7
$\begingroup$

Im researching electric fields in periodic arrays of charges, and encountered this summation that I can't find any published work on.

Has anybody encountered a solution to $\sum_{m,n=-\infty,\infty}\frac{1}{a+m^2+n^2}$ where a is a real constant?

When constant a=0 it converges, see mathworld.wolfram.com/DoubleSeries.html.

The 1D version also converges, see the "Rational Functions" section of here: en.wikipedia.org/wiki/List_of_mathematical_series#cite_note-7 Therefore I believe the 2D sum that this post focuses on will also converge.

$\endgroup$
7
  • $\begingroup$ This sum looks divergent. For large $m$ and $n$ we can approximate the sum by an integral: set $m=r\cos\phi$ and $n=r\sin\phi$, so that $m^2+n^2=r^2$ and the sum is approximately $\int_0^{\infty}\frac{1}{a+r^2}2\pi r\,d r$, which diverges as $r\to\infty$. $\endgroup$
    – Roman
    Nov 26 '21 at 20:53
  • $\begingroup$ The 1D sum converges, see the "Rational Functions" section of here: en.wikipedia.org/wiki/List_of_mathematical_series#cite_note-7 Therefore I believe the 2D sum will also converge. Also, the 2D summation (without constant a) converges too, see mathworld.wolfram.com/DoubleSeries.html $\endgroup$
    – dturn805
    Nov 26 '21 at 21:03
  • 2
    $\begingroup$ Yes it converges for dimension $d<2$, but not for $d\ge2$. The double sum you are linking does not converge: $\sum_{i,j=-\infty}^{\infty}'\frac{1}{(i^2+j^2)^s}=4\beta(s)\zeta(s)$ only works for $s>1$, not for your case $s=1$. $\endgroup$
    – Roman
    Nov 26 '21 at 21:14
  • $\begingroup$ Thank you Roman! $\endgroup$
    – dturn805
    Nov 27 '21 at 6:03
  • $\begingroup$ As it stands (with all terms being positive) the summation diverges logaritmically, but you can get conditional convergence if you add a term (-1)^(m+n) in the numerator. The physical meaning of that term is that (as I suppose) the electric charges of the points in the lattice are of alternating signs (cations and anions) $\endgroup$
    – Danel
    Dec 1 '21 at 7:04
14
$\begingroup$

This kind of sum can be studied by integral transformation. Notice that $\int_0^1t^{x-1}dt=\frac{1}{x}$ if $\text{Re}(x)>0$:

Integrate[t^(x - 1), {t, 0, 1}]
(*    1/x if Re[x] > 0    *)

and therefore a $d$-dimensional lattice sum is $$ S_d(a)=\sum_{n_1,n_2,\ldots,n_d=-\infty}^{\infty} \frac{1}{a+n_1^2+n_2^2+\ldots+n_d^2}= \sum_{n_1,n_2,\ldots,n_d=-\infty}^{\infty} \int_0^1 t^{a+n_1^2+n_2^2+\ldots+n_d^2-1}dt = \int_0^1t^{a-1}\left(\sum_{n=-\infty}^{\infty}t^{n^2}\right)^ddt $$

The infinite sum is Jacobi's elliptic theta function $\vartheta_3(0,t)=\vartheta_3(t)$:

Sum[t^(n^2), {n, -∞, ∞}]
(*    EllipticTheta[3, 0, t]    *)

which makes the lattice sum $$ S_d(a)= \int_0^1t^{a-1}\left[\vartheta_3(t)\right]^ddt $$

S[d_?NumericQ, a_?NumericQ] :=
  NIntegrate[t^(a - 1) EllipticTheta[3, t]^d, {t, 0, 1}]

This integral converges only for $d<2$ because the elliptic theta function is $\theta_3(t)\approx\sqrt{\frac{\pi}{1-t}}$ for $t\to1$:

Limit[Sqrt[1 - t] * EllipticTheta[3, t], t -> 1, Direction -> "FromBelow"]
(*    Sqrt[π]    *)

more general denominator

For the more general sum $$ S_{d,p}(a)=\sum_{n_1,n_2,\ldots,n_d=-\infty}^{\infty}\frac{1}{(a+n_1^2+n_2^2+\ldots+n_d^2)^p} $$ we can use the integral relationship $$ \int_0^1\frac{(-\ln t)^{p-1}}{\Gamma(p)}t^{x-1}dt=\frac{1}{x^p} $$

