FOURIER EXPANSION HURWITZ ZETA FUNCTION

The goal of today´s post is to prove the following Fourier expansion for the Hurwitz zeta Function

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

Recall the Series definition of the Hurwitz Zeta function

$\zeta(s,a)=\sum_{n=0}^\infty \frac{1}{(n+a)^s}$ (1)

valid for $\sigma>1$ . Where $s=\sigma+i t$

Integral representation of the Hurwitz zeta function

$\zeta(s,a)=\frac{1}{\Gamma(s)}\int_0^\infty \frac{e^{-ax}x^{s-1}}{1-e^{-x}}\,dx$ (2)

To prove (2), we start from the Gamma function

$\begin{aligned} &\Gamma(s)=\int_0^\infty e^{-x}x^{s-1}\,dx\\ &\Gamma(s)=(n+a)^s\int_0^\infty e^{-x(n+a)}x^{s-1}\,dx & (x \to (n+a)x)\\ &\frac{\Gamma(s)}{(n+a)^s}=\int_0^\infty e^{-x(n+a)}x^{s-1}\,dx\\ &\sum_{n=0}^\infty\frac{\Gamma(s)}{(n+a)^s}=\sum_{n=0}^\infty\int_0^\infty e^{-x(n+a)}x^{s-1}\,dx\\ &\Gamma(s)\zeta(s,a)=\int_0^\infty e^{-ax}\left(\sum_{n=0}^\infty e^{-nx}\right)x^{s-1}\,dx\\ &\Gamma(s)\zeta(s,a)=\int_0^\infty \frac{e^{-ax}x^{s-1}}{1-e^{-x}}\,dx\\ &\zeta(s,a)=\frac{1}{\Gamma(s)}\int_0^\infty \frac{e^{-ax}x^{s-1}}{1-e^{-x}}\,dx \qquad \blacksquare\\ \end{aligned}$

Contour integral representation of the Hurwitz zeta function

We now derive a contour integral representation for the Hurwitz Zeta function. The contour is the classic Hankel contour which is a loop around the negative real axis. It starts at −∞, encircles the origin once in the positive direction without enclosing any of the points ±2πi,±4πi,⋯ and returns to −∞ acording to the picture below

For $0<a\leq 1$ the function defined by the contour integral

$\operatorname{I}(s,a)=\frac{1}{2 \pi i}\int_{H}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz$ (3)

is entire. For $\sigma>1$ we have:

$\zeta(s,a)=\Gamma(1-s)\operatorname{I}(s,a)$ (4)

Proof:

Over the bottom edge of the contour Letting $z=xe^{-i \pi}$

$\begin{aligned} I_{C_L}&=\frac{1}{2 \pi i}\int_{C_L}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz\\ &=\frac{1}{2 \pi i}\int_R^\epsilon \frac{(xe^{-i \pi})^{s-1}e^{axe^{-i \pi}}}{1-e^{xe^{-i \pi}}}\,e^{-i \pi}dx\\ &=\frac{-e^{-i \pi s}}{2 \pi i}\int_\epsilon^R \frac{x^{s-1}e^{-ax}}{1-e^{-x}}\,dx \end{aligned}$

On the upper edge we have $z=xe^{i \pi}$

$\begin{aligned} I_{C_U}&=\frac{1}{2 \pi i}\int_{C_U}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz\\ &=\frac{1}{2 \pi i}\int_\epsilon^R \frac{(xe^{i \pi})^{s-1}e^{axe^{i \pi}}}{1-e^{xe^{i \pi}}}\,e^{i \pi}dx\\ &=\frac{e^{i \pi s}}{2 \pi i}\int_\epsilon^R \frac{x^{s-1}e^{-ax}}{1-e^{-x}}\,dx \end{aligned}$

Letting $z =\epsilon e^{i \theta}$

$\begin{aligned} I_{C_\epsilon}&=\int_{C_\epsilon}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz\\ &=\int_{-\pi}^\pi \frac{\left( \epsilon e^{i \theta}\right)^{s-1}e^{a\epsilon e^{i \theta}}}{1-e^{\epsilon e^{i \theta}}}\,i \epsilon e^{i \theta}d \theta\\ &=i \epsilon^s \int_{-\pi}^\pi \frac{e^{is \theta}e^{a\epsilon\left(\cos(\theta)+i \sin(\theta) \right) }}{1-e^{\epsilon e^{i \theta}}}\,d \theta\\ &=i \epsilon^s \int_{-\pi}^\pi \frac{e^{is \theta}e^{a\epsilon\left(\cos(\theta)+i \sin(\theta) \right) }}{1-e^{\epsilon e^{i \theta}}}\frac{\epsilon e^{i \theta}}{\epsilon e^{i \theta}}d \theta\\ &=i \epsilon^s \int_{-\pi}^\pi \frac{\epsilon e^{i \theta}}{1-e^{\epsilon e^{i \theta}}}\frac{e^{is \theta}e^{a\epsilon\left(\cos(\theta)+i \sin(\theta) \right) }}{\epsilon e^{i \theta}}d \theta\\ & \leq \Bigg|i \epsilon^s \int_{-\pi}^\pi \frac{\epsilon e^{i \theta}}{1-e^{\epsilon e^{i \theta}}}\frac{e^{is \theta}e^{a\epsilon\left(\cos(\theta)+i \sin(\theta) \right) }}{\epsilon e^{i \theta}}d \theta \Bigg|\\ & \leq \epsilon^s \int_{-\pi}^\pi \Bigg|\frac{\epsilon e^{i \theta}}{1-e^{\epsilon e^{i \theta}}}\Bigg| \, \Bigg|\frac{e^{is \theta}e^{a\epsilon\left(\cos(\theta)+i \sin(\theta) \right) }}{\epsilon e^{i \theta}}\Bigg|\,d \theta\\ & \leq \epsilon^{s-1} \int_{-\pi}^\pi \Bigg|\frac{\epsilon e^{i \theta}}{1-e^{\epsilon e^{i \theta}}}\Bigg| \, e^{a\epsilon \cos(\theta)}\,d \theta\\ & \leq \epsilon^{s-1} C \int_{-\pi}^\pi e^{a\epsilon \cos(\theta)}\,d \theta\\ & \text{letting} \,\, \epsilon \to 0 \\ I_{C_\epsilon} &\leq \lim_{\epsilon \to 0}\epsilon^{s-1} C \int_{-\pi}^\pi e^{a\epsilon \cos(\theta)}\,d \theta\\ &= \lim_{\epsilon \to 0}\epsilon^{s-1} C \int_{-\pi}^\pi \lim_{\epsilon \to 0} e^{a\epsilon \cos(\theta)}\,d \theta\\ &= \lim_{\epsilon \to 0}\epsilon^{s-1} 2 \pi C \to 0 \qquad \text{for }\operatorname{Re}(s)>1 \end{aligned}$

To show the boundedness of $\frac{\epsilon e^{i \theta}}{1-e^{\epsilon e^{i \theta}}}$ we proceed as following

$\begin{aligned} \frac{\epsilon e^{i \theta}}{1-e^{\epsilon e^{i \theta}}}&=\frac{z}{1-e^{z}}\\ &=\frac{z}{1-\left(\sum_{n=0}^\infty \frac{z^n}{n!} \right)}\\ &=\frac{z}{1-\left(1+z+\frac{z^2}{2!}+\cdots \right)}\\ &=-\frac{z}{z+\frac{z^2}{2!}+\cdots }\\ &=-\frac{z}{z}\cdot\frac{1}{1+\frac{z}{2!}+\cdots }\\ &=-\frac{1}{1+\frac{z}{2!}+\cdots }\\ &=-\frac{1}{1+\frac{\epsilon e^{i \theta}}{2!}+\cdots }\\ & \text{as} \,\,\epsilon \to 0 \,\,\text{the limit tends to}\,\,-1 \end{aligned}$

Taking limit of $R \to \infty$ and $\epsilon \to 0$ we obtain

$\begin{aligned} \frac{1}{2 \pi i}\int_{H}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz&=\frac{e^{i \pi s}}{2 \pi i}\int_0^\infty \frac{x^{s-1}e^{-ax}}{1-e^{-x}}\,dx-\frac{e^{-i \pi s}}{2 \pi i}\int_0^\infty \frac{x^{s-1}e^{-ax}}{1-e^{-x}}\,dx\\ &= \frac{\sin(\pi s)}{\pi}\int_0^\infty \frac{x^{s-1}e^{-ax}}{1-e^{-x}}\,dx\\ &=\frac{\sin(\pi s)}{\pi}\Gamma(s)\zeta(s,a)\\ &=\frac{\zeta(s,a)}{\Gamma(1-s)}\\ \end{aligned}$

