Analytic Continuation of the Riemann Zeta Function

Posted November 28. 2020. 819 words. 4 min read.
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In the study of analytic number theory, the Riemannn zeta function ζ(s)\zeta(s) is a frequently used tool to study number-theoretic objects. Originally, ζ(s)\zeta(s) is defined as

ζ(s)=∑n=1∞1ns(1)\zeta(s)=\sum_{n=1}^\infty{1\over n^s}\tag1

To determine its convegence, let's consider Riemann-Stieltjes integration:

ζN(s)=∑n=1N1ns=∫1−ε∞d⌊t⌋ts=⌊t⌋ts∣1−εN+s∫1N⌊t⌋ts+1dt=N1−s+s∫1N⌊t⌋t+1dt\begin{aligned} \zeta_N(s) &=\sum_{n=1}^N{1\over n^s}=\int_{1-\varepsilon}^\infty{\mathrm d\lfloor t\rfloor\over t^s} \\ &=\left.{\lfloor t\rfloor\over t^s}\right|_{1-\varepsilon}^N+s\int_1^N{\lfloor t\rfloor\over t^{s+1}}\mathrm dt \\ &=N^{1-s}+s\int_1^N{\lfloor t\rfloor\over t^{+1}}\mathrm dt \end{aligned}

It can be easily verified that this expression converges absolutely and uniformly when ℜ(s)>1\Re(s)>1, which allows us to make some manipulations with it. Let's have a look

An Identity due to Poisson's Summation Formula

Define

ψ(x)=∑n=1∞e−n2πx\psi(x)=\sum_{n=1}^\infty e^{-n^2\pi x}

Then by Poisson's summation formula we have

2ψ(x)+1=∑n∈Ze−n2πx=1x∑n∈Ze−n2π/x2\psi(x)+1=\sum_{n\in\mathbb Z}e^{-n^2\pi x}={1\over\sqrt x}\sum_{n\in\mathbb Z}e^{-n^2\pi/x}

which leads to

ψ(x)=1xψ(1x)+12x−12(2)\psi(x)={1\over\sqrt x}\psi\left(\frac1x\right)+{1\over2\sqrt x}-\frac12\tag2

Integral Representation for ζ(s)\zeta(s)

Let's perform a Mellin transform on this function so that

∫0∞xs/2−1ψ(x)dx=∫0∞xs/2−1∑n=1∞e−n2πxdx=π−s/2Γ(s2)∑n=1∞1ns\begin{aligned} \int_0^\infty x^{s/2-1}\psi(x)\mathrm dx &=\int_0^\infty x^{s/2-1}\sum_{n=1}^\infty e^{-n^2\pi x}\mathrm dx \\ &=\pi^{-s/2}\Gamma\left(\frac s2\right)\sum_{n=1}^\infty{1\over n^s} \end{aligned}

Now, by (1) we obtain this identity:

∫0∞xs/2−1ψ(x)dx=π−s/2Γ(s2)ζ(s)(3)\int_0^\infty x^{s/2-1}\psi(x)\mathrm dx=\pi^{-s/2}\Gamma\left(\frac s2\right)\zeta(s)\tag3

As a result, we can study the properties of the Riemann zeta function by digging deeper into the integral on the left hand side.

Analytic Continuation of the Integral

First, let's split the integral into two parts

∫0∞xs/2−1ψ(x)dx=∫01xs/2−1ψ(x)dx+∫1∞xs/2−1ψ(x)dx\int_0^\infty x^{s/2-1}\psi(x)\mathrm dx=\int_0^1x^{s/2-1}\psi(x)\mathrm dx+\int_1^\infty x^{s/2-1}\psi(x)\mathrm dx

