没有额外参数的函数的GSL数值积分

Question

我有一个问题，我需要对一个单变量函数进行数值积分，而不要对要积分的变量进行多个附加输入。积分是从零到无穷大。

我说没有额外的参数，因为我已经定义了一个类，其中额外的参数是私有成员变量。然后将运算符定义为仅接受积分变量（因此，单变量）。对于此类，我想使用GSL数字积分库（gsl / gsl_integration.h）进行积分。有没有一种方法可以使用GSL在类内部定义此集成的成员函数？

#include <cmath>
#include <Rmath.h>
#include <algorithm>
#include <iterator>
#include <RcppArmadillo.h>
#include <progress.hpp>
#include <progress_bar.hpp>
#include <RcppGSL.h>
#include <gsl/gsl_integration.h>
#include <Rdefines.h>
// [[Rcpp::depends(RcppArmadillo, RcppProgress, RcppGSL)]]
using namespace arma;

class ObservedLik
{
private:
    const int& Tk;
    const arma::vec& resid;
    const arma::mat& ZEREZ_S;
    const double& nu;
    const double& maxll;
public:
    ObservedLik(const int& Tk_,
                const arma::vec& resid_,
                const arma::mat& ZEREZ_S_,
                const double& nu_,
                const double& maxll_) : Tk(Tk_), resid(resid_), ZEREZ_S(ZEREZ_S_), nu(nu_), maxll(maxll_) {}

    double operator()(const double& lam) const {
        double loglik = -M_LN_SQRT_2PI * static_cast<double>(Tk) + (0.5 * nu - 1.0) * lam - 0.5 * nu * lam + 0.5 * nu * (std::log(nu) - M_LN2) - R::lgammafn(0.5 * nu);
        double logdet_val;
        double logdet_sign;
        log_det(logdet_val, logdet_sign, ZEREZ_S);
        loglik -= 0.5 * (logdet_val + arma::accu(resid % arma::solve(ZEREZ_S, resid)));
        /***********************************
        subtract by maximum likelihood value
        for numerical stability
        ***********************************/
        return std::exp(loglik  - maxll);
    }

    double integrate() {
        /* do the integration here */
        gsl_integration_workspace * w
            = gsl_integration_workspace_alloc (1000);

        double result, error;

        gsl_function F; 
        F.function = &f; // make this the operator()
        F.params = &alpha; // I don't need this part

        gsl_integration_qagiu (&F, 0.0, 0, 1, 0, 1e-7, 1000,
                        w, &result, &error);
        return result;
    }
};

Answer 1

我找到了解决方案。解决方案是脱离GSL，使用Boost库。 Boost数学库中有一个Gauss-Kronrod正交函数，因此可以完成此工作。

#include <cmath>
#include <Rmath.h>
#include <algorithm>
#include <iterator>
#include <RcppArmadillo.h>
#include <progress.hpp>
#include <progress_bar.hpp>
#include <boost/math/quadrature/gauss_kronrod.hpp>
#include "dic_nmr.h"
// [[Rcpp::depends(RcppArmadillo, RcppProgress, BH)]]
using namespace arma;
using namespace boost::math::quadrature;

class ObservedLik
{
private:
    const int& Tk;
    const arma::vec& resid;
    const arma::mat& ZEREZ_S;
    const double& nu;
    const double& maxll;
public:
    ObservedLik(const int& Tk_,
            const arma::vec& resid_,
            const arma::mat& ZEREZ_S_,
            const double& nu_,
            const double& maxll_) : Tk(Tk_), resid(resid_), ZEREZ_S(ZEREZ_S_), nu(nu_), maxll(maxll_) {}

    double integrate_()(double lam) const {
        double loglik = -M_LN_SQRT_2PI * static_cast<double>(Tk) + (0.5 * nu - 1.0) * lam - 0.5 * nu * lam + 0.5 * nu * (std::log(nu) - M_LN2) - R::lgammafn(0.5 * nu);
        double logdet_val;
        double logdet_sign;
        log_det(logdet_val, logdet_sign, ZEREZ_S);
        loglik -= 0.5 * (logdet_val + arma::accu(resid % arma::solve(ZEREZ_S, resid)));
        /***********************************
        subtract by maximum likelihood value
        for numerical stability
        ***********************************/
        return std::exp(loglik  - maxll);
    }

    double integrate() {
        /* do the integration here */

        double error;
        double Q = gauss_kronrod<double, 31>::integrate(integrate_, 0.0, std::numeric_limits<double>::infinity(), 5, 1e-14, &error);

        return Q;
    }
};