Riemann Zeta Function

The Riemann zeta function can be defined by the integral

$$\zeta(x) \equiv {1\over\Gamma(x)} \int_0^\infty {u^{x-1}\over e^u-1} \,du,$$ (1)

where $x>1$. If $x$ is an Integer $n$, then

$${u^{n-1}\over e^u-1} = {e^{-u}u^{n-1}\over 1-e^{-u}} = e^{-u... ... \sum_{k=0}^\infty e^{-ku} = \sum_{k=1}^\infty e^{-ku}u^{n-1},$$ (2)

so

$$\int_0^\infty {u^{n-1}\over e^u-1}\,du = \sum_{k=1}^\infty \int^\infty_0 e^{-ku}u^{n-1}\,du.$$ (3)

Let $y\equiv ku$, then $dy = k \,du$ and

$$\zeta(n) = {1\over\Gamma(n)} \sum_{k=1}^\infty \int^\infty_0 e^{-ku}u^{n-1}\,du$$

$$= {1\over\Gamma(n)}\sum_{k=1}^\infty \int^\infty_0 e^{-y}\left({y\over k}\right)^{n-1}{dy\over k}$$

$$= {1\over\Gamma(n)}\sum_{k=1}^\infty {1\over k^n} \int^\infty_0 e^{-y}y^{n-1}\,dy,$$ (4)

where $\Gamma(n)$ is the Gamma Function. Integrating the final expression in (4) gives $\Gamma(n)$, which cancels the factor $1/\Gamma(n)$ and gives the most common form of the Riemann zeta function,

$$\zeta(n) = \sum_{k=1}^\infty {1\over k^n}.$$ (5)

At $n=1$, the zeta function reduces to the Harmonic Series (which diverges), and therefore has a singularity. In the Complex Plane, trivial zeros occur at $-2$, $-4$, $-6$, ..., and nontrivial zeros at

$$s\equiv\sigma+it$$ (6)

for $0\leq\sigma\leq 1$. The figures below show the structure of $\zeta(z)$ by plotting $\vert\zeta(z)\vert$ and $1/\vert\zeta(z)\vert$.

The Riemann Hypothesis asserts that the nontrivial Roots of $\zeta(s)$ all have Real Part $\sigma=\Re[s]=1/2$, a line called the ``Critical Strip.'' This is known to be true for the first $200,000,001$ roots (Brent et al. 1982). The above plot shows $\vert\zeta(1/2+it)\vert$ for $t$ between 0 and 60. As can be seen, the first few nontrivial zeros occur at $t=14.134725$, 21.022040, 25.010858, 30.424876, 32.935062, 37.586178, ... (Wagon 1991, pp. 361-362 and 367-368).

The Riemann zeta function can also be defined in terms of Multiple Integrals by

$$\zeta(n)=\underbrace{\int_0^1\cdots\int_0^1}_n {\prod_{i=1}^n dx_i\over 1-\prod_{i=1}^n x_i}.$$ (7)

The Riemann zeta function can be split up into

$$\zeta({\textstyle{1\over 2}}+it)=z(t)e^{-i\vartheta(t)},$$ (8)

where $z(t)$ and $\vartheta(t)$ are the Riemann-Siegel Functions. An additional identity is

$$\lim_{s\to 1} \zeta(s)-{1\over s-1}=\gamma,$$ (9)

where $\gamma$ is the Euler-Mascheroni Constant.

The Riemann zeta function is related to the Dirichlet Lambda Function $\lambda(\nu)$ and Dirichlet Eta Function $\eta(\nu)$ by

$${\zeta(\nu)\over 2^\nu}={\lambda(\nu)\over 2^\nu-1}={\eta(\nu)\over 2^{\nu}-2}$$ (10)

and

$$\zeta(\nu)+\eta(\nu)=2\lambda(\nu)$$ (11)

(Spanier and Oldham 1987). It is related to the Liouville Function $\lambda(n)$ by

$${\zeta(2s)\over\zeta(s)}=\sum_{n=1}^\infty {\lambda(n)\over n^s}$$ (12)

(Lehman 1960, Hardy and Wright 1979). Furthermore,

$${\zeta^2(s)\over\zeta(2s)}=\sum_{n=1}^\infty {2^{\omega(n)}\over n^s},$$ (13)

where $\omega(n)=\sigma_0(n)$ is the number of different prime factors of $n$ (Hardy and Wright 1979).

