When integrating using the method of residues, one needs to check if the integrand satisfies the Cauchy-Riemann equations first.
However, it is possible to have an integral that doesn't satisfy Cauchy-Riemann and yet the method of residues still works! To see an example of such an integral, see the Appendix of this document:
http://www.eleceng.adelaide.edu.au/personal/dabbott/conference/UPN_abbott1996b.pdf
The questions are:
a) Is this type of exception generally known?
b) Can you explain what is going on and why