ \end{aligned}$

Theorem 2:

If k and h are integers with $1 \leq h \leq k$ , then for all s we have

$\zeta\left(1-s,\frac{h}{k}\right)=\frac{2\Gamma(s)}{(2 \pi k)^s}\sum_{r=1}^k\cos\left(\frac{ \pi s}{2}}-\frac{2 \pi r h}{k}\right)\zeta\left(s,\frac{r}{k}\right)$

Proof:

Plugging (4) in (2) we obtain:

$\begin{aligned} \zeta(1-s,a) &=\frac{\Gamma(s)}{(2 \pi)^s}\left(e^{-\frac{i \pi s}{2}}F\left(\frac{h}{k},s\right)+e^{\frac{i \pi s}{2}}F\left(-\frac{h}{k},s\right)\right)\\ &=\frac{\Gamma(s)}{(2 \pi)^s}\left(e^{-\frac{i \pi s}{2}}\frac{1}{k^s}\sum_{r=1}^k e^{2 \pi i r \frac{h}{k}}\zeta\left(s,\frac{r}{k}\right)+e^{\frac{i \pi s}{2}}\frac{1}{k^s}\sum_{r=1}^k e^{-2 \pi i r \frac{h}{k}}\zeta\left(s,\frac{r}{k}\right)\right)\\ &=\frac{2\Gamma(s)}{(2 \pi k)^s}\sum_{r=1}^k\cos\left(\frac{ \pi s}{2}}-\frac{2 \pi r h}{k}\right)\zeta\left(s,\frac{r}{k}\right) \qquad \blacksquare \end{aligned}$

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