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# The Prolonged Riemann Speculation and Ramanujan’s Sum

## Riemann Speculation and Ramanujan’s Sum Clarification

The purpose of this text is to supply the definitions and theorems which are needed to know these two Riemann hypotheses, i.e. why is a strip within the advanced airplane referred to as important, what are trivial and non-trivial zeros, and what’s a group character, and so on. We are going to collect a few info across the features concerned, particularly a number of useful equations.

The prolonged Riemann speculation is a generalization of the Riemann speculation that grew to become essential when fashionable laptop science started to make use of number-theoretical outcomes for encryption mechanisms. We are going to elaborate on these relations to utilized arithmetic in one other article. The principle topic of this text would be the Riemann zeta perform which is type of a particular case of an L-function and what we and most different individuals imply if we briefly say zeta-function or ##zeta##-function.

### Zeta Features

• Ethereal zeta perform, associated to the zeros of the Ethereal perform
• Arakawa–Kaneko zeta perform
• Arithmetic zeta perform
• Artin–Mazur zeta perform of a dynamical system
• Barnes zeta perform or double zeta perform
• Beurling zeta perform of Beurling generalized primes
• Dedekind zeta perform of a quantity subject
• Duursma zeta perform of error-correcting codes
• Epstein zeta perform of a quadratic type
• Goss zeta perform of a perform subject
• Hasse–Weil zeta perform of a spread
• Peak zeta perform of a spread
• Hurwitz zeta perform, a generalization of the Riemann zeta perform
• Igusa zeta perform
• Ihara zeta perform of a graph
• Jacobi zeta perform
• L-function, a “twisted” zeta perform
• Lefschetz zeta perform of a morphism
• Lerch zeta perform, a generalization of the Riemann zeta perform
• Native zeta perform of a characteristic-p selection
• Matsumoto zeta perform
• Minakshisundaram–Pleijel zeta perform of a Laplacian
• Motivic zeta perform of a motive
• A number of zeta perform, or Mordell–Tornheim zeta perform of a number of variables
• Prime zeta perform, just like the Riemann zeta perform, however solely summed over primes
• Riemann zeta perform, the archetypal instance
• Ruelle zeta perform
• Selberg zeta perform of a Riemann floor
• Shimizu L-function
• Shintani zeta perform
• Subgroup zeta perform
• Weierstraß zeta perform
• Witten zeta perform of a Lie group
• Zeta perform of an incidence algebra, a perform that maps each interval of a poset to the fixed worth 1. Regardless of not resembling a holomorphic perform, the particular case for the poset of integer divisibility is expounded as a proper Dirichlet sequence to the Riemann zeta perform.
• Zeta perform of an operator or spectral zeta perform

### Definitions

start{align*}
zeta(s)&:=sum_{n=1}^infty dfrac{1}{n^s}=1+dfrac{1}{2^s}+dfrac{1}{3^s}+dfrac{1}{4^s}+dfrac{1}{5^s}+cdots = prod_{ptext{ prime}}dfrac{1}{1-p^{-s}}quad (sin mathbb{C})
Gamma(z)&:=lim_{n to infty}dfrac{n!n^z}{zcdot (z+1)cdot ldotscdot (z+n)}=int_0^{infty }t^{z-1}e^{-t},dt[6pt]
Gamma(0)&=1, , ,Gamma(1)=1, , ,Gamma(n)=(n-1)!, , ,Gamma(1/2)=sqrt{pi}, , ,Gamma(z+1)=zGamma(z)
finish{align*}

Let ##mathbb{P}## be the set of all primes and ##pin mathbb{P}.## Then
start{align*}
left(1-dfrac{1}{p^s}proper)zeta(s)&=sum_{n=1}^infty dfrac{1}{n^s}-sum_{n=1}^infty dfrac{1}{(pn)^s} =sum_{stackrel{n=1}{nnotin pmathbb{Z}}}^infty dfrac{1}{n^s}
=sum_{stackrel{n=1}{pnmid n}}^infty dfrac{1}{n^s}
prod_{pinmathbb{P}}left(1-dfrac{1}{p^s}proper)zeta(s)&=sum_{stackrel{n=1}{nnotequiv 0mod p,forall ,pin mathbb{P}}}^infty dfrac{1}{n^s}=sum_{nin {1}}dfrac{1}{n^s}=1
zeta(s)&=1/prod_{pinmathbb{P}}left(1-dfrac{1}{p^s}proper)=
prod_{pinmathbb{P}}left(1/left(1-dfrac{1}{p^s}proper)proper)=prod_{pinmathbb{P}}dfrac{1}{1-p^{-s}}
finish{align*}

