Approximaths

Zeta summation I

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The Zeta function is defined in the complex-plan for Re(z) > 1 as:

\displaystyle \zeta(z)=\frac{1}{1^z}+\frac{1}{2^z}+\frac{1}{3^z}+\frac{1}{4^z}+...=\sum_{n=1}^{\infty}\frac{1}{n^z}

For example:

\displaystyle \zeta(2)=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...=\frac{\pi^2}{6}

If we would like to use the Zeta function in order to perform a summation with Re(z)\leq 1 we should use the analytic continuation of the Zeta function.

If f(z) is single-valued in some region of the complex-plan, the derivative of f(z) is defined as:

\displaystyle f'(z)=\lim_{\Delta z\to0}\frac{f(z+\Delta z)-f(z)}{\Delta z}

If this limit exists we say that f(z) is differentiable at z. If the derivative exists in all points of a region of the complex-plan we say that f(z) is analytic in this region. If the function is analytic inside some circle of convergence C_0 it can be represented by the Taylor series:

\displaystyle a_0+a_1(z-z_0)+a_2(z-z_0)^2+...

where z_0 is the center of the circle C_0. By choosing a new point z_1 in C_0 which is the center of a new circle C_1 the function can be represented by the Taylor series:

\displaystyle b_0+b_1(z-z_1)+b_2(z-z_1)^2+...

The figure above illustrates analytic continuation. I like to imagine analytic functions as exquisite porcelain—delicately fragile, yet rigid, able to be extended through analytic continuation.

In order to analytically continue the Zeta function we first consider the following function:

\displaystyle f(\xi)=e^{-\pi \xi^2}

where \xi\in\mathbb{R}. We would like to show that this function is equivalent to its Fourier transform

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

if \xi=0:

\displaystyle f(0)=1=\int_{-\infty}^{\infty}e^{-\pi x^2}\,dx

if \xi<0 :

\displaystyle \int_{-\infty}^{\infty} e^{-\pi x^2} e^{-2\pi i x \xi}\,dx = \int_{-\infty}^{\infty} e^{-\pi u^2} e^{2\pi i u (-\xi)}\,du = \int_{-\infty}^{\infty} e^{-\pi u^2} e^{-2\pi i u |\xi|}\,du = e^{-\pi \xi^2}

The demonstration will be completed in the next post.

I wish you a merry Christmas 2025 !

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