Functional Equation for Riemann Zeta Function

Theorem

For all $\Re(s) > 0$,

$\displaystyle \pi^{-s/2}\Gamma(s/2)\zeta(s) = -\frac 1 {s(1-s)} + \int_1^\infty \left[ x^{s/2-1} + x^{-(s+1)/2} \right] \omega(x)\ dx$

where $\Gamma$ is the gamma function, $\displaystyle \omega(x) = \sum_{n=1}^\infty e^{-\pi n^2 x}$ and $\zeta$ is the Riemann zeta function.

Thus:

$\xi(s) = \xi(1-s)$

where $\xi$ is the completed Riemann zeta function.

Proof

We have

$\displaystyle \Gamma \left({\frac s 2}\right) = \int_0^\infty t^{s/2-1} e^{-t}\ dt$

First let us substitute $t = \pi n^2 x$ to give:

$\displaystyle \pi^{-s/2}\Gamma \left({\frac s 2}\right) n^{-s} = \int_0^\infty x^{s/2-1} e^{-\pi n^2 x}\ dx$

Now by definition, on $\Re(s) > 1$, we have

$\displaystyle \zeta(s) = \sum_{n \geq 1} n^{-s}$

so, summing over $n \geq 1$ we obtain:

$\displaystyle \pi^{-s/2}\Gamma \left({\frac s 2}\right) \zeta(s) = \int_0^\infty x^{s/2-1} \omega(x)\ dx$

where $\displaystyle \omega(x) = \sum_{n \geq 1} e^{-\pi n^2 x}$.

Next we split the integral at $x = 1$ as follows:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \pi^{-s/2}\Gamma \left({\frac s 2}\right) \zeta(s)\)	\(=\)	\(\displaystyle \int_0^1 x^{s/2-1} \omega(x)\ dx + \int_1^\infty x^{s/2-1} \omega(x)\ dx\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \int_1^\infty x^{s/2-1} \omega(x)\ dx + \int_1^\infty x^{-s/2-1} \omega(x^{-1})\ dx\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	substituting $x \mapsto x^{-1}$ in the first integral.

Now we define the Jacobi theta function:

$\displaystyle \theta(x) = \sum_{n \in \Z}e^{-\pi n^2 x} = 2\omega(x) + 1$

Since $e^{-x^2}$ is a fixed point of the Fourier transform, by Properties of the Fourier Transform we have

$\displaystyle \mathcal F [e^{-\pi t^2 x}](u) = x^{-1/2} e^{-\pi u^2/x}$

where $\mathcal F$ denotes the Fourier transform.

Therefore, by the Poisson Summation Formula we have:

$\displaystyle \theta(x) = \sqrt{x} \theta(x^{-1})$

whence:

$\displaystyle \omega(x^{-1}) = -\frac 12 + \frac 12 \sqrt{x} + \sqrt{x}\omega(x)$

Therefore:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int_1^\infty x^{-s/2-1} \omega(x^{-1})\ dx\)	\(=\)	\(\displaystyle \int_1^\infty x^{-s/2-1} \left( -\frac 12 + \frac 12 \sqrt{x} + \sqrt{x}\omega(x) \right)\ dx\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle -\frac 1 s + \frac 1{s-1} + \int_1^\infty x^{-(s+1)/2} \omega(x)\ dx\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So we have

$\displaystyle \pi^{-s/2}\Gamma(s/2)\zeta(s) = -\frac 1 {s(1-s)} + \int_1^\infty \left[ x^{s/2-1} + x^{-(s+1)/2} \right] \omega(x)\ dx$

as required. The functional equation:

$\xi(s) = \xi(1-s)$

follows upon observing that this integral is invariant under $s \mapsto 1-s$.

$\blacksquare$