From ##e^{-t}=displaystyle{lim_{n to infty}}left(1-dfrac{t}{n}proper)^n## we get
start{align*}
int_0^infty e^{-t},t^{z-1},dt&=lim_{n to infty}int_0^nleft(1-dfrac{t}{n}proper)^nt^{z-1},dt
&stackrel{t=ns}{=}lim_{n to infty}int_0^1 (1-s)^nn^{z}s^{z-1},ds
&=lim_{n to infty}n^z left(left[dfrac{s^z}{z}(1-s)^{n}right]_0^1 +dfrac{n}{z}int_0^1 (1-s)^{n-1}s^z,dsright)
&=lim_{n to infty}n^zleft(dfrac{n}{z}cdot dfrac{n-1}{z+1}cdotldotscdot dfrac{1}{z+n-1}int_0^1s^{z+n-1},dsright)
&=Gamma(z)
finish{align*}

start{align*}
Gamma(z+1)&=int_0^{infty }t^{z}e^{-t},dt=-left[t^ze^{-t}right]_0^infty + zint_0^{infty }t^{z-1}e^{-t},dt=zGamma(z)
finish{align*}

The actual values of the ##Gamma##-function are apparent aside from ##Gamma(1/2).##
start{align*}
left(int_mathbb{R}e^{-t^2},dtright)^2&=left(int_mathbb{R}e^{-x^2},dx proper)cdot left(int_mathbb{R}e^{-y^2},dy proper)=int_mathbb{R}int_mathbb{R}e^{-x^2-y^2},dx,dy
&=int_0^{infty}int_0^{2pi}re^{-r^2},dvarphi ,dr=2pi int_0^infty re^{-r^2},dr=-pileft[e^{-r^2}right]_0^infty =pi
finish{align*}
Therefore
\$\$
Gamma(1/2)=int_0^infty t^{-1/2}e^{-t},dt=2int_0^infty e^{-r^2},dr=int_mathbb{R} e^{-r^2},dr= sqrt{pi}.
\$\$

We are going to want the essential useful equation for the zeta-function (Riemann, 1859) which we are going to show subsequent. The proof requires some preparations.

### Poisson Summation Method

Let ##f, : ,mathbb{R}longrightarrow mathbb{C}## be a repeatedly differentiable perform with ##f(x)=O(|x|^{-2})## for ##|x|to infty .## Let ##hat{f}, : ,mathbb{R}longrightarrow mathbb{C}## be the Fourier remodel of ##f,## i.e.
\$\$
hat{f}(t)=int_{-infty }^{infty }f(x)e^{-2pi i x t},dx.
\$\$
Then \$\$
sum_{nin mathbb{Z}} f(n)=sum_{nin mathbb{Z}}hat{f}(n).
\$\$

Moreover, we get for ##lambda >0## and ##f_lambda (x):=f(lambda x)##
\$\$
hat{f_lambda }(t)=dfrac{1}{lambda }hat{f}left(dfrac{t}{lambda }proper).
\$\$

##F(x):=sum_{nin mathbb{Z}} f(x+n) , : ,mathbb{R}longrightarrow mathbb{C}## is a repeatedly differentiable perform with interval ##1## and might as such be developed right into a Fourier-series
\$\$
F(x)=sum_{nin mathbb{Z}} c_ne^{2pi i n x}.
\$\$
This sequence converges uniformly since ##F## is repeatedly differentiable. Due to this fact
start{align*}
c_n &=int_0^1 F(x)e^{-2pi i n x},dx=sum_{okay=-infty }^infty int_0^1 f(x+okay)e^{-2pi i n x},dx
&=sum_{okay=-infty }^infty int_{okay}^{okay+1}f(x)e^{-2pi i n x},dx=int_{-infty }^infty f(x)e^{-2pi i n x},dx=hat{f}(n)
finish{align*}
Thus
\$\$
sum_{nin mathbb{Z}}f(x+n)=sum_{nin mathbb{Z}}hat{f}(n)e^{-2pi i n x}
\$\$
and the assertion follows for ##x=0.##  The method for ##hat{f_lambda }## follows from
start{align*}
hat{f_lambda }(t)&=int_{-infty }^infty f(lambda x)e^{-2pi i x t},dx=frac{1}{lambda }int_{-infty }^infty f(lambda x)e^{-2pi i (lambda x) frac{t}{lambda }},d(lambda x)=frac{1}{lambda }hat{f}left(frac{t}{lambda }proper)
finish{align*}

