Month: January 2026

Zeta summation V

In a previous post we have computed the meromorphic continuation into the entire complex plane of the Zeta function:

$\displaystyle \zeta(s) = \pi^{s/2}\frac{\xi(s)}{\Gamma(s/2)}$

This analytic continuation allows us to sum the Zeta function in the whole complex plane. For special (‘known’) values of the Xi function we can calculate the Zeta function. For example for

s = 2 we have $\xi(2) = \frac{\pi}{6}$ and $\Gamma(2/2) = \Gamma(1) = 1$ and therefore:

$\displaystyle \zeta(2) = \frac{\pi^2}{6}$

$\displaystyle \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}$

For s = 4 we have $\xi(4) = \frac{\pi^2}{90}$ and $\Gamma(4/2) = \Gamma(2) = 1$ :

$\displaystyle \zeta(4) = \frac{\pi^4}{90}$

$\displaystyle \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + ... = \frac{\pi^4}{90}$

For s = 6 we have $\xi(6) = \frac{2\pi^3}{945}$ and $\Gamma(6/2) = \Gamma(3) = 2$ :

$\displaystyle \zeta(6) = \frac{\pi^6}{945}$

$\displaystyle \frac{1}{1^6} + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + ... = \frac{\pi^6}{945}$

So, Zeta of all the positive even integers has the form

$\displaystyle\zeta(2n) = (-1)^{n+1} \frac{B_{2n} (2\pi)^{2n}}{2(2n)!}$

Where $B_{2n}$ are Bernoulli numbers.

January 29, 2026
1 + 2 + 3 + 4 + …

To calculate the value of the Riemann zeta function at $s = -1$ , we use the functional equation (presented here):

$\displaystyle \zeta(s) = \pi^{s/2} \frac{\xi(s)}{\Gamma(s/2)}$

The Dirichlet series for the Riemann zeta function is defined for $\Re(s) > 1$ as:

$\displaystyle \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$

For $s = -1$ , the formal series (which is divergent in the classical sense) is:

$\displaystyle \zeta(-1) = \sum_{n=1}^{\infty} \frac{1}{n^{-1}}= \sum_{n=1}^{\infty} n = 1 + 2 + 3 + 4 + \dots$

Substituting $s = -1$ into the formula, we obtain:

$\displaystyle \zeta(-1) = \pi^{-1/2} \frac{\xi(-1)}{\Gamma(-1/2)}$

Using the recurrence property of the Gamma function, $\Gamma(z) = \frac{\Gamma(z+1)}{z}$ , and the fact that $\Gamma(1/2) = \sqrt{\pi}$ :

$\displaystyle \Gamma(-1/2) = \frac{\Gamma(-1/2+1)}{-1/2} = \frac{\Gamma(1/2)}{-1/2} = -2\sqrt{\pi}$

We exploit the functional equation $\xi(s) = \xi(1-s)$ , which implies:

$\displaystyle \xi(-1) = \xi(1 - (-1)) = \xi(2)$

From the definition of $\xi(s)$ , we have:

$\displaystyle \xi(2) = \pi^{-2/2} \Gamma(2/2) \zeta(2)$

Using $\Gamma(1) = 1$ and the solution of the Basel problem presented in the previous post $\zeta(2) = \frac{\pi^2}{6}$ :

$\displaystyle \xi(-1) = \pi^{-1} \cdot 1 \cdot \frac{\pi^2}{6} = \frac{\pi}{6}$

Combining these results into our original expression:

$\displaystyle \zeta(-1) =\frac{1}{\sqrt{\pi}} \cdot \frac{\frac{\pi}{6}}{-2\sqrt{\pi}}$

$\displaystyle = \frac{\pi}{-12 \cdot (\sqrt{\pi} \cdot \sqrt{\pi})}$

$\displaystyle = \frac{\pi}{-12\pi}$

$\displaystyle \boxed{\zeta(-1) = -\frac{1}{12}}$

Though surprising at first, turns out to be extremely useful in physics — most notably in string theory and the calculation of the Casimir effect, where the infinite sum 1 + 2 + 3 + 4 + … = −1/12 naturally appears when regularizing the vacuum energy between two conducting plates.

January 22, 2026
Zeta summation: Basel problem

For $s= 2$ , the zeta series is:

$\displaystyle \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^{2}}= 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots$

We would like to evaluate this series using the representation presented the previous post in we obtain

$\displaystyle \zeta(s) = \pi^{s/2} \frac{\xi(s)}{\Gamma(s/2)}$

For $s = 2$ ,

$\displaystyle \zeta(2) = \pi^{2/2} \frac{\xi(2)}{\Gamma(2/2)} = \pi \, \xi(2)$

since $\Gamma(1) = 1$ . We need to evaluate

$\displaystyle \xi(2) = \int_0^\infty \frac{\vartheta(u) - 1}{2} \, du.$

We split the integral as follows:

$\displaystyle \int_0^1 \frac{\vartheta(u) - 1}{2} \, du + \int_1^\infty \frac{\vartheta(u) - 1}{2} \, du$

For the second part, we make the change of variable $u = 1/v$ , so $du = -dv/v^2$ . Using the functional equation of the theta function presented in a previous post.