A generalized Riemann zeta function $\zeta(s,a)$ known as the Hurwitz Zeta Function can also be defined such that

$$\zeta(s)\equiv \zeta(s,0).$$ (14)

The Riemann zeta function may be computed analytically for Even $n$ using either Contour Integration or Parseval's Theorem with the appropriate Fourier Series. An interesting formula involving the product of Primes was first discovered by Euler in 1737,

$$\zeta(x)(1-2^{-x}) = \left({1 + {1\over 2^x} + {1\over 3^x} +\ldots}\right)\left({1-{1\over 2^x}}\right)$$

$$= \left({1 + {1\over 2^x} + {1\over 3^x} +\ldots}\right)-\left({{1\over 2^x} + {1\over 4^x} + {1\over 6^x} + \ldots}\right)$$ (15)

$$\zeta(x)(1-2^{-x})(1-3^{-x}) = \left({1 + {1\over 3^x} + {1\... ...({{1\over 3^x} + {1\over 9^x} + {1\over 15^x} + \ldots}\right)$$ (16)

$$\zeta (x)(1-2^{-x})(1-3^{-x})\cdots (1-p^{-x})\cdots = \zeta(x) \prod_{n=2}^\infty (1-p^{-x}) = 1.$$ (17)

Here, each subsequent multiplication by the next Prime $p$ leaves only terms which are Powers of $p^{-x}$. Therefore,

$$\zeta(x) = \left[{\,\prod_{p=2}^\infty (1-p^{-x})}\right]^{-1},$$ (18)

where $p$ runs over all Primes. Euler's product formula can also be written

$$\zeta(s)=(1-2^{-s})^{-1}\prod_{q\equiv 1\atop ({\rm mod\ }4)... ...-s})^{-1}\prod_{r\equiv 3\atop{\rm (mod\ }4)} (1-r^{-s})^{-1}.$$ (19)

Two sum identities involving $\zeta(n)$ are

$$\sum_{n=2}^\infty [\zeta (n)-1] = 1$$ (20)

$$\sum_{n=2}^\infty (-1)^n[\zeta (n)-1] = {\textstyle{1\over 2}}.$$ (21)

The Riemann zeta function is related to the Gamma Function $\Gamma(z)$ by

$$\Gamma\left({s\over 2}\right)\pi^{-s/2}\zeta(s)=\Gamma\left({1-s\over 2}\right)\pi^{-(1-s)/2} \zeta(1-s).$$ (22)

$\zeta(n)$ was proved to be transcendental for all even $n$ by Euler. Apéry (1979) proved $\zeta(3)$ to be Irrational with the aid of the $k^{-3}$ sum formula below. As a result, $\zeta(3)$ is sometimes called Apéry's Constant.

$$\zeta(2) = 3\sum_{k=1}^\infty {1\over k^2{2k\choose k}}$$ (23)

$$\zeta(3) = {5\over 2}\sum_{k=1}^\infty {(-1)^{k-1}\over k^3{2k\choose k}}$$ (24)

$$\zeta(4) = {36\over 17}\sum_{k=1}^\infty {1 \over k^4{2k\choose k}}$$ (25)

(Guy 1994, p. 257). A relation of the form

$$\zeta(5) = Z_5\sum_{k=1}^\infty {(-1)^{k-1}\over k^5{2k\choose k}}$$ (26)

has been searched for with $Z_5$ a Rational or Algebraic Number, but if $Z_5$ is a Root of a Polynomial of degree 25 or less, then the Euclidean norm of the coefficients must be larger than $2\times 10^{37}$ (Bailey, Bailey and Plouffe). Therefore, no such sums are known for $\zeta(n)$ are known for $n\geq 5$.