### Instance

The Fourier remodel of ##f(x):=e^{-pi x^2}## is ##hat{f}(t)=displaystyle{int_{-infty }^infty e^{-pi x^2}e^{-2pi i x t},dx}=f(t).##

start{align*}
hat{f}(0)&= int_{-infty }^infty e^{-pi x^2},dx=dfrac{1}{sqrt{pi}} int_mathbb{R}e^{-y^2},dystackrel{(s.a.)}{=}dfrac{Gamma(1/2)}{sqrt{pi}}=1
hat{f}(t)&=e^{-pi t^2}int_mathbb{R}e^{pi t^2 -2pi i x t -pi x^2},dx=e^{-pi t^2}int_mathbb{R}e^{-pi(x+it)^2},dx
&stackrel{(s=x+it)}{=}e^{-pi t^2}int_mathbb{R}e^{-pi s^2},ds=e^{-pi t^2}hat{f}(0)=e^{-pi t^2}=f(t)
finish{align*}
With a purpose to see why the substitution works, we apply Cauchy’s integral theorem on ##zlongmapsto e^{-pi z^2}## alongside the closed integration path start{align*}
int_{-R}^{R}e^{-pi(x+ i t)^2},dx&=int_{-R+it}^{R+it}e^{-pi z^2},dz= underbrace{int_{-R+it}^{-R}e^{-pi z^2},dz}_{to 0 textual content{ for }Rto infty} + int_{-R}^{R}e^{-pi s^2},ds + underbrace{int_{R}^{R+it}e^{-pi z^2},dz}_{to 0 textual content{ for }Rto infty}
finish{align*}

The classical Jacobian theta-function is outlined as ##vartheta (z,tau ):=sum_{n= -infty }^{infty }e^{pi i n^{2}tau +2pi inz}.## The sequence is convergent in ##mathbb {C} occasions mathbb {H} ## the place ##mathbb {H} ={xin mathbb{C} mid mathfrak{I(tau)}>0}.## We have an interest within the theta-function with a hard and fast worth ##z## such that ##vartheta(z,cdot)## is a holomorphic perform on ##mathbb{H}.## Let ##z= 0## and ##theta, : ,mathbb{R}^+longrightarrow mathbb{C}##
\$\$
theta(x):=sum_{n= -infty }^{infty }e^{-pi n^{2}x}
\$\$

### Purposeful Equation of the Theta-Collection

start{align*}
theta(x)&=dfrac{1}{sqrt{x}};thetaleft(dfrac{1}{x}proper); textual content{ for all } x>0;textual content{ and }; theta(x)=O(1/sqrt{x},),textual content{ for }x searrow 0
finish{align*}

We’ve got ##hat{f_1}(x)=f_1(x)## for ##f_1, : ,mathbb{R}longrightarrow mathbb{C}, , ,f_1(x):=e^{-pi x^2}## by the instance above. For ##lambda >0## we get with ##f_lambda (x):=e^{-pi x^2lambda }##
\$\$
hat{f_lambda }(x)=dfrac{1}{sqrt{lambda }}hat{f}left(dfrac{x}{lambda }proper)= dfrac{1}{sqrt{lambda }}e^{-pi x^2/lambda }
\$\$
such that the Poisson summation with ##f=f_1## yields
\$\$
theta(lambda )=sum_{nin mathbb{Z}}e^{-pi n^2lambda }=sum_{nin mathbb{Z}}f_lambda (n)=sum_{nin mathbb{Z}}hat{f_lambda}(n)=dfrac{1}{sqrt{lambda }}sum_{nin mathbb{Z}}e^{-pi n^2/lambda }=dfrac{1}{sqrt{lambda }};thetaleft(dfrac{1}{lambda }proper)
\$\$
Lastly, ##lim_{lambda to infty}theta(lambda )=1,## i.e. with ##x=1/lambda ##
\$\$
lim_{xsearrow 0}theta(x)=lim_{lambdatoinfty}theta(1/lambda )=lim_{lambdatoinfty}sqrt{lambda };theta(lambda )
=lim_{lambdatoinfty}sqrt{lambda }=lim_{x searrow 0}1/sqrt{x}.
\$\$

