integrate {integrate}R Documentation

Adaptive Integration in 1-20 Dimensions

Description

integrate() integrates a function of 1 variable over a specified interval, i.e. it computes

integral[l .. u] functn(t) dt

where l =lower, u =upper.

adapt() allows 1 to 20 variables and integrates over a rectangular box.

Usage

adapt(ndim, lower, upper, minpts, maxpts, functn, eps,...)
integrate(functn, lower,upper,minpts=100,maxpts=500,eps=0.01,...)

Arguments

ndim the number of dimensions of the function/integral
lower vector of at least length ndim of the lower bounds on the integral
upper vector of at least length ndim of the upper bounds on the integral
minpts the minimum number of function evaluations
maxpts the maximum number of function evaluations or NULL
functn the name of an R function. functn should take a single vector argument and possibly some parameters and return the function value at that point. functn must return a single numeric value.
eps The desired accuracy for the relative error.
... Other parameters to be passed to functn

Value

A list of class "integrate" with components

value the estimated integral
relerr the estimated relative error; < eps argument if
minpts the actual number of function evaluations
ifail an error indicator. If ifail is not equal to 0, the function warns the user of the error condition.

Note

This is modified from Mike Meyer's S code. The functions just call A.C. Genz's fortran ADAPT subroutine to do all of the calculations. A work array is allocated within the C/Fortran code.

The Fortran function has been modified to use double precision, for compatibility with R. It only works in two or more dimensions; for one-dimensional integrals we integrate over a strip of unit width.

Setting maxpts to NULL asks the function to keep doubling maxpts until the desired precision is achieved or R runs out of room.

Examples

integrate(dnorm,-1.96,1.96)
##-        value       relerr       minpts       lenwrk        ifail
##-    0.9500042 0.0003456134          125           73            0

normloc <- function(x,mu) dnorm(x-mu)
integrate(normloc,-1.96,1.96, mu=0) # passing 'mu' arg. to  normloc()
integrate(normloc,-2.96,0.96, mu=-1)

## Example of a three dimensional spherical normal distribution:
ir2pi <- 1/sqrt(2*pi)
fred <- function(z) { ir2pi^length(z) * exp(-0.5 * sum(z * z))}
adapt(3, c(-2,-2,-2),c(2,2,2),100,500,fred,.01)
##-        value       relerr       minpts       lenwrk        ifail
##-    0.8691984 2.579652e-05          345           54            0
##
## i.e., it took 345 function evaluations to find the integral.
      adapt(3, c(-2,-2,-2),c(2,2,2),100,NULL,fred,.00001)
##       value       relerr       minpts       lenwrk        ifail 
##   0.8696159 1.902895e-06         5073          804            0
##  ie 5073 points needed (4 doublings).

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