Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function

Theorem
For all $\Re(s) > 0$:


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

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

Thus:
 * $\xi \left({s}\right) = \xi \left({1 - s}\right)$

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

Proof
By definition of the Gamma function:


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

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


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

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:


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

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

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

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:

So we have:


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

as required.

The functional equation:


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

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