### Supplementary Theorem of the Gamma Operate, Euler 1749

\$\$
Gamma(z)Gamma(1-z) = dfrac{pi}{sin pi z};textual content{ for all }zin mathbb{C}-mathbb{Z}
\$\$

\$\$
phi(z):=Gamma(z)Gamma(1-z)=dfrac{Gamma(1+z)}{z}cdot dfrac{Gamma(2-z)}{1-z}
\$\$
is meromorphic on ##mathbb{C}## with first order singularities in ##z=nin mathbb{Z},## and bounded on ##S_1:={zin mathbb{C},|,0leq mathfrak{R}(z)leq 1text{ and }|mathfrak{I}(z)|geq 1}.## Moreover, let
\$\$
psi(z):=sin(pi z)phi(z)=sin(pi z)Gamma(z)Gamma(1-z)
\$\$
that’s holomorphic all over the place as a result of ##sin(pi n)=0## with first order zeros.
start{align*}
phi(z+1)&=phi(-z)=-dfrac{Gamma(1-z)}{z}cdot dfrac{Gamma(2+z)}{1+z}=-dfrac{Gamma(1-z)}{z}cdot dfrac{zGamma(z)(z+1)}{1+z}
&=-Gamma(1-z)Gamma(z)=-phi(z)[12pt]
psi(z+1)&=sin(pi(z+1))phi(z+1)=sin(pi z)phi(z)=psi(z)
finish{align*}
Thus ##|psi(z)|leq |phi(z)|leq Ce^z;(*)## for all ##zin mathbb{C}## since ##phi(z)## is bounded on ##S_1.## Nevertheless, ##psi(z)neq 0## all over the place, therefore ##psi(z)=e^{g(z)}## for some on ##mathbb{C}## holomorphic perform ##g(z).## This implies, given ##(*),## that ##g(z)=a+bz## for some constants ##a,bin mathbb{C}.## It follows that ##b=0## as a result of ##psi(-z)=sin(-pi z)phi(-z)=sin(pi z)phi(z)=psi(z)## and ##psi(z)## is fixed. From
\$\$
psi(z)=dfrac{sin(pi z)}{z} Gamma(z+1)Gamma(1-z)
\$\$
we get ##psi(0)=pi## which proves the assertion, and offers a second proof for ##Gamma(1/2) = sqrt{pi}.##

### Legendre’s Doubling Method, 1809

\$\$
Gammaleft(dfrac{z}{2}proper)cdotGammaleft(dfrac{z+1}{2}proper)=dfrac{sqrt{pi}}{2^{z-1}}cdot Gamma(z);textual content{ for all }zin mathbb{C}-{0,-1,-2,ldots}
\$\$

Set ##G(z):=2^zGammaleft(dfrac{z}{2}proper)Gammaleft(dfrac{z+1}{2}proper).## Then
start{align*}
G(z+1)&=2^{z+1}Gammaleft(dfrac{z+1}{2}proper)Gammaleft(dfrac{z}{2}+1right)=2^{z+1}Gammaleft(dfrac{z+1}{2}proper)dfrac{z}{2}Gammaleft(dfrac{z}{2}proper)=zG(z)
finish{align*}
By the concept of Bohr-Mollerup, we get ##G(z) = G(1) cdot Gamma(z) = 2sqrt{pi};Gamma(z).##

Cp. first drawback in September 2021.