$\zeta(s,a)=\frac{\Gamma(1-s)}{2 \pi i}\int_{H}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz$

$\zeta(s,a)=\Gamma(1-s)\operatorname{I}(s,a)$

We can modify the above contour so that it encloses the poles of the integrand at $\pm 2\pi i, \pm 4\pi i , \cdots , \pm 2N\pi i$ . And the integral can be evaluated as the sum of the residues leading us to the following series expansion:

(5)

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

Proof:

Let $f(z)=\frac{z^{s-1}e^{az}}{1-e^{z}}$ around a closed contour according to the picture below:

The radius of the circle is given by $R=(2N+1)\pi$ .

The integrand has simple poles at $\pm 2\pi i, \pm 4\pi i , \cdots , \pm 2N\pi i$ . Hence, by the residues theorem we have:

$\frac{1}{2 \pi i}\int_{C}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz=\frac{1}{2 \pi i}\int_{C_R}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz+\frac{1}{2 \pi i}\int_{H}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz=\sum_{n=1}^N \left(R_n+R_n^\prime \right)$ (6)

Where $R_n+R_n^\prime$ are the residues over the negative and positive integers respectively.

Lets start by calculating the residues. We will use the following expression in the calculation

$\begin{aligned} \operatorname{Res}_{z=2\pi i n}F(z)&=\lim_{z \to 2\pi i n}(z-2 \pi i n)\frac{f(z)}{g(z)}\\ &=\lim_{z \to 2\pi i n}\frac{f(z)}{\frac{g(z)-g(2 \pi i n)}{z - 2\pi i n}}\\ &=\frac{f(2 \pi i n)}{g^\prime(2 \pi i n)}\\ \end{aligned}$

For positive n we have

$\begin{aligned} \lim_{z \to 2 \pi i n}(z-2 \pi i n)\frac{z^{s-1}e^{az}}{1-e^{z}}&=\lim_{z \to 2 \pi i n}\frac{z^{s-1}e^{az}}{-e^{z}}\\ &=\frac{(2 \pi i n)^{s-1}e^{2 a \pi i n}}{-e^{2 \pi i n}}\\ &=-(2 \pi ne^{ \frac{i\pi}{2} })^{s-1}e^{2 a \pi i n}\\ &=-\frac{e^{\frac{i\pi s}{2} }}{i}(2 \pi n)^{s-1}e^{2 a \pi i n}\\ &=i(2 \pi n)^{s-1}e^{\frac{i \pi s }{2}+2 a \pi i n}\\ \end{aligned}$

For negative n

$\begin{aligned} \lim_{z \to -2 \pi i n}(z+2 \pi i n)\frac{z^{s-1}e^{az}}{1-e^{z}}&=\lim_{z \to -2 \pi i n}\frac{z^{s-1}e^{az}}{-e^{z}}\\ &=\frac{(-2 \pi i n)^{s-1}e^{-2 a \pi i n}}{-e^{-2 \pi i n}}\\ &=-i(2 \pi n)^{s-1}e^{-\frac{i\pi s}{2} }e^{-2 a \pi i n}\\ &=-i(2 \pi n)^{s-1}e^{-\frac{i\pi s}{2}-2 a \pi i n}\\ \end{aligned}$

Hence letting $N \to \infty$ , by (6) we have that

$\begin{aligned} \lim_{N \to \infty}\sum_{n=1}^N \left(R_n+R_n^\prime \right) &=\lim_{N \to \infty}\sum_{n=1}^N \left(i(2 \pi n)^{s-1}e^{\frac{i \pi }{2}+2 a \pi i n}-i(2 \pi n)^{s-1}e^{-\frac{i\pi}{2}-2 a \pi i n}\right)\\ &=\frac{2}{(2 \pi)^{1-s}}\sum_{n=1}^\infty \frac{\sin\left(\frac{\pi s}{2}+2a \pi n \right)}{n^{1-s}}\\ &=\frac{2 \sin\left(\frac{\pi s}{2}\right)}{(2 \pi)^{1-s}}\sum_{n=1}^\infty \frac{\cos\left(2a \pi n \right)}{n^{1-s}}+\frac{2 \cos\left(\frac{\pi s}{2}\right)}{(2 \pi)^{1-s}}\sum_{n=1}^\infty \frac{\sin\left(2a \pi n \right)}{n^{1-s}} \qquad \blacksquare\\ \end{aligned}$

Now lets show that the integral over the big arc doesn´t contribute to the integral.

