这个积分可以在Matlab或Mathematica中以数字方式完成吗？

Question

我希望能够完全以数字方式完成下面的积分。

其中，和，和是常量，为简单起见，这些常量都可以设置为1。

x上的积分可以通过手工或使用Mathematica进行分析，然后y上的积分可以使用NIntegrate以数字方式完成，但这两种方法给出了不同的答案。

从分析：

In[160]:= ex := 2 (1 - Cos[x])

In[149]:= ey := 2 (1 - Cos[y])

In[161]:= kx := 1/(1 + Exp[ex])

In[151]:= ky := 1/(1 + Exp[ey])

In[162]:= Fn1 := 1/(2 \[Pi]) ((Cos[(x + y)/2])^2)/(ex - ey)

In[163]:= Integrate[Fn1, {x, -Pi, Pi}]

Out[163]= -(1/(4 \[Pi]))
 If[Re[y] >= \[Pi] || \[Pi] + Re[y] <= 0 || 
   y \[NotElement] Reals, \[Pi] Cos[y] - Log[-Cos[y/2]] Sin[y] + 
   Log[Cos[y/2]] Sin[y], 
  Integrate[Cos[(x + y)/2]^2/(Cos[x] - Cos[y]), {x, -\[Pi], \[Pi]}, 
   Assumptions -> ! (Re[y] >= \[Pi] || \[Pi] + Re[y] <= 0 || 
       y \[NotElement] Reals)]]

In[164]:= Fn2 := -1/(
  4 \[Pi]) ((\[Pi] Cos[y] - Log[-Cos[y/2]] Sin[y] + 
     Log[Cos[y/2]] Sin[y]) (1 - ky) ky )/(2 \[Pi])

In[165]:= NIntegrate[Fn2, {y, -Pi, Pi}]

Out[165]= -0.0160323 - 2.23302*10^-15 I

数值方法1：

In[107]:= Fn4 := 
 1/(4 \[Pi]^2) ((Cos[(x + y)/2])^2) (1 - ky) ky/(ex - ey)

In[109]:= NIntegrate[Fn4, {x, -Pi, Pi}, {y, -Pi, Pi}]

During evaluation of In[109]:= 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. >>

During evaluation of In[109]:= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 18 recursive bisections in x near {x,y} = {0.0000202323,2.16219}. NIntegrate obtained 132827.66472461013` and 19442.543606302774` for the integral and error estimates. >>

Out[109]= 132828.

数字2：

In[113]:= delta = .001;
pw[x_, y_] := Piecewise[{{1, Abs[Abs[x] - Abs[y]] > delta}}, 0]

In[116]:= Fn5 := (Fn4)*pw[Cos[x], Cos[y]]

In[131]:= NIntegrate[Fn5, {x, -Pi, Pi}, {y, -Pi, Pi}]

During evaluation of In[131]:= 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. >>

During evaluation of In[131]:= 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 0.013006903336304906` and 0.0006852739534086272` for the integral and error estimates. >>

Out[131]= 0.0130069

因此，这两种数值方法都没有给出-0.0160323。我理解为什么 - 第一种方法在分母引起的无穷大方面遇到麻烦，第二种方法有效地删除了引起问题的积分部分。但我希望能够整合另一个积分（一个比x，y和z更难的积分），这些积分无法通过分析进行简化。上面的积分为我提供了一种测试任何新方法的方法，因为我知道答案应该是什么。

Answer 1

除非我记下错误，否则这应该可以胜任：

n[x_] := 1/(1 + Exp[eps[x]])
eps[x_] := 2(1 - Cos[x])
.25/(2*Pi)^2*NIntegrate[
    Cos[(x + y)/
     2]^2 ((1 - n[y]) n[y] - (1 - n[x]) n[x])/(Cos[y] - Cos[x]),
    {x, -Pi, Pi},
    {y, -Pi, Pi},
    Exclusions -> {Cos[x] == Cos[y]}
  ]

给予0.0130098。

好的，我想最快解释的方法是这样的：

从第一个eq的第一行到第二行我刚刚将y重命名为x（积分区域是对称的，所以没关系）。然后我为I添加了两个（等效的）表达式以获得2I，我在数字上整合的是那个。关键是分子和分母在相同的点消失，所以实际上不需要Exclusions选项。请注意，在我上面的方法草图中，为了简洁，我放弃了1/(4*Pi^2)（或者由于懒惰，取决于观点）。