如何克服数值积分中的奇点（在 Matlab 或 Mathematica 中）

Question

I want to numerically integrate the following:

where

and a , b and β are constants which for simplicity, can all be set to 1 .

Neither Matlab using dblquad , nor Mathematica using NIntegrate can deal with the singularity created by the denominator. Since it's a double integral, I can't specify where the singularity is in Mathematica.

I'm sure that it is not infinite since this integral is based in perturbation theory and without the

has been found before (just not by me so I don't know how it's done).

Any ideas?

Answer 1

(1) It would be helpful if you provide the explicit code you use. That way others (read: me) need not code it up separately.

(2) If the integral exists, it has to be zero. This is because you negate the n(y)-n(x) factor when you swap x and y but keep the rest the same. Yet the integration range symmetry means that amounts to just renaming your variables, hence it must stay the same.

(3) Here is some code that shows it will be zero, at least if we zero out the singular part and a small band around it.

a = 1;
b = 1;
beta = 1;
eps[x_] := 2*(a-b*Cos[x])
n[x_] := 1/(1+Exp[beta*eps[x]])
delta = .001;
pw[x_,y_] := Piecewise[{{1,Abs[Abs[x]-Abs[y]]>delta}}, 0]

We add 1 to the integrand just to avoid accuracy issues with results that are near zero.

NIntegrate[1+Cos[(x+y)/2]^2*(n[x]-n[y])/(eps[x]-eps[y])^2*pw[Cos[x],Cos[y]],
  {x,-Pi,Pi}, {y,-Pi,Pi}] / (4*Pi^2)

I get the result below.

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::eincr: 
   The global error of the strategy GlobalAdaptive has increased more than 
    2000 times. The global error is expected to decrease monotonically after a
     number of integrand evaluations. Suspect one of the following: the
     working precision is insufficient for the specified precision goal; the
     integrand is highly oscillatory or it is not a (piecewise) smooth
     function; or the true value of the integral is 0. Increasing the value of
     the GlobalAdaptive option MaxErrorIncreases might lead to a convergent
     numerical integration. NIntegrate obtained 39.4791 and 0.459541
     for the integral and error estimates.

Out[24]= 1.00002

This is a good indication that the unadulterated result will be zero.

(4) Substituting cx for cos(x) and cy for cos(y), and removing extraneous factors for purposes of convergence assessment, gives the expression below.

((1 + E^(2*(1 - cx)))^(-1) - (1 + E^(2*(1 - cy)))^(-1))/
 (2*(1 - cx) - 2*(1 - cy))^2

A series expansion in cy, centered at cx, indicates a pole of order 1. So it does appear to be a singular integral.

Daniel Lichtblau

Answer 2

The integral looks like a Cauchy Principal Value type integral (ie it has a strong singularity). That's why you can't apply standard quadrature techniques.

Have you tried PrincipalValue->True in Mathematica's Integrate?

Answer 3

In addition to Daniel's observation about integrating an odd integrand over a symmetric range (so that symmetry indicates the result should be zero), you can also do this to understand its convergence better (I'll use latex, writing this out with pen and paper should make it easier to read; it took a lot longer to write than to do, it's not that complicated):

First, epsilon(x)-\epsilon(y)\propto\cos(y)-\cos(x)=2\sin(\xi_+)\sin(\xi_-) where I have defined \xi_\pm=(x\pm y)/2 (so I've rotated the axes by pi/4). The region of integration then is \xi_+ between \pi/\sqrt{2} and -\pi/\sqrt{2} and \xi_- between \pm(\pi/\sqrt{2}-\xi_-) . Then the integrand takes the form \frac{1}{\sin^2(\xi_-)\sin^2(\xi_+)} times terms with no divergences. So, evidently, there are second-order poles, and this isn't convergent as presented.

Perhaps you could email the persons who obtained an answer with the cos term and ask what precisely it is they did. Perhaps there's a physical regularisation procedure being employed. Or you could have given more information on the physical origin of this (some sort of second order perturbation theory for some sort of bosonic system?), had that not been off-topic here...

Answer 4

May be I am missing something here, but the integrand f[x,y]=Cos^2[(x+y)/2]*(n[x]-n[y])/(eps[x]-eps[y]) with n[x]=1/(1+Exp[Beta*eps[x]]) and eps[x]=2(ab*Cos[x]) is indeed a symmetric function in x and y: f[x,-y]= f[-x,y]=f[x,y]. Therefore its integral over any domain [-u,u]x[-v,v] is zero. No numerical integration seems to be needed here. The result is just zero.