Madelung's constant

Problem 1. Prove that

$$\sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^{i+j+k}}{\sqrt{i^2+j^2+k^2}}=12\pi \sum_{\substack{m,n=1\\ m,n\;\mathrm{odd}}}^{\infty} \mathrm{sech}^2\bigg(\frac{\pi}{2}\sqrt{m^2+n^2}\bigg)$$

where the prime indicates that the pole $(i,j,k)=(0,0,0)$ is left out.

Solution 1. It follows by symmetry that

$$\sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^{i+j+k}}{\sqrt{i^2+j^2+k^2}} = \sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^{i+j+k}(i^2+j^2+k^2)}{\sqrt{(i^2+j^2+k^2)^3}}\\ = 3\sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^{i+j+k}\cdot i^2}{\sqrt{(i^2+j^2+k^2)^3}}\\ = 3\sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^i i^2(-1)^{j+k}}{\sqrt{(i^2+j^2+k^2)^3}}$$

Now using the Mellin transform

$$\mathcal{M}\{f(x)\}(s)=\int_0^{\infty} x^{s-1}f(x)\,dx$$

Notice that for $p>0$, we have $$\mathcal{M}\{e^{-px}\}(s)=\int_0^{\infty} x^{s-1}e^{-px}\,dx=\int_0^{\infty} \bigg(\frac{u}{p}\bigg)^{s-1}e^{-u}\,\frac{du}{p}=\frac{1}{p^s}\int_0^{\infty} u^{s-1}e^{-u}\,du=\frac{1}{p^s}\Gamma(s)$$

using the change of variable $u=px$. Therefore we obtain

$$\mathcal{M}\{q^{n^2+j^2+k^2}\}\bigg(\frac{3}{2}\bigg)=\frac{1}{(n^2+j^2+k^2)^{\frac{3}{2}}}\Gamma\bigg(\frac{3}{2}\bigg)=\frac{1}{\sqrt{(n^2+j^2+k^2)^3}}\Gamma\bigg(\frac{3}{2}\bigg)$$

where $n,j$, and $k$ are arbitrary integers and $q=e^{-x}$. Hence we rewrite the summation as

$$\Gamma\bigg(\frac{3}{2}\bigg)\sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^{i+j+k}}{\sqrt{i^2+j^2+k^2}} = 3\sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^i i^2(-1)^{j+k}}{\sqrt{(i^2+j^2+k^2)^3}}\Gamma\bigg(\frac{3}{2}\bigg)\\ = 3\sum_{n=-\infty}^{\infty} (-1)^nn^2\cdot\mathcal{M}\bigg\{\sum_{j,k=-\infty}^{\infty} (-1)^{j+k}q^{n^2+j^2+k^2}\bigg\}\bigg(\frac{3}{2}\bigg)\\ =3\cdot\mathcal{M}\bigg\{\sum_{n=-\infty}^{\infty} (-1)^nn^2q^{n^2}\sum_{j=-\infty}^{\infty} (-1)^jq^{j^2}\cdot\sum_{k=-\infty}^{\infty} (-1)^kq^{k^2}\bigg\}\bigg(\frac{3}{2}\bigg)\\ = 3\cdot\mathcal{M}\bigg\{\sum_{n=-\infty}^{\infty} (-1)^nn^2q^{n^2}\theta^2_4(q)\bigg\}\bigg(\frac{3}{2}\bigg) $$

where $\theta_4(q)=\sum_{n=-\infty}^{\infty} (-1)^nq^{n^2}$ is the basic Jacobi theta function. Now using Lemma 3 by setting $s=\frac{\pi}{x}$ and since $\Gamma(\frac{3}{2})=\frac{\sqrt{\pi}}{2}$, and using Lemma 4 by setting $a=l$ and $b=\pi\sqrt{m^2+n^2}$, we obtain

$$\sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^{i+j+k}}{\sqrt{i^2+j^2+k^2}} = \frac{6}{\sqrt{\pi}}\cdot\mathcal{M}\bigg\{\sum_{n=-\infty}^{\infty} (-1)^nn^2q^{n^2}\cdot\frac{\pi}{x}\theta^2_2(e^{-\frac{\pi^2}{x}})\bigg\}\bigg(\frac{3}{2}\bigg)\\ = \frac{6}{\sqrt{\pi}}\sum_{n=-\infty}^{\infty} (-1)^nn^2\cdot\mathcal{M}\bigg\{e^{-n^2x}\cdot\frac{\pi}{x}\theta^2_2(e^{-\frac{\pi^2}{x}})\bigg\}\bigg(\frac{3}{2}\bigg)\\ = 12\sqrt{\pi}\sum_{n=1}^{\infty} (-1)^nn^2\int_0^{\infty} x^{-\frac{1}{2}}e^{-n^2x}\theta^2_2(e^{-\frac{\pi^2}{x}})\,dx\\ = 12\sqrt{\pi}\sum_{n=1}^{\infty} (-1)^nn^2\int_0^{\infty} x^{-\frac{1}{2}}e^{-n^2x}\sum_{j=-\infty}^{\infty} e^{-\frac{\pi^2}{x}(j+\frac{1}{2})^2}\cdot\sum_{k=-\infty}^{\infty} e^{-\frac{\pi^2}{x}(k+\frac{1}{2})^2}\,dx\\ = 48\sqrt{\pi}\sum_{l=1}^{\infty} (-1)^ll^2\sum_{\substack{m,n=1\\ m,n\;\mathrm{odd}}}^{\infty}\int_0^{\infty} x^{-\frac{1}{2}}e^{-l^2x}e^{-\frac{\pi^2}{4x}(m^2+n^2)}\,dx\\ = 48\sqrt{\pi}\sum_{\substack{m,n=1\\ m,n\;\mathrm{odd}}}^{\infty}\sum_{l=1}^{\infty} (-1)^ll^2\cdot\frac{\sqrt{\pi}}{l}e^{-\pi\sqrt{m^2+n^2}l}\\ = 48\pi\sum_{\substack{m,n=1\\ m,n\;\mathrm{odd}}}^{\infty}\sum_{l=1}^{\infty} (-1)^lle^{-\pi\sqrt{m^2+n^2}l} $$

Now using the fact that for $a>0$, we have

$$4\sum_{n=1}^{\infty} (-1)^{n+1}ne^{-an}=\frac{4e^{-a}}{(1+e^{-a})^2}=\mathrm{sech}^2\bigg(\frac{a}{2}\bigg)$$

since by geometric series, we have $\sum_{n=0}^{\infty} (-1)^{n}e^{-an}=\sum_{n=0}^{\infty} (-e^{-a})^n=\frac{1}{1+e^{-a}}$, hence differentiate both side with respect to $a$ yields $\sum_{n=0}^{\infty} (-1)^{n+1}ne^{-an}=\frac{e^{-a}}{(1+e^{-a})^2}$. Notice that $\mathrm{cosh}(x)=\frac{e^x+e^{-x}}{2}$, hence we obtain

$\mathrm{sech}^2(\frac{a}{2})=(\frac{2}{e^{\frac{a}{2}}+e^{-\frac{a}{2}}})^2=(\frac{2e^{-\frac{a}{2}}}{1+e^{-a}})^2=\frac{4e^{-a}}{(1+e^{-a})^2}$, therefore Problem 1 follows

$$\sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^{i+j+k+1}}{\sqrt{i^2+j^2+k^2}}=12\pi \sum_{\substack{m,n=1\\ m,n\;\mathrm{odd}}}^{\infty} \mathrm{sech}^2\bigg(\frac{\pi}{2}\sqrt{m^2+n^2}\bigg)\quad\Box$$