$\displaystyle \vartheta(u) = u^{-1/2} \vartheta(1/u), \qquad u>0.$

This can be rewritten as

$\displaystyle \vartheta(1/v) = \sqrt{v} \, \vartheta(v)$

one can show (after standard calculations) that the two parts combine to yield a convergent integral. The final evaluation, which relies on the Poisson summation formula applied to the Gaussian, gives

$\displaystyle \xi(2) = \frac{\pi}{6}$

We obtain

$\displaystyle \zeta(2) = \pi \cdot \xi(2) = \pi \cdot \frac{\pi}{6} = \frac{\pi^2}{6}$

Finally

$\displaystyle 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots = \frac{\pi^2}{6}$

The Basel problem was sloved by Euler in 1735, who showed that the sum equals $\displaystyle \frac{\pi^2}{6}$ .

January 15, 2026
Zeta summation IV

In the previous post we derived the functional equation of the theta function

$\displaystyle \vartheta(t) = t^{-1/2} \vartheta(1/t), \qquad t>0.$

With $t > 0$ . We observe that:

$\displaystyle \vartheta(t) \leq C t^{-1/2} \text{ as } t \to 0$

note that

$\displaystyle \sum_{n \geq 1}e^{-\pi n^2 t} \leq \sum_{n \geq 1}e^{-\pi n t} \leq Ce^{-\pi t}, t \geq 1$

Remember the definition of the Gamma function. For $s >0$ the Gamma function is defined by:

$\displaystyle \Gamma(s) = \int_{0}^{\infty} e^{-t} t^{s-1}\,dt$

$\displaystyle \Gamma(s/2) = \int_{0}^{\infty} e^{-t} t^{\frac{s}{2}-1}\,dt$

Define $u:= \frac{t}{\pi n^2} \implies du = \frac{1}{\pi n^2}\,dt$ and $dt = \pi n^2 \,du$

$\displaystyle \Gamma(s/2) = \int_{0}^{\infty} e^{-\pi n^2 u} (\pi n^2 u)^{\frac{s}{2}-1} \pi n^2\,du$

$\displaystyle \Gamma(s/2) = \pi^{s/2} n^{s}\int_{0}^{\infty} e^{-\pi n^2 u} u^{\frac{s}{2}-1}\,du$

$\displaystyle \pi^{-s/2} \Gamma(s/2) n^{-s} = \int_{0}^{\infty} e^{-\pi n^2 u} u^{\frac{s}{2}-1}\,du \qquad (n \geq 1)$

Observe that for the $\vartheta(t)$ defined above:

$\displaystyle \vartheta(u) = \sum_{n \in \mathbb{Z}} e^{-\pi n^2 u}$

$\displaystyle = 2 \sum_{n = 1}^{\infty} e^{-\pi n^2 u} + 1$

$\displaystyle \frac{\vartheta(u) -1}{2} = \sum_{n = 1}^{\infty} e^{-\pi n^2 u}$

It follows that:

$\displaystyle \int_{0}^{\infty} u^{\frac{s}{2}-1} \biggl(\frac{\vartheta(u) -1}{2}\biggr) \,du = \int_{0}^{\infty} u^{\frac{s}{2}-1} \sum_{n = 1}^{ \infty} e^{-\pi n^2 u} \,du$

$\displaystyle = \sum_{n = 1}^{\infty} \int_{0}^{\infty} u^{\frac{s}{2}-1} e^{-\pi n^2 u}\,du$

$\displaystyle = \sum_{n = 1}^{\infty} \int_{0}^{\infty} e^{-\pi n^2 u} u^{\frac{s}{2}-1} \,du$

$\displaystyle = \sum_{n = 1}^{\infty} \pi^{-s/2} \Gamma(s/2) n^{-s}$

$\displaystyle = \pi^{-s/2} \Gamma(s/2) \sum_{n = 1}^{\infty} n^{-s}$

$\displaystyle = \pi^{-s/2} \Gamma(s/2) \zeta(s)$

Therefore:

$\displaystyle \zeta(s) = \frac{\int_{0}^{\infty} u^{\frac{s}{2}-1} \biggl(\frac{\vartheta(u) -1}{2}\biggr) \,du}{\pi^{-s/2} \Gamma(s/2) }$

The define the Xi function as :

$\displaystyle \xi(s) := \int_{0}^{\infty} u^{\frac{s}{2}-1} \biggl(\frac{\vartheta(u) -1}{2}\biggr) \,du$

We finally have

$\boxed{\displaystyle \zeta(s) = \pi^{s/2}\frac{\xi(s)}{\Gamma(s/2)}}$

which is the meromorphic continuation into the complex plane of the Zeta function with a simple pole at s = 1.

January 8, 2026
Zeta summation III

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:

$\displaystyle \sum_{n\in\mathbb{Z}} f(n) = \sum_{n\in\mathbb{Z}} \mathcal{F}(f)(n).$

In the present case, this yields

$\displaystyle \sum_{n\in\mathbb{Z}} e^{-\pi t n^2} = \sum_{n\in\mathbb{Z}} t^{-1/2} e^{-\pi n^2/t}.$

The left-hand side defines the theta function

$\displaystyle \vartheta(t) := \sum_{n\in\mathbb{Z}} e^{-\pi t n^2}.$

We immediately obtain the functional equation

$\displaystyle \boxed{\vartheta(t) = t^{-1/2} \vartheta(1/t),\qquad t>0.}$

January 1, 2026