The zeta function is defined for $\Re[s]>1$, but can be analytically continued to $\Re[s]>0$ as follows:

$\sum_{n=1}^\infty (-1)^n n^{-s}+\sum_{n=1}^\infty n^{-s} = 2\sum_{n=2, 4, \ldots}^\infty n^{-s}$
$= 2\sum_{k=1}^\infty (2k)^{-s} = 2^{1-s}\sum_{k=1}^\infty k^{-s}\quad$	(27)

$$\sum_{n=1}^\infty (-1)^nn^{-s}+\zeta(s)=2^{1-s}\zeta(s)$$ (28)

$$\zeta(s) = {1\over 1-2^{1-s}} \sum_{n=1}^\infty (-1)^n n^{-s}.$$ (29)

The Derivative of the Riemann zeta function is defined by

$$\zeta'(s)=-s\sum_{k=1}^\infty k^{-s}\ln k = -\sum_{k=2}^\infty {\ln k\over k^s}.$$ (30)

As $s\to 0$,

$$\zeta'(0)=-{\textstyle{1\over 2}}\ln(2\pi).$$ (31)

For Even $n\equiv 2k$,

$$\zeta(n)={2^{n-1}\vert B_n\vert\pi^n\over n!},$$ (32)

where $B_n$ is a Bernoulli Number. Another intimate connection with the Bernoulli Numbers is provided by

$$B_n = (-1)^{n+1}n\zeta(1-n).$$ (33)

No analytic form for $\zeta(n)$ is known for Odd $n\equiv 2k+1$, but $\zeta(2k+1)$ can be expressed as the sum limit

$$\zeta(2k+1) =({\textstyle{1\over 2}}\pi)^{2k+1}\lim_{t\to\in... ...1}^\infty \left[{\cot\left({i\over 2t+1}\right)}\right]^{2k+1}$$ (34)

(Stark 1974). The values for the first few integral arguments are

$$\zeta(0) \equiv -{\textstyle{1\over 2}}$$

$$\zeta(1) = \infty$$

$$\zeta(2) = {\pi^2\over 6}$$

$$\zeta(3) = 1.202 056 903 2\ldots$$

$$\zeta(4) = {\pi^4\over 90}$$

$$\zeta(5) = 1.036 927 755 1\ldots$$

$$\zeta(6) = {\pi^6\over 945}$$

$$\zeta(7) = 1.008 349 277 4\ldots$$

$$\zeta(8) = {\pi^8\over 9450}$$

$$\zeta(9) = 1.002 008 392 8\ldots$$

$$\zeta(10) = {\pi^{10}\over 93{,}555}.$$

Euler gave $\zeta(2)$ to $\zeta(26)$ for Even $n$, and Stieltjes (1993) determined the values of $\zeta(2)$, ..., $\zeta(70)$ to 30 digits of accuracy in 1887. The denominators of $\zeta(2n)$ for $n=1$, 2, ... are 6, 90, 945, 9450, 93555, 638512875, ... (Sloane's A002432).

Using the LLL Algorithm, Plouffe (inspired by Zucker 1979, Zucker 1984, and Berndt 1988) has found some beautiful infinite sums for $\zeta(n)$ with Odd $n$. Let

$$S_\pm(n)\equiv \sum_{k=1}^\infty {1\over k^n(e^{2\pi k}\pm 1)},$$ (35)

then

$$\zeta(3) = {\textstyle{7\over 180}}\pi^3-2S_-(3)$$ (36)

$$\zeta(5) = {\textstyle{1\over 294}}\pi^5-{\textstyle{72\over 35}}S_-(5)-{\textstyle{2\over 35}}S_+(5)$$ (37)

$$\zeta(7) = {\textstyle{19\over 56700}}\pi^7-2S_-(7)$$ (38)

$$\zeta(9) = {\textstyle{125\over 3704778}}\pi^9 - {\textstyle{992\over 495}}S_-(9) - {\textstyle{2\over 495}}S_+(9)$$ (39)

$$\zeta(11) = {\textstyle{1453\over 425675250}}\pi^{11} - 2S_-(11)$$ (40)

$$\zeta(13) = {\textstyle{89\over 257432175}}\pi^{13} - {\textstyle{16512\over 8255}}S_-(13) - {\textstyle{2\over 8255}}S_+(13)$$ (41)

$$\zeta(15) = {\textstyle{13687\over 390769879500}}\pi^{15} - 2S_-(15)$$ (42)

