Zeta summation III

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Recall that in the previous section our goal was to analytically extend the Riemann zeta function.

To this end, we showed that the Gaussian function $e^{-\pi x^2}$ is invariant under the Fourier transform:

$\displaystyle \mathcal{F}(e^{-\pi x^2})(\xi) = \int_{-\infty}^{\infty} e^{-\pi x^2} e^{-2\pi i x \xi}\,dx = e^{-\pi \xi^2}.$

We recall that the Fourier transform is defined by

$\displaystyle \mathcal{F}(f)(\xi) = \int_{-\infty}^{\infty} f(x)\,e^{-2\pi i x \xi}\,dx.$

For $a>0$ , a change of variables shows that

$\displaystyle \mathcal{F}(f(ax))(\xi) = \frac{1}{a}\,\mathcal{F}\!\left(f\right)\!\left(\frac{\xi}{a}\right).$

Applying this to the Gaussian $f(x)=e^{-\pi x^2}$ with $a=\sqrt{t}$ , we obtain

$\displaystyle \mathcal{F}\!\left(e^{-\pi t x^2}\right)(\xi) = t^{-1/2} e^{-\pi \xi^2/t}.$

Both functions belong to the class $\mathcal{S}$ of functions (The Schwartz class consists of infinitely differentiable functions on $\mathbb{R}^n$ that, together with all their derivatives, decay faster than any polynomial at infinity (rapidly decreasing functions)) and Poisson summation applies: