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Revision en21, by bfsof123, 2024-05-31 12:46:06

In this blog we prove a Ramanujan-type identity:

$$$\sum\limits_{n_1 \in \mathbb{Z}}(-1)^{n_1}\sum\limits_{(n_2, n_3) \in \mathbb{Z}^2}\frac{1}{\sqrt{n_1^2 + (n_2+0.5)^2+(n_3+0.5)^2}\sinh{(\pi\sqrt{n_1^2 + (n_2+0.5)^2+(n_3+0.5)^2})}} = 1$$$. (1)

Magic code

First, we consider a 4D lattice sum:

$$$S(n_1, n_2, n_3, n_4) := \sum\limits_{(n_1, n_2, n_3, n_4) \in \mathbb{Z}^4} \frac{(-1)^{n_1+n_4}}{n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2 + n_4^2}$$$. We will show later that $$$S(n_1, n_2, n_3, n_4)$$$ equals to $$$\pi$$$ (code shown below).

Pi code

For $$$\lambda > 0$$$, $$$\int_{0}^\infty e^{-\lambda x}dx = \frac{1}{\lambda}$$$, therefore $$$S(n_1, n_2, n_3, n_4) = \sum\limits_{(n_1, n_2, n_3, n_4) \in \mathbb{Z}^4} \int_{0}^\infty (-1)^{n_1+n_4} \exp((-n_1^2-(n_2+0.5)^2-(n_3+0.5)^2-n_4^2)x)dx$$$. (2)

By the Poisson summation formula of the theta function (see this),

$$$\sum\limits_{n \in \mathbb{Z}}(-1)^n \exp(-n^2 x) = \sqrt{\frac{\pi}{x}} \sum\limits_{n \in \mathbb{Z}}(-1)^n\exp(\frac{-(n+0.5)^2\pi^2}{x})$$$ (3)

By combining (2) and (3), $$$S$$$ can be written as a Mellin-transform-type integral:

$$$S(n_1, n_2, n_3, n_4) = \sum\limits_{(n_1, n_2, n_3, n_4) \in \mathbb{Z}^4} \sqrt{\pi}\int_{0}^\infty (-1)^{n_1} x^{-0.5} \exp(-(n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2)x - \frac{(n_4+0.5)^2\pi^2}{x})dx$$$. (4)

We then use a famous integral representation of the Modified Bessel functions of the second kind (i.e., $$$K$$$):

$$$M_s[\exp(-ax-\frac{b}{x})] := \int_{0}^\infty x^{s-1}\exp(-ax-\frac{b}{x})dx = 2(\frac{b}{a})^{\frac{s}{2}}K_s(2\sqrt{ab})$$$. (5)

In (5), $$$M_s$$$ denotes the Mellin transform with parameter $$$s$$$. And we note that when $$$s=0.5$$$, $$$K_{0.5}(z)$$$ has a closed-form: $$$K_{0.5}(z) = \sqrt\frac{\pi}{2} \frac{e^{-z}}{\sqrt{z}}$$$.

Then, $$$\int_{0}^\infty x^{-0.5}\exp(-ax-\frac{b}{x})dx = 2(\frac{b}{a})^{0.25} K_{0.5}(2\sqrt{ab}) = 2(\frac{b}{a})^{0.25}\sqrt{\frac{\pi}{2}}\frac{\exp(-2\sqrt{ab})}{\sqrt{2}(ab)^{0.25}} = \sqrt{\frac{\pi}{a}}\exp(-2\sqrt{ab})$$$. (6)

We set $$$a = n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2$$$ and $$$b = (n_4+0.5)^2\pi^2$$$, and

$$$S_{n_1, n_2, n_3, n_4} = \sum\limits_{(n_1, n_2, n_3) \in \mathbb{Z}^3} (-1)^{n_1}\frac{\pi}{\sqrt{n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2}}\sum\limits_{n_4} \exp(-\pi |2n_4+1| \sqrt{n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2})$$$

Tags math, modular form, ramanujan, lattice sum, sinh, codeforces

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