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On Stack Overflow the question was asked how to numerically integrate a function over a infinite range in Rcpp, e.g. by using RcppNumerical. As an example, the integral
\[ \int_{-\infty}^{\infty} \mathrm{d}x \exp\left(-\frac{(x-\mu)^4}{2}\right) \]
was given. Using RcppNumerical is straight forward. One defines a class that extends Numer::Func
for the function and an interface function that calls Numer::integrate
on it:
// [[Rcpp::depends(RcppEigen)]] // [[Rcpp::depends(RcppNumerical)]] #include <RcppNumerical.h> class exp4: public Numer::Func { private: double mean; public: exp4(double mean_) : mean(mean_) {} double operator()(const double& x) const { return exp(-pow(x-mean, 4) / 2); } }; // [[Rcpp::export]] Rcpp::NumericVector integrate_exp4(const double &mean, const double &lower, const double &upper) { exp4 function(mean); double err_est; int err_code; const double result = Numer::integrate(function, lower, upper, err_est, err_code); return Rcpp::NumericVector::create(Rcpp::Named("result") = result, Rcpp::Named("error") = err_est); }
This works fine for finite ranges:
integrate_exp4(4, 0, 4) ## result error ## 1.077900e+00 9.252237e-08
However, it produces NA
for infinite ones:
integrate_exp4(4, -Inf, Inf) ## result error ## NaN NaN
This is disappointing, since base R’s integrate()
handles this without problems:
exp4 <- function(x, mean) exp(-(x - mean)^4 / 2) integrate(exp4, 0, 4, mean = 4) ## 1.0779 with absolute error < 1.3e-07 integrate(exp4, -Inf, Inf, mean = 4) ## 2.155801 with absolute error < 7.9e-06
In this particular case the problem can be easily solved in two different ways. First, the integral can be expressed in terms of the Gamma function:
\[ \int_{-\infty}^{\infty} \mathrm{d}x \exp\left(-\frac{(x-\mu)^4}{2}\right) = 2^{-\frac{3}{4}} \Gamma\left(\frac{1}{4}\right) \approx 2.155801 \]
Second, the integrand is almost zero almost everywhere:
It is therefore sufficient to integrate over a small region around mean
to get a reasonable approximation for the integral over the infinite range:
integrate_exp4(4, 1, 7) ## result error ## 2.155801e+00 9.926448e-13
However, the trick to approximate the integral over an infinite range with an integral over a (possibly large) finite range does not work for functions that approach zero more slowly. The help page for integrate()
has a nice example for this effect:
## a slowly-convergent integral integrand <- function(x) {1/((x+1)*sqrt(x))} integrate(integrand, lower = 0, upper = Inf) ## 3.141593 with absolute error < 2.7e-05 ## don't do this if you really want the integral from 0 to Inf integrate(integrand, lower = 0, upper = 10) ## 2.529038 with absolute error < 3e-04 integrate(integrand, lower = 0, upper = 100000) ## 3.135268 with absolute error < 4.2e-07 integrate(integrand, lower = 0, upper = 1000000, stop.on.error = FALSE) ## failed with message 'the integral is probably divergent'
How does integrate()
handle the infinite range and can we replicate this in Rcpp? The help page states:
If one or both limits are infinite, the infinite range is mapped onto a finite interval.
This is in fact done by a different function from R’s C-API: Rdqagi()
instead of Rdqags()
. In principle one could call Rdqagi()
via Rcpp, but this is not straightforward. Fortunately, there are at least two other solutions.
The GNU Scientific Library provides a function to integrate over the infinte interval \((-\infty, \infty)\), which can be used via the RcppGSL package:
// [[Rcpp::depends(RcppGSL)]] #include <RcppGSL.h> #include <gsl/gsl_integration.h> double exp4 (double x, void * params) { double mean = *(double *) params; return exp(-pow(x-mean, 4) / 2); } // [[Rcpp::export]] Rcpp::NumericVector normalize_exp4_gsl(double &mean) { gsl_integration_workspace *w = gsl_integration_workspace_alloc (1000); double result, error; gsl_function F; F.function = &exp4; F.params = &mean; gsl_integration_qagi(&F, 0, 1e-7, 1000, w, &result, &error); gsl_integration_workspace_free (w); return Rcpp::NumericVector::create(Rcpp::Named("result") = result, Rcpp::Named("error") = error); } normalize_exp4_gsl(4) ## result error ## 2.155801e+00 3.718126e-08
Alternatively, one can apply the transformation used by GSL (and probably R) also in conjunction with RcppNumerical. To do so, one has to substitute \(x = (1-t)/t\) resulting in
\[ \int_{-\infty}^{\infty} \mathrm{d}x f(x) = \int_0^1 \mathrm{d}t \frac{f((1-t)/t) + f(-(1-t)/t)}{t^2} \]
Now one could write the code for the transformed function directly, but it is of course nicer to have a general solution, i.e. use a class template that can transform any function in the desired fashion
// [[Rcpp::depends(RcppEigen)]] // [[Rcpp::depends(RcppNumerical)]] #include <RcppNumerical.h> class exp4: public Numer::Func { private: double mean; public: exp4(double mean_) : mean(mean_) {} double operator()(const double& x) const { return exp(-pow(x-mean, 4) / 2); } }; // [[Rcpp::plugins(cpp11)]] template<class T> class trans_func: public T { public: using T::T; double operator()(const double& t) const { double x = (1-t)/t; return (T::operator()(x) + T::operator()(-x))/pow(t, 2); } }; // [[Rcpp::export]] Rcpp::NumericVector normalize_exp4(const double &mean) { trans_func<exp4> f(mean); double err_est; int err_code; const double result = Numer::integrate(f, 0, 1, err_est, err_code); return Rcpp::NumericVector::create(Rcpp::Named("result") = result, Rcpp::Named("error") = err_est); } normalize_exp4(4) ## result error ## 2.155801e+00 1.439771e-06
Note that the exp4
class is identical to the one from the initial example. This means one can use the same class to calculate integrals over a finite range and after transformation over an infinite range.
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