Alternatively, take into account with ##x=z/2## (with a view to keep away from too many powers of ##z/2##)
start{align*}
2^zGammaleft(dfrac{z}{2}proper)Gammaleft(dfrac{z+1}{2}proper)&=
lim_{n to infty}dfrac{2^{2x} n^xn!n^{x+1/2}n!}{x(x+1)cdots(x+n)(x+1/2)(x+3/2)cdots(x+n+1/2)}[6pt]
&=lim_{n to infty} dfrac{2^{z+2n+2}(n!)^2n^{z+1/2}}{z(z+2)cdots(z+2n)(z+1)(z+3)cdots(z+2n+1)}[6pt]
&=lim_{n to infty}dfrac{2^{2(n+1)}(n!)^2n^{1/2}}{(2n+1)!}cdotdfrac{(2n+1)^z(2n+1)!}{z(z+1)cdots(z+2n+1)}cdotleft(dfrac{2n}{2n+1}proper)^z[6pt]
&=lim_{n to infty}dfrac{2^{2(n+1)}(n!)^2n^{1/2}}{(2n)!}cdot Gamma(z)
finish{align*}
From ##z=1## we get ##2Gamma(1/2)Gamma(1)=2sqrt{pi}=lim_{n to infty}dfrac{2^{2(n+1)}(n!)^2n^{1/2}}{(2n)!}cdotGamma(1)## and
\$\$
2^{z-1}Gammaleft(dfrac{z}{2}proper)Gammaleft(dfrac{z+1}{2}proper)=sqrt{pi}cdotGamma(z)
\$\$

The Legendre doubling method is a particular case of (with out proof)

### Gauß’s Multiplication Method, 1812

start{align*}
Gammaleft(dfrac{z}{n}proper)cdotGammaleft(dfrac{z+1}{n}proper)cdotsGammaleft(dfrac{z+n-1}{n}proper)&=dfrac{(2pi)^{(n-1)/2}}{n^{(z-1/2)}}cdot Gamma(z)[6pt]
textual content{for }nin mathbb{N},zin mathbb{C}-{0,-1,-2ldots}&
finish{align*}

Gregory Chudnovsky has proven in 1975, that each quantity ##Gamma (1/6)##, ##Gamma (1/4)##, ##Gamma (1/3)##, ##Gamma (2/3)##, ##Gamma (3/4)## and ## Gamma (5/6)## is transcendent and algebraically impartial of ##pi.## Then again, it’s unknown whether or not ##Gamma(1/5)## is even irrational.

##rho(t):=displaystyle{sum_{n=1}^infty e^{-pi n^2 t}= dfrac{1}{2}(theta(t)-1) stackrel{tsearrow 0}{longrightarrow }Oleft(1/sqrt{t}proper)}.## Thus the next integral exists, and for ##sin mathbb{C}## with ##mathfrak{R}(s)>1,## we get (##t=pi n^2 u##)
start{align*}
Gammaleft(dfrac{s}{2}proper)zeta(s)&= sum_{n=1}^infty Gammaleft(dfrac{s}{2}proper) dfrac{1}{n^s}
=sum_{n=1}^infty dfrac{1}{n^s}int_0^infty t^{s/2}e^{-t}dfrac{dt}{t}= pi^{s/2} int_0^infty u^{s/2}sum_{n=1}^inftyleft(e^{-pi n^2 u}proper)dfrac{du}{u}.
finish{align*}

Lastly, we are able to show the

### Purposeful Equation of the ##zeta##-Operate

Set ##xi(s):=pi^{-s/2}Gamma(s/2)zeta(s).## This at prior on the half airplane ##{mathfrak{R}(s)>1}## holomorphic perform may be prolonged to a perform that’s meromorphic on the whole advanced airplane. The prolonged perform has first order singularities at ##sin {0,1},##
and is holomorphic elsewhere. Furthermore,
\$\$
xi(s)=xi(1-s).
\$\$
The ##zeta##-function can due to this fact be meromorphic prolonged on ##mathbb{C}## with a single pole at ##s=1.## Furthermore,
\$\$
zeta(1-s)=2(2pi)^{-s}Gamma(s)cosleft(dfrac{pi s}{2}proper)zeta(s).
\$\$

