Contour integral along a disc / an arc.

Description

Evaluates the contour integral along the disc given the centre and the radius of a disc. The function is first parametrised so the real and imaginary part can be evaluated seperately using the R built-in integrate function.

Usage

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disc.integrate(f, z0, R = 0, gamma = c(0, 2 * pi), 
 rel.tol = .Machine$double.eps^0.5, subdivisions = 100L)

Arguments

f

A function whose integral to be evaluated.

z0

The centre of the disc, usually an isolated singularity.

R

The radius of the disc (hence a non-negative real number). If the default of zero is used, a sequence of very small radius (starting from 1e-12) will be used sequentially until a solution is sought. Also see details below for using the default radius.

gamma

A vector of two real numbers indicating the argument of the starting and end point of the disc. Therefore, the default is set as c(0,2*pi).

rel.tol

The relative tolerance in calculating the integral of the real and imaginary part.

subdivisions

The number of subdivisions in calculating the integral of the real and imaginary part.

Details

This function works well regardless of the type of the isolated singularity (i.e. removable singularity, pole & essential singularity).
Note that, due to the rounding error, the value of the integral evaluated at a removable singularity is not exactly but very close to zero (see Examples).
In the case where a branch cut or point is wihtin the disc and the radius is well defined, this function still works well. However, if the radius is not well defined, one should consider a contour integral shrunk around to a singularity and/or use line.integrate to evaluate the integral alone the branch line.
In the case where a singularity is on the branch line (see Examples below), the function residue should be used instead.

Value

value

The output of evaluating the integral.

abs.error

The absolute error.

record

Only if R=0. It shows the value of the integral evaluated with different values of radius.

Author(s)

Char Leung

See Also

residue,poles,line.integrate

Examples

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## integral evaluated at a removable singularity
disc.integrate(function(z){sin(z)/z},0)
## evaluating an integral in the presence of a branch cut
disc.integrate(function(z){sqrt(z)},0,R=1) #still works
#evaluate the integral along the branch cut 
-line.integrate(function(z){sqrt(z)},0,1)+line.integrate(function(z){sqrt(z)},1,0) 
## Incorrect result: evaluating an integral at a pole on a branch cut
disc.integrate(function(z){log(z)/(1+z)},-1,R=1) #Incorrect
residue(function(z){log(z)/(1+z)},-1) #Correct (using residue)