Then, applying (2) to ∫01\int_0^1 side gives

∫01xs/2−1ψ(x)dx=∫01xs/2−1[1xψ(1x)+12x−12]dx=∫01x(s−1)/2−1ψ(1x)dx⏟t=x−1+12∫01x(s−1)/2−1dx−12∫01xs/2−1dx=∫1∞t(1−s)/2−1ψ(t)dt+1s(s−1)\begin{aligned} \int_0^1x^{s/2-1}\psi(x)\mathrm dx &=\int_0^1x^{s/2-1}\left[{1\over\sqrt x}\psi\left(\frac1x\right)+{1\over2\sqrt x}-\frac12\right]\mathrm dx \\ &=\underbrace{\int_0^1x^{(s-1)/2-1}\psi\left(\frac1x\right)\mathrm dx}_{t=x^{-1}} \\ &+\frac12\int_0^1x^{(s-1)/2-1}\mathrm dx-\frac12\int_0^1x^{s/2-1}\mathrm dx \\ &=\int_1^\infty t^{(1-s)/2-1}\psi(t)\mathrm dt+{1\over s(s-1)} \end{aligned}

Now, plugging this result to the original equation gives

∫0∞xs/2−1ψ(x)dx=1s(s−1)+∫1∞[xs/2+x(1−s)/2]ψ(x)dxx\int_0^\infty x^{s/2-1}\psi(x)\mathrm dx={1\over s(s-1)}+\int_1^\infty[x^{s/2}+x^{(1-s)/2}]\psi(x){\mathrm dx\over x}

As we can observe that the right hand side does not change when we replace ss with 1−s1-s. Hence, by (3) we have

π−s/2Γ(s2)ζ(s)=π(s−1)/2Γ(1−s2)ζ(1−s)\pi^{-s/2}\Gamma\left(\frac s2\right)\zeta(s) =\pi^{(s-1)/2}\Gamma\left({1-s\over2}\right)\zeta(1-s)

Now, in order to achieve the optimal simplicity, we multiply both side with Γ(1−s2)\Gamma\left(1-\frac s2\right):

Γ(s2)Γ(1−s2)ζ(s)=πs−1/2Γ(1−s2)Γ(1−s2+12)ζ(s)\Gamma\left(\frac s2\right)\Gamma\left(1-\frac s2\right)\zeta(s)=\pi^{s-1/2}\Gamma\left({1-s\over2}\right)\Gamma\left({1-s\over2}+\frac12\right)\zeta(s)

By Euler's reflection formula, we have

Γ(s2)Γ(1−s2)=πcsc⁡(πs2)\Gamma\left(\frac s2\right)\Gamma\left(1-\frac s2\right)=\pi\csc\left(\pi s\over2\right)

and by Legendre's Duplication formula, we deduce

Γ(1−s2)Γ(1−s2+12)=2sπ1/2Γ(1−s)\Gamma\left({1-s\over2}\right)\Gamma\left({1-s\over2}+\frac12\right)=2^s\pi^{1/2}\Gamma(1-s)

Plugging these results back gives us

πcsc⁡(πs2)ζ(s)=2sπsΓ(1−s)ζ(1−s)\pi\csc\left(\pi s\over2\right)\zeta(s)=2^s\pi^s\Gamma(1-s)\zeta(1-s)

Now, if we were to perform more manipulations, we get

ζ(s)=2sπs−1sin⁡(πs2)Γ(1−s)ζ(1−s)\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)

which is known as the functional equation for ζ(s)\zeta(s).

Conclusion

In this blog post, we begin with the Dirichlet series definition of ζ(s)\zeta(s), and then we try to connect zeta function with an integral representation. Subsequently, we use Poisson's summation formula to obtain its analytic continuation. However, this analytic continuation has other impacts. If we look back to the equation

ζ(s)=2sπs−1sin⁡(πs2)Γ(1−s)ζ(1−s),\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s),

we can observe that for ℜ(s)<0\Re(s)<0 the right hand side becomes zero whenever s=−2k≠0s=-2k\ne0. Hence, we call such ss's as the trivial zeros of ζ(s)\zeta(s). However, there are also other occasions in which the right hand side is zero. Alternatively, we call that kind of zeros the nontrivial zeros of ζ(s)\zeta(s).

On going through these definition, we can now have a good basic grasp of the Riemann hypothesis:

Riemann hypothesis: All nontrivial zeros of ζ(s)\zeta(s) lie on the line ℜ(s)=12\Re(s)=\frac12.

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We sometimes leave a dangling `else` just as a threat to the compiler that is better run that if statement or else.

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