Assuming the primary half, we get
start{align*}
zeta(s-1)&=pi^{(1-s)/2}Gammaleft(dfrac{1-s}{2}proper)^{-1}xi(1-s)=pi^{(1-s)/2}Gammaleft(dfrac{1-s}{2}proper)^{-1}pi^{-s/2}Gammaleft(dfrac{s}{2}proper)zeta(s)
&=pi^{-s}sqrt{pi};Gammaleft(dfrac{1-s}{2}proper)^{-1}Gammaleft(dfrac{s}{2}proper)zeta(s)
&=pi^{-s}sqrt{pi};dfrac{sinleft(dfrac{pi(1-s)}{2}proper)}{pi}Gammaleft(dfrac{1+s}{2}proper)Gammaleft(dfrac{s}{2}proper)zeta(s)
&=pi^{-s}cosleft(dfrac{pi s}{2}proper)dfrac{1}{2^{s-1}}Gamma(s)zeta(s)
finish{align*}
We get from the earlier comment that
\$\$
xi(s) = pi^{-s/2}Gamma(s/2)zeta(s)=int_0^infty t^{s/2}underbrace{left(sum_{n=1}^infty e^{-pi n^2t}proper)}_{=rho(t)=(1/2)(theta(t)-1)}dfrac{dt}{t}quad (mathfrak{R}(s)>1)
\$\$
Observe that
\$\$
rholeft(dfrac{1}{t}proper)=dfrac{1}{2}left(thetaleft(dfrac{1}{t}proper)-1right)=dfrac{1}{2}left(sqrt{t};theta(t)-1right)=sqrt{t};rho(t)-dfrac{1}{2}+dfrac{1}{2}sqrt{t}
\$\$
Then
start{align*}
int_0^1 t^{s/2}rho(t)dfrac{dt}{t}&=int_0^1 t^{s/2}left(dfrac{1}{sqrt{t}};rholeft(dfrac{1}{t}proper)+dfrac{1}{2sqrt{t}}-dfrac{1}{2}proper)dfrac{dt}{t}
&=int_0^1 t^{(s-1)/2}rholeft(dfrac{1}{t}proper)dfrac{dt}{t}+dfrac{1}{2}int_0^1 (t^{(s-1)/2}-t^{s/2})dfrac{dt}{t}
&=int_1^infty u^{(1-s)/2}rho(u)dfrac{du}{u}+dfrac{1}{s-1}-dfrac{1}{s}
xi(s)&=int_0^1t^{s/2}rho(t)dfrac{dt}{t}+int_1^infty t^{s/2}rho(t)dfrac{dt}{t}
&= dfrac{1}{s-1}-dfrac{1}{s}+int_1^infty left(t^{s/2}+t^{(1-s)/2}proper) rho(t) dfrac{dt}{t}
finish{align*}
##rho(t)## tends exponentially towards ##0## with growing ##t,## i.e. the integral exists for all ##sin mathbb{C}.## It’s a holomorphic perform in ##s.## Thus, ##xi(s)## is meromorphic prolonged. The final illustration of ##xi(s)## is invariant beneath ##s mapsto 1-s,## which proves the assertion.

### Zeros and Poles of the ##zeta##-Operate

We’ve got simply confirmed that the one pole of the ##zeta##-Operate is at ##x=1.##

To think about the zeros, we are going to distinguish between three circumstances.

1. ##mathfrak{R}(s)>1.##
On this case, we conclude by the product method that
\$\$
zeta(s)=prod_{p textual content{ prime}}dfrac{1}{1-p^{-s}} neq 0
\$\$
2. ##mathfrak{R}(s)<0.##
On this case, we have now ##mathfrak{R}(1-s)>1##, and the useful equation
\$\$
zeta(s)=2(2pi)^{s-1}Gamma(1-s)cosleft(dfrac{pi (1-s)}{2}proper)zeta(1-s)=0
\$\$
if and provided that the cosine turns into zero as a result of ##mathfrak{R}(1-s)>1.## Nevertheless, the advanced cosine has solely actual zeros at ##(2k+1)pi/2.## Due to this fact ##1-s=2k+1## or ##zeta(s)=zeta(-2k)=0## if and provided that ##kin {1,2,3,ldots}.##The zeros ##-2,-4,-6,ldots## are referred to as trivial zeros of the ##zeta##-function.
3. ##0leq mathfrak{R}(s)leq 1.##
This space of the advanced quantity airplane known as the important strip. All remaining zeros should be throughout the important strip. We all know that if ##zeta(a+ib)=0## then ##zeta(1-a-ib)=0,## i.e. potential zeros happen mirrored at ##mathfrak{R}(s)=1/2,## which we name the important line, and mirrored at the actual axis. All ##10^{13}## zeros which were discovered to this point lie on the important line.