$$\zeta(17) = {\textstyle{397549\over 112024529867250}}\pi^{17} - {\textstyle{261632\over 130815}}S_-(17)- {\textstyle{2\over 130815}}S_+(17)$$ (43)

$$\zeta(19) = {\textstyle{7708537\over 21438612514068750}}\pi^{19} - 2S_-(19)$$ (44)

$$\zeta(21) = {\textstyle{68529640373\over 1881063815762259253125}}\pi^{21} - {\textstyle{4196352\over 2098175}}S_-(21)- {\textstyle{2\over 2098175}}S_+(21)$$ (45)

The inverse of the Riemann Zeta Function $1/\zeta(p)$ is the asymptotic density of $p$th-powerfree numbers (i.e., Squarefree numbers, Cubefree numbers, etc.). The following table gives the number $Q_p(n)$ of $p$th-powerfree numbers $\leq n$ for several values of $n$.

$p$	$1/\zeta(p)$	$Q_p(10)$	$Q_p(100)$	$Q_p(10^3)$	$Q_p(10^4)$	$Q_p(10^5)$	$Q_p(10^6)$
2	0.607927	7	61	608	6083	60794	607926
3	0.831907	9	85	833	8319	83190	831910
4	0.923938	10	93	925	9240	92395	923939
5	0.964387	10	97	965	9645	96440	964388
6	0.982953	10	99	984	9831	98297	982954

The value for $\zeta(2)$ can be found using a number of different techniques (Apostol 1983, Choe 1987, Giesy 1972, Holme 1970, Kimble 1987, Knopp and Schur 1918, Kortram 1996, Matsuoka 1961, Papadimitriou 1973, Simmons 1992, Stark 1969, Stark 1970, Yaglom and Yaglom 1987). The problem of finding this value analytically is sometimes known as the Basler Problem (Castellanos 1988). Yaglom and Yaglom (1987), Holme (1970), and Papadimitrou (1973) all derive the result from de Moivre's Identity or related identities.

Consider the Fourier Series of $f(x)=x^{2n}$

$$f(x)={\textstyle{1\over 2}}a_0+\sum_{m=1}^\infty a_m\cos(mx)+\sum_{m=1}^\infty b_m\sin(mx),$$ (46)

which has coefficients given by

$$a_0 = {1\over \pi} \int_{-\pi}^\pi f(x)\,dx = {2\over \pi} \int_0^\pi x^{2n} \,dx$$

$$= {2\over \pi} \left[{x^{2n+1}\over 2n+1}\right]^\pi_0 = {2\pi^{2n}\over 2n+1}$$ (47)

$$a_m = {1\over \pi} \int_{\pi}^\pi x^{2n}\cos (mx)\,dx$$

$$= {2\over \pi} \int_0^\pi x^{2n}\cos (mx)\,dx$$ (48)

$$b_m = {1\over \pi} \int_{-\pi}^\pi x^{2n}\sin(mx)\,dx=0,$$ (49)

where the latter is true since the integrand is Odd. Therefore, the Fourier Series is given explicitly by

$$x^{2n}={\pi^{2n}\over 2n+1}+\sum_{m=1}^\infty a_m\cos(mx).$$ (50)

Now, $a_m$ is given by the Cosine Integral

$$a_m = {2\over \pi} (-1)^{n+1}(2n)!\left[{\sin(mx)\sum_{k=0}^n {(-1)^k\over (2k)!m^{2n-2k+1}} x^{2k}}\right.$$

$$\phantom{=} +\left.{\cos(mx)\sum_{k=1}^n {(-1)^{k+1}\over (2k-3)!m^{2n-2k+2}} x^{2k-1}}\right]_0^\pi.$$ (51)

But $\cos(m\pi)=(-1)^m$, and $\sin(m\pi)=\sin 0 =0$, so

$$a_m = {2\over \pi} (-1)^{n+1}(2n)!(-1)^m \sum_{k=1}^n {(-1)^{k+1} \over (2k-3)!m^{2n-2k+2}} \pi^{2k-1}$$

$$= (-1)^{m+n} 2(2n)! \sum_{k=1}^n {(-1)^k\over (2k-3)!m^{2n-2k+2}} \pi^{2k-2}.$$ (52)

Now, if $n=1$,

$$a_m = (-1)^{m+1}2(2!) \sum_{k=1}^1 {(-1)^k\over (2k-3)!m^{4-2k}} \pi^{2k-2}$$

$$= 4(-1)^{m+1} {(-1)\over (-1)!m^2} \pi^0 ={4(-1)^m\over m^2},$$ (53)

so the Fourier Series is

$$x^2 = {\pi^2\over 3} + 4\sum_{m=1}^\infty {(-1)^m \cos(mx)\over m^2}.$$ (54)

Letting $x \equiv \pi$ gives $\cos(m\pi)=(-1)^m$, so

$$\pi^2 = {\pi^2\over 3} + 4\sum_{m=1}^\infty {1\over m^2},$$ (55)

and we have

$$\zeta(2)= \sum_{m=1}^\infty {1\over m^2} = {\pi^2\over 6}.$$ (56)

Higher values of $n$ can be obtained by finding $a_m$ and proceeding as above.

The value $\zeta(2)$ can also be found simply using the Root Linear Coefficient Theorem. Consider the equation $\sin z=0$ and expand sin in a Maclaurin Series

$$\sin z=z-{z^3\over 3!}+{z^5\over 5!}+\ldots = 0$$ (57)

$$0=1-{z^2\over 3!}+{z^4\over 5!}+\ldots = 1-{w\over 3!}+{w^2\over 5!}+\ldots,$$ (58)

where $w\equiv z^2$. But the zeros of $\sin(z)$ occur at $\pi$, $2\pi$, $3\pi$, ..., so the zeros of $\sin w=\sin \sqrt{z}$ occur at $\pi^2$, $(2\pi)^2$, .... Therefore, the sum of the roots equals the Coefficient of the leading term

$${1\over \pi^2}+{1\over (2\pi)^2}+{1\over (3\pi^2)}+\ldots={1\over 3!}={1\over 6},$$ (59)

which can be rearranged to yield

$$\zeta(2)={\pi^2\over 6}.$$ (60)

Yet another derivation (Simmons 1992) evaluates the integral using the integral

$$I = \int_0^1\int_0^1 {dx\,dy\over 1-xy}=\int_0^1 \int_0^1 (1+xy+x^2y^2+\ldots)\,dx\,dy$$

$$= \int_0^1 [(x+{\textstyle{1\over 2}}x^2 y+{\textstyle{1\over 3}} x^3y^2+\ldots)]^1_0\,dy$$

$$= \int_0^1 (1+{\textstyle{1\over 2}}y+{\textstyle{1\over 3}} y^2+\ldots)\,dy$$

$$= \left[{y+{y^2\over 2^2}+{y^3\over 3^2}+\ldots}\right]_0^1=1+{1\over 2^2}+{1\over 3^2}+\ldots.$$ (61)

To evaluate the integral, rotate the coordinate system by $\pi/4$ so

$$x = u\cos\theta-v\sin\theta={\textstyle{1\over 2}}\sqrt{2}\,(u-v)$$ (62)

$$y = u\sin\theta+v\cos\theta={\textstyle{1\over 2}}\sqrt{2}\,(u+v)$$ (63)

and

$$xy = {\textstyle{1\over 2}}(u^2-v^2)$$ (64)

$$1-xy = {\textstyle{1\over 2}}(2-u^2+v^2).$$ (65)

Then

$$I=4\int_0^{\sqrt{2}\,/2} \int_0^u {du\,dv\over 2-u^2+v^2}+4\... ...}}\int_0^{\sqrt{2}\,-u} {du\,dv\over 2-u^2+v^2}\equiv I_1+I_2.$$ (66)

Now compute the integrals $I_1$ and $I_2$.

$$I_1 = 4\int_0^{\sqrt{2}\,/2} \left[{\int_0^u {dv\over 2-u^2+v^2}}\right]\,du$$

$$= 4\int_0^{\sqrt{2}\,/2} \left[{{1\over\sqrt{2-u^2}} \tan^{-1}\left({v\over \sqrt{2-u^2}}\right)}\right]_0^u\,du$$

$$= 4\int_0^{\sqrt{2}\,/2} {1\over\sqrt{2-u^2}} \tan^{-1}\left({u\over\sqrt{2-u^2}}\right)\,du.$$ (67)

