myintegrate

0th

Percentile

Complex integration

Integration of complex valued functions along the real axis (myintegrate()), along arbitrary paths (integrate.contour()), and following arbitrary straight line segments (integrate.segments()). Also, evaluation of a function at a point using the residue theorem (residue()).

Keywords
math
Usage
myintegrate(f, lower,upper, ...)
integrate.contour(f,u,udash, ...)
integrate.segments(f,points, close=TRUE, ...)
residue(f, z0, r, O=z0, ...)
Arguments
f

function, possibly complex valued

lower,upper

Lower and upper limits of integration in myintegrate(); real numbers (for complex values, use integrate.contour() or integrate.segments())

u

Function mapping \([0,1]\) to the contour. For a closed contour, require that \(u(0)=u(1)\)

udash

Derivative of u

points

In function integrate.segments(), a vector of complex numbers. Integration will be taken over straight segments joining consecutive elements of points

close

In function integrate.segments(), a Boolean variable with default TRUE meaning to integrate along the segment from points[n] to points[1] in addition to the internal segments

r,O,z0

In function residue() returns f(z0) by integrating \(f(z)/(z-z0)\) around a circle of radius r and center O

...

Extra arguments passed to integrate()

Aliases
  • myintegrate
  • integrate.contour
  • integrate.segments
  • residue
Examples
# NOT RUN {
f1 <- function(z){sin(exp(z))}
f2 <- function(z,p){p/z}

myintegrate(f1,2,3)  # that is, along the real axis


integrate.segments(f1,c(1,1i,-1,-1i),close=TRUE)   # should be zero

# (following should be pi*2i; note secondary argument):
integrate.segments(f2,points=c(1,1i,-1,-1i),close=TRUE,p=1)



# To integrate round the unit circle, we need the contour and its
# derivative:

 u <- function(x){exp(pi*2i*x)}
 udash <- function(x){pi*2i*exp(pi*2i*x)}

# Some elementary functions, for practice:

# (following should be 2i*pi; note secondary argument 'p'):
integrate.contour(function(z,p){p/z},u,udash,p=1)      
integrate.contour(function(z){log(z)},u,udash)         # should be -2i*pi
integrate.contour(function(z){sin(z)+1/z^2},u,udash)   # should be zero



# residue() is a convenience wrapper integrating f(z)/(z-z0) along a
# circular contour:

residue(function(z){1/z},2,r=0.1)  # should be 1/2=0.5



# Now, some elliptic functions:
g <- c(3,2+4i)
Zeta <- function(z){zeta(z,g)}
Sigma <- function(z){sigma(z,g)}
WeierstrassP <- function(z){P(z,g)}

jj <- integrate.contour(Zeta,u,udash) 
abs(jj-2*pi*1i)                              # should be zero
abs(integrate.contour(Sigma,u,udash))        # should be zero
abs(integrate.contour(WeierstrassP,u,udash)) # should be zero




# Now integrate f(x) = exp(1i*x)/(1+x^2) from -Inf to +Inf along the
# real axis, using the Residue Theorem.  This tells us that integral of
# f(z) along any closed path is equal to pi*2i times the sum of the
# residues inside it.  Take a semicircular path P from -R to +R along
# the real axis, then following a semicircle in the upper half plane, of
# radius R to close the loop.  Now consider large R.  Then P encloses a
# pole at +1i [there is one at -1i also, but this is outside P, so
# irrelevent here] at which the residue is -1i/2e.  Thus the integral of
# f(z) = 2i*pi*(-1i/2e) = pi/e along P; the contribution from the
# semicircle tends to zero as R tends to infinity; thus the integral
# along the real axis is the whole path integral, or pi/e.

# We can now reproduce this result analytically.  First, choose an R:
R <- 400

# now define P.  First, the semicircle, u1:
u1     <- function(x){R*exp(pi*1i*x)}
u1dash <- function(x){R*pi*1i*exp(pi*1i*x)}

# and now the straight part along the real axis, u2:
u2     <- function(x){R*(2*x-1)}
u2dash <- function(x){R*2}

# Better define the function:
f <- function(z){exp(1i*z)/(1+z^2)}

# OK, now carry out the path integral.  I'll do it explicitly, but note
# that the contribution from the first integral should be small:

answer.approximate <-
    integrate.contour(f,u1,u1dash) +
    integrate.contour(f,u2,u2dash) 

# And compare with the analytical value:
answer.exact <- pi/exp(1)
abs(answer.approximate - answer.exact)


# Now try the same thing but integrating over a triangle, using
# integrate.segments().  Use a path P' with base from -R to +R along the
# real axis, closed by two straight segments, one from +R to 1i*R, the
# other from 1i*R to -R:

abs(integrate.segments(f,c(-R,R,1i*R))- answer.exact)


# Observe how much better one can do by integrating over a big square
# instead:

abs(integrate.segments(f,c(-R,R,R+1i*R, -R+1i*R))- answer.exact)


# Now in the interests of search engine findability, here is an
# application of Cauchy's integral formula, or Cauchy's formula.  I will
# use it to find sin(0.8):

u     <- function(x){exp(pi*2i*x)}
udash <- function(x){pi*2i*exp(pi*2i*x)}

g <- function(z){sin(z)/(z-0.8)}

a <- 1/(2i*pi)*integrate.contour(g,u,udash)


abs(a-sin(0.8))


# }
Documentation reproduced from package elliptic, version 1.4-0, License: GPL-2

Community examples

Looks like there are no examples yet.