与变量限制集成

Question

我需要评估内部上限可变的双积分：
integral2 between -5 and 5 ( integral1 between 0 and y f(x)dx )dy。

我坚持计算外环，这取决于内环。我的代码运行了很长时间但返回零。

如何计算具有可变限制的积分？

首先我创建了一个函数doubleIntegrate。首先，该函数保存具有梯形规则系数的数组。

double NumericIntegrationDouble::doubleIntegrate(double (*doubleFunc 
(const double &x), double dy, const double &innerLowBound, const double 
&outerLowBound) 
{

double innerValue = 0.0;
double outerValue = 0.0;

// arrays which store function values for the inner (X) and the outer (Y) integration loop 

// vector filled with coefficients for the inner poop (trapezoidal rule)
std::vector<double> vecCoeffsX(numberOfIntervalsDouble+1, 2);
vecCoeffsX[0] = 1;                         // fist coeff = 1
vecCoeffsX[vecCoeffsX.size()-1] = 1;       // last coeff = 1
std::vector<double> funcValuesX(numberOfIntervalsDouble+1); 


// vector filled with coefficients for the inner poop (trapezoidal rule)
std::vector<double> vecCoeffsY(numberOfIntervalsDouble+1, 2);
vecCoeffsY[0] = 1;                        // same as above
vecCoeffsY[vecCoeffsY.size()-1] = 1;      // same as above
std::vector<double> funcValuesY(numberOfIntervalsDouble+1)


// Then i created a loop in a loop where dy and dy stands for step size of integration. The variables xi and yi stand for the current x and y value. 

// outer integration loop dy
for(int i=0; i<=numberOfIntervalsDouble; i++)
{
    double yi = outerLowBound + dy*i;
    funcValuesY[i] = (*doubleFunc)(yi);

    // inner integration loop dx
    for(int j=0; j<=numberOfIntervalsDouble; j++)
    {
        double dx = abs(yi - innerLowBound) / (double)numberOfIntervalsDouble;
        double xi = innerLowBound + j*dx;
        funcValuesX[j] = (*doubleFunc)(xi);
        double multValueX = std::inner_product(vecCoeffsX.begin(), vecCoeffsX.end(), funcValuesX.begin(), 0.0);
        double innerValue = 0.5 * dx * multValueX;
        suminnerValue = suminnerValue + innerValue;
    }
    //auto multValueY = std::inner_product(vecCoeffsY.begin(), vecCoeffsY.end(), funcValuesY.begin(), 0.0);

    outerValue = 0.5 * dy * suminnerValue;
}
return outerValue;
}