Make the substitution

$$u = \sqrt{2}\,\sin\theta$$ (68)

$$\sqrt{2-u^2} = \sqrt{2}\,\cos\theta$$ (69)

$$du = \sqrt{2}\,\cos\theta\,d\theta,$$ (70)

so

$$\tan^{-1}\left({u\over\sqrt{2-u^2}}\right)=\tan^{-1}\left({\sqrt{2}\,\sin\theta\over \sqrt{2}\,\cos\theta}\right)=\theta$$ (71)

and

$$I_1=4\int_0^{\pi/6} {1\over\sqrt{2}\,\cos\theta} \theta\sqrt{2}\,\cos\theta\,d\theta = 2[\theta^2]^{\pi/6}_0={\pi^2\over 18}.$$ (72)

$I_2$ can also be computed analytically,

$$I_2 = 4\int_{\sqrt{2}\,/2}^{\sqrt{2}} \left[{\int_0^{\sqrt{2}\,-u} {dv\over 2-u^2+v^2}}\right]\,du$$

$$= 4\int_{\sqrt{2}\,/2}^{\sqrt{2}} \left[{{1\over\sqrt{2-u^2}}\tan^{-1}\left({v\over \sqrt{2-u^2}}\right)}\right]_0^{\sqrt{2}\,-u} \,du$$

$$= 4\int^{\sqrt{2}}_{\sqrt{2}\,/2} {1\over\sqrt{2-u^2}} \tan^{-1}\left({\sqrt{2}-u\over \sqrt{2-u^2}}\right)\,du.$$ (73)

But

$\tan^{-1}\left({\sqrt{2}-u\over\sqrt{2-u^2}}\right)=\tan^{-1}\left({\sqrt{2}-\sqrt{2}\,\sin\theta\over\sqrt{2}\,\cos\theta}\right)$
$\quad =\tan\left({1-\sin\theta\over\cos\theta}\right)=\tan^{-1}\left({\cos\theta\over 1+\sin\theta}\right)$
$\quad =\tan^{-1}\left[{\sin({\textstyle{1\over 2}}\pi-\theta)\over 1+\cos({\textstyle{1\over 2}}\pi-\theta)}\right]$
$\quad = \tan^{-1}\left\{{2\sin[{\textstyle{1\over 2}}({\textstyle{1\over 2}}\pi... ...over 2\cos^2[{\textstyle{1\over 2}}({\textstyle{1\over 2}}\pi-\theta)]}\right\}$
$\quad = {\textstyle{1\over 2}}({\textstyle{1\over 2}}\pi-\theta),$	(74)

so

$$I_2 = 4\int_{\pi/6}^{\pi/2} {1\over\sqrt{2}\,\cos\theta}({\textstyle{1\over 4}}\pi-{\textstyle{1\over 2}}\theta)\sqrt{2}\,\cos\theta\,d\theta$$

$$= 4\left[{{\textstyle{1\over 4}}\pi\theta-{\textstyle{1\over 4}}\theta^2}\right]_{\pi/6}^{\pi/2}$$

$$= 4\left[{\left({{\pi^2\over 8}-{\pi^2\over 16}}\right)-\left({{\pi^2\over 24}-{\pi^2\over 144}}\right)}\right]= {\pi^2\over 9}.$$ (75)

Combining $I_1$ and $I_2$ gives

$$\zeta(2)=I_1+I_2={\pi^2\over 18}+{\pi^2\over 9}={\pi^2\over 6}.$$ (76)

See also Abel's Functional Equation, Debye Functions, Dirichlet Beta Function, Dirichlet Eta Function, Dirichlet Lambda Function, Harmonic Series, Hurwitz Zeta Function, Khintchine's Constant, Lehmer's Phenomenon, Psi Function, Riemann Hypothesis, Riemann P-Series, Riemann-Siegel Functions, Stieltjes Constants, Xi Function

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Riemann Zeta Function

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