π“œ

Math Notes

University Mathematics

ProofComplete

Proof of Functional Equation

We prove the functional equation using the Poisson summation formula and theta functions, Riemann's original approach.

TheoremFunctional Equation (Restated)

For all s∈Cs \in \mathbb{C}, sβ‰ 0,1s \neq 0, 1:

ΞΎ(s)=ΞΎ(1βˆ’s)\xi(s) = \xi(1-s)

where ΞΎ(s)=12s(sβˆ’1)Ο€βˆ’s/2Ξ“(s/2)ΞΆ(s)\xi(s) = \frac{1}{2} s(s-1) \pi^{-s/2} \Gamma(s/2) \zeta(s).

ProofProof via Jacobi Theta Function

Step 1: Define the Theta Function

Define:

ΞΈ(x)=βˆ‘n=βˆ’βˆžβˆžeβˆ’Ο€n2xforΒ x>0\theta(x) = \sum_{n=-\infty}^{\infty} e^{-\pi n^2 x} \quad \text{for } x > 0

This is the Jacobi theta function, which satisfies the transformation property:

ΞΈ(1/x)=x θ(x)\theta(1/x) = \sqrt{x} \, \theta(x)

This can be proved using Poisson summation or by recognizing ΞΈ\theta as a modular form.

Step 2: Connect to Gamma and Zeta

For β„œ(s)>1\Re(s) > 1:

Ο€βˆ’s/2Ξ“(s/2)ΞΆ(s)=∫0∞xs/2βˆ’1βˆ‘n=1∞eβˆ’Ο€n2xdx\pi^{-s/2} \Gamma(s/2) \zeta(s) = \int_0^{\infty} x^{s/2-1} \sum_{n=1}^{\infty} e^{-\pi n^2 x} dx

Exchanging sum and integral (justified by convergence):

=βˆ‘n=1∞∫0∞eβˆ’Ο€n2xxs/2βˆ’1dx=βˆ‘n=1∞1ns∫0∞eβˆ’Ο€yys/2βˆ’1dy= \sum_{n=1}^{\infty} \int_0^{\infty} e^{-\pi n^2 x} x^{s/2-1} dx = \sum_{n=1}^{\infty} \frac{1}{n^s} \int_0^{\infty} e^{-\pi y} y^{s/2-1} dy

Therefore:

Ο€βˆ’s/2Ξ“(s/2)ΞΆ(s)=∫0∞xs/2βˆ’1ΞΈ(x)βˆ’12dx\pi^{-s/2} \Gamma(s/2) \zeta(s) = \int_0^{\infty} x^{s/2-1} \frac{\theta(x) - 1}{2} dx

Step 3: Split the Integral

Split at x=1x = 1:

=∫1∞xs/2βˆ’1ΞΈ(x)βˆ’12dx+∫01xs/2βˆ’1ΞΈ(x)βˆ’12dx= \int_1^{\infty} x^{s/2-1} \frac{\theta(x) - 1}{2} dx + \int_0^1 x^{s/2-1} \frac{\theta(x) - 1}{2} dx

Step 4: Apply Theta Transformation

In the second integral, substitute x↦1/xx \mapsto 1/x and use ΞΈ(1/x)=xΞΈ(x)\theta(1/x) = \sqrt{x} \theta(x):

∫01xs/2βˆ’1ΞΈ(x)βˆ’12dx=∫1∞xβˆ’s/2βˆ’1ΞΈ(1/x)βˆ’12β‹…x2dx\int_0^1 x^{s/2-1} \frac{\theta(x) - 1}{2} dx = \int_1^{\infty} x^{-s/2-1} \frac{\theta(1/x) - 1}{2} \cdot x^2 dx=∫1∞xβˆ’s/2+1/21/xΞΈ(1/x)βˆ’12dx=∫1∞x(1βˆ’s)/2βˆ’1ΞΈ(x)βˆ’12dx= \int_1^{\infty} x^{-s/2+1/2} \frac{\sqrt{1/x}\theta(1/x) - 1}{2} dx = \int_1^{\infty} x^{(1-s)/2-1} \frac{\theta(x) - 1}{2} dx

Step 5: Combine

We get:

Ο€βˆ’s/2Ξ“(s/2)ΞΆ(s)=∫1∞(xs/2βˆ’1+x(1βˆ’s)/2βˆ’1)ΞΈ(x)βˆ’12dx+boundaryΒ terms\pi^{-s/2} \Gamma(s/2) \zeta(s) = \int_1^{\infty} \left(x^{s/2-1} + x^{(1-s)/2-1}\right) \frac{\theta(x) - 1}{2} dx + \text{boundary terms}

Handling boundary terms carefully (from the pole at s=1s=1) gives:

ΞΎ(s)=ΞΎ(1βˆ’s)\xi(s) = \xi(1-s)
β– 
Remark

Alternative proofs use:

  • Mellin transform: Direct Mellin transform of ΞΈ(x)\theta(x)
  • Contour integration: Residue calculus on a modified zeta integral
  • Modular forms: Recognizing ΞΈ\theta as a modular form of weight 1/21/2
ExampleModern Perspective

The functional equation reflects the self-duality of the Gaussian eβˆ’Ο€x2e^{-\pi x^2} under Fourier transform. This connects to the Poisson summation formula:

βˆ‘n∈Zf(n)=βˆ‘k∈Zf^(k)\sum_{n \in \mathbb{Z}} f(n) = \sum_{k \in \mathbb{Z}} \hat{f}(k)

and reveals deep connections to automorphic forms and the Langlands program.