Certain integral over torus

Question

Let $F(t_1,s_1,t_2,s_2)=$ $$\big((2+\cos t_1)\cos s_1 - (2+\cos t_2)\cos s_2\big)^2 + \big((2+\cos t_1)\sin s_1-(2+\cos t_2)\sin s_2\big)^2 + (\sin t_1 -\sin t_2 )^2.$$ I am interested in computing the following integral:

$$\int_0^{2\pi}\int_0^{2\pi}\int_0^{2\pi}\int_0^{2\pi}\frac{(2+\cos t_1)(2+\cos t_2)\,dt_1\,ds_1\,dt_2\,ds_2}{F(t_1,s_1,t_2,s_2)}$$ with Mathematica using the following

NIntegrate
 [((2+Cos[t])(2+Cos[u]))/((2+Cos[t])Cos[s] - (2+Cos[u])Cos[v])^2
 + ((2+Cos[t])Sin[s]-(2+Cos[u])Sin[v])^2+(Sin[t]-Sin[u])^2, 
    {t,0, 2Pi}, {s, 0, 2Pi}, {u, 0, 2Pi}, {v,0, 2Pi}]

, but so far no success.

To me it gives the result 1422.46, together with the warning that this result may not be reliable. Do you have a way to verify it? — glS, Nov 23 '15 at 05:13
Actually its the other way around.i was going to check with Mathematica to see if it converges or not — BigM, Nov 23 '15 at 05:17

score 4 · Answer 1 · answered Nov 23 '15 at 05:28

4

I am by no means expert in numerical integration techniques, but using the simplicistic rule of thumb high dimensional integrals -> Monte Carlo techniques work better I tried using a Monte Carlo method for the integration (you can find here an explanation of all numerical techniques employed by Mathematica). Using this and incrementing MaxPoint to around 10⁷ seems to compute without warnings:

f[t1_, s1_, t2_, 
  s2_] := ((2 + Cos[t1]) Cos[s1] - (2 + Cos[t2] Cos[s2]))^2 +
  ((2 + Cos[t1]) Sin[s1] - (2 + Cos[t2] Sin[s2]))^2 +
  (Sin[t1] - Sin[t2])^2
NIntegrate[
 (2 + Cos[t1]) (2 + Cos[t2])/f[t1, s1, t2, s2],
 {t1, 0, 2 Pi}, {s1, 0, 2 Pi}, {t2, 0, 2 Pi}, {s2, 0, 2 Pi},
 Method -> {"MonteCarlo", "MaxPoints" -> 10^7}
 ]

Which gives a value $\approx 1400$, though varying a bit in each run.

answered Nov 23 '15 at 05:28

glS

7,623
1
21
61

Very interesting.this integral came up while I was studying certain potentials over tori. its counterpart diverges over sphere ,so I thought this one also must diverge but it seems i doesn't. – BigM Nov 23 '15 at 05:32
@BigM please take this result with a grain of salt though. While it seems to converge, it would probably be best to find an analytical confirmation. You could try that on math.SE (but I don't know if there is a simple technique to assess convergence in such a case). – glS Nov 23 '15 at 05:51
That sounds like a great idea. I'll post it on math.SE. – BigM Nov 23 '15 at 05:54
2

I wouldn't call 4 D "high dimensional", but MC is definitely a safe option in any number of dimensions. If the result is stable across runs and for different values of points, it is likely reliable. You should be careful for $t_1 = t_2$ and $s_1=s_2$, because the integrand diverges, so I'd sample that region more thoroughly to be safer. – Davidmh Nov 23 '15 at 09:22

Michael E2 · Answer 2 · 2015-11-23T14:06:20.343

The integrand has singularities occasionally when the coordinates are a multiple of Pi/4. If we subdivide the domain at multiples of Pi/4, we seem to get divergence.

integrand = ((2 + Cos[t]) (2 + Cos[u]))/(
   ((2 + Cos[t]) Cos[s] - (2 + Cos[u]) Cos[v])^2 +
    ((2 + Cos[t]) Sin[s] - (2 + Cos[u]) Sin[v])^2 + 
    (Sin[t] - Sin[u])^2);

NIntegrate[integrand,
 {t, 0, π/4, π/2, (3 π)/4, π, (5 π)/4, (3 π)/2, (7 π)/4, 2 Pi},
 {s, 0, π/4, π/2, (3 π)/4, π, (5 π)/4, (3 π)/2, (7 π)/4, 2 Pi},
 {u, 0, π/4, π/2, (3 π)/4, π, (5 π)/4, (3 π)/2, (7 π)/4, 2 Pi},
 {v, 0, π/4, π/2, (3 π)/4, π, (5 π)/4, (3 π)/2, (7 π)/4, 2 Pi}]

NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. >>

NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 36 recursive bisections in t near {t,s,u,v} = {0.392699,3.01917,0.662973,3.01917}. NIntegrate obtained 5.391691817606044`*^12 and 5.934913000757257`*^13 for the integral and error estimates. >>

(*  5.39169*10^12  *)

Over one of the subregions, it seems to diverge, too.

NIntegrate[integrand,
 {t, 0, π/4}, {s, 0, π/4}, {u, 0, π/4}, {v, 0, π/4}]

NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. >>

NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 36 recursive bisections in t near {t,s,u,v} = {0.392699,0.662973,0.662973,0.662973}. NIntegrate obtained 5.390318262293057`*^12 and 5.933648796990452`*^13 for the integral and error estimates. >>

(*  5.39032*10^12  *)

ybeltukov · Answer 3 · 2015-11-23T11:55:08.330

Local adaptive method gives 1423.97 without errors

NIntegrate[(2 + Cos[t1]) (2 + Cos[t2])/f[t1, s1, t2, s2], {t1, 0, 
  2 Pi}, {s1, 0, 2 Pi}, {t2, 0, 2 Pi}, {s2, 0, 2 Pi}, Method -> "LocalAdaptive"]
(* 1423.97 *)

Global adaptive method also converges to this value with option Method -> {"GlobalAdaptive", "MaxErrorIncreases" -> bigNumber} but it is much slower.

This value has expected order of magnitude because your function is of the order of 1 and the integration volume is

(2 π)^4 // N
(* 1558.55 *)

Certain integral over torus

3 Answers3