The Riemann speculation says, that each one non-trivial zeros of the Riemann zeta perform lie on the important line. ### The Prolonged Riemann Speculation

A advanced Dirichlet character of a finite cyclic group is a bunch homomorphism ##chi: left(mathbb{Z}_nright)^occasions rightarrow mathbb{C}^occasions## into the multiplicative group of the advanced numbers
start{align*}
chi, : ,,(a,n)=1&longrightarrow mathbb{C}-{0}
chi(a)cdot chi(b)=chi(acdot b)
finish{align*}
the place ##(a,n)## denotes the best widespread divisor of ##a## and ##n.## We affiliate with it a periodic perform modulo ##n## by
start{align*}
chi, &: ,mathbb{Z} longrightarrow mathbb{C}
chi(a)=&start{circumstances}
chi(a) &textual content{ if }(a,n)=1
0 &textual content{ if }(a,n)>1 finish{circumstances}
finish{align*}
which we additionally name Dirichlet character modulo n. They play a key function in Dirichlet’s theorem on arithmetic progressions:

For any two constructive coprime integers ##a## and ##d## there are infinitely many primes of the shape ##a + nd.##

Examples are the trivial character ##chi_0(a)=1## for all ##(a,n)=1,## and the Jacobi-character
\$\$
chi(a)=left(dfrac{a}{p}proper)=
start{circumstances}
1 &textual content{ if } anotequiv 0 mod p textual content{ and there may be an integer x such that}aequiv x^2 (p)
-1 &textual content{ if }anotequiv 0 mod p textual content{ and there’s no such x}
0 &textual content{ if }aequiv 0 mod p
finish{circumstances}
\$\$
prolonged to a homomorphism for ##n=prod p## (cp. Jacobi image). Each Dirichlet character has a (Dirichlet) ##L##-function, or ##L##-series outlined by
\$\$
L_chi(z):=sum_{a=1}^infty dfrac{chi(a)}{a^z}
\$\$
Dirichlet sequence converge completely and regionally uniform within the half-plane ##{mathfrak{R}(z)>1}## since
\$\$
left|dfrac{chi(a)}{a^z}proper|=left|dfrac{chi(a)}{a^{mathfrak{R}(z)} cdot e^{ i log(a) mathfrak{I}(z) }} proper|=dfrac{1}{a^{mathfrak{R}(z)}} textual content{ or } 0
\$\$
and the homomorphism property forces ##chi(a)## to be a root of unity (or ##0##). ##L##-series may be holomorphic prolonged on ##{mathfrak{R}(z)>0},## aside from the trivial character, which has a primary order pole at ##z=1.## The Dirichlet perform ##L_chi(z)## has the Riemann property, if all zeros throughout the important strip are already on the important line ##{mathfrak{R}(z)=frac{1}{2}}##.

The prolonged Riemann speculation says that each one zeros of all Dirichlet L-functions to advanced Dirichlet characters of finite cyclic teams throughout the important strip lie already on the important line.

We get for the trivial character ##chi_0## modulo ##n##
\$\$
L_{chi_0}(z)=sum_{(a,n)=1}dfrac{1}{a^z}=zeta(z)cdot prod_{p|ntext{ prime}}left(1-dfrac{1}{p^z}proper)
\$\$
which has precisely the identical zeros in ##{mathfrak{R}(z)>0}## because the ##zeta##-function. The prolonged Riemann speculation thus implies the abnormal Riemann speculation.

### Srinivasa Ramanujan, 1887-1920 “The divergent sequence are the invention of the satan, and it’s a disgrace to base on them any demonstration by any means.” (Niels Henrik Abel, 1832)

\$\$
\$\$
Therefore
\$\$
1+2+3+ldots=zeta(-1)=zeta(1-2)=2(2pi)^{-2}Gamma(2)cosleft(dfrac{2pi}{2}proper)zeta(2)=-dfrac{1}{12}
\$\$