Integrate[(-Log[t])^(p - 1)/Gamma[p] t^(x - 1), {t, 0, 1}]
(*    x^(-p) if Re[p] > 0 && Re[x] > 0    *)

Therefore, $$ S_{d,p}(a)=\sum_{n_1,n_2,\ldots,n_d=-\infty}^{\infty}\int_0^1\frac{(-\ln t)^{p-1}}{\Gamma(p)}t^{a+n_1^2+n_2^2+\ldots+n_d^2-1}dt\\ =\int_0^1 \frac{(-\ln t)^{p-1}}{\Gamma(p)} t^{a-1}\left[\vartheta_3(t)\right]^ddt $$

S[d_?NumericQ, p_?NumericQ, a_?NumericQ] /; d < 2*p && a > 0 := 
  NIntegrate[(-Log[t])^(p - 1)/Gamma[p] t^(a - 1) EllipticTheta[3, t]^d,
             {t, 0, 1}]

This integral converges for $d<2p$: as $t\to1^-$ we have $(-\ln t)\approx(1-t)$ and $\vartheta_3(t)\approx\sqrt{\frac{\pi}{1-t}}$, and so the integrand is asymptotically proportional to $(1-t)^{p-1-d/2}$; the integral converges if $p-1-d/2>-1$.

numerical stability

A note on convergence: the above numerical integral is rather difficult to evaluate for small $a\ll1$. To improve stability, we separate out the singular part of the integrand around $t\approx0$ by setting $\vartheta_3(t\approx0)\approx1$ and

J[d_, p_, a_, t_] = (-Log[t])^(p - 1)/Gamma[p] t^(a - 1) EllipticTheta[3, t]^d;
J0[d_, p_, a_, t_] = (-Log[t])^(p - 1)/Gamma[p] t^(a - 1);

The function J0 carries the same singularity as the integrand J but can be integrated analytically:

Assuming[p > 0 && a > 0, Integrate[J0[d, p, a, t], {t, 0, 1}]]
(*    a^-p    *)

Therefore, the numerical integral can be written as

S1[d_?NumericQ, p_?NumericQ, a_?NumericQ] /; d < 2*p && a > 0 && p > 0 :=
  a^-p + NIntegrate[J[d, p, a, t] - J0[d, p, a, t], {t, 0, 1}]

which can be evaluated for even very small values of $a$.

You are interested in the case $d=2$ and $p=3/2$:

Plot[S1[2, 3/2, a], {a, 0, 5}]

enter image description here

$\endgroup$
10
  • $\begingroup$ Really nice analysis of the problem. Does the integral converge for $d\in[1,2)$? Or only for $d=1$? $\endgroup$
    – Joe
    Nov 27 '21 at 17:42
  • $\begingroup$ @Joe it converges for $d<2$, so $d\in(-\infty,2)$. Not sure what $d<0$ means in practice; but the integral seems fine with it. $\endgroup$
    – Roman
    Nov 27 '21 at 18:56
  • $\begingroup$ Interesting. Does $\lim_{d\rightarrow 2^-}$ exist? That could be relevant to OP's physical problem, given that $S_2(a)$ is undefined $\endgroup$
    – Joe
    Nov 27 '21 at 19:10
  • $\begingroup$ @Joe it diverges as $d\to2^-$. $\endgroup$
    – Roman
    Nov 27 '21 at 19:32
  • 1
    $\begingroup$ We indeed encounter such sums in QFT. You must be very careful with your regulator to get something sensible, especially in 2D where you get logs rather than something that will be removed by dimensional regularization. If you have a 2D square lattice with N items on a side, with periodic boundary conditions, you probably want $\left(\sum_{|m|,|n|\leq N/2} \frac{1}{n^2+m^2-x}\right)-2\pi \log(N/2)$; see eq 58 in arxiv.org/abs/1912.04425 $\endgroup$
    – evanb
    Nov 28 '21 at 12:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.