$\begin{aligned} I_{C_R}&=\int_{C_R}\frac{ z^{s-1}e^{az}}{1-e^z}\,dz\\ &=\int_\pi^{-\pi} \frac{(Re^{i \theta})^{s-1}e^{aRe^{i \theta}}}{1-e^{Re^{i \theta}}}\,i Re^{i \theta} d \theta\\ &=-iR^{s}\int_{-\pi}^\pi \frac{e^{i s \theta}e^{aR\cos(\theta)+iaR\sin(\theta)}}{1-e^{R e^{i \theta}}}\,d\theta\\ & \leq \Bigg|-iR^{s}\int_{-\pi}^\pi \frac{e^{i s \theta}e^{aR\cos(\theta)+iaR\sin(\theta)}}{1-e^{R e^{i \theta}}}\,d\theta \Bigg|\\ & \leq R^{\operatorname{Re}(s)} \int_{-\pi}^\pi \Bigg| \frac{e^{aR\cos(\theta)+iaR\sin(\theta)}}{1-e^{R e^{i \theta}}}\Bigg|\,\Bigg|e^{i (\operatorname{Re}(s)+i\operatorname{Im}(s)) \theta}\Bigg|\,d \theta\\ & \leq R^{\operatorname{Re}(s)}K \int_{-\pi}^\pi \,e^{-\operatorname{Im}(s)) \theta}\,d \theta\\ & \text{if} \operatorname{Re}(s)<0 \,\,\text{as} \, R \to \infty\,\, \text{the integral vanishes!} \end{aligned}$

For $0<a\leq 1$ we have

$\begin{aligned} \frac{ e^{az}}{1-e^z}&=\frac{ e^{(a-1)z}}{e^{-z}-1}\\ &=-\frac{ e^{(a-1)z}}{1-e^{-z}}\\ & \text{And for }\,\, 0<a\leq 1 \,\,\text{this quotient is bounded} \end{aligned}$

Therefore, by (4) and (6) we obtain the desired result

If we let $s \to 1-s$ we obtain

$\zeta(1-s,a)=\frac{2\Gamma(s) }{(2 \pi)^{s}} \left(\cos\left(\frac{\pi s}{2}\right)\sum_{n=1}^\infty \frac{\cos\left(2a \pi n \right)}{n^{s}}+\sin\left(\frac{\pi s}{2}\right)\sum_{n=1}^\infty \frac{\sin\left(2a \pi n \right)}{n^{s}}\right)$ (6)

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

Letting $a \to 1-a$ in (6) we obtain

$\zeta(1-s,1-a)=\frac{2\Gamma(s) }{(2 \pi)^{s}} \left(\cos\left(\frac{\pi s}{2}\right)\sum_{n=1}^\infty \frac{\cos\left(2a \pi n \right)}{n^{s}}-\sin\left(\frac{\pi s}{2}\right)\sum_{n=1}^\infty \frac{\sin\left(2a \pi n \right)}{n^{s}}\right)$ (7)

Adding (6) and (7) we get

$\zeta(1-s,a)+\zeta(1-s,1-a)=\frac{4\Gamma(s) }{(2 \pi)^{s}}\cos\left(\frac{\pi s}{2}\right)\sum_{n=1}^\infty \frac{\cos\left(2a \pi n \right)}{n^{s}}$ (8)

Subtracting (7) from (6) we obtain

$\zeta(1-s,a)-\zeta(1-s,1-a)=\frac{4\Gamma(s) }{(2 \pi)^{s}}\sin\left(\frac{\pi s}{2}\right)\sum_{n=1}^\infty \frac{\sin\left(2a \pi n \right)}{n^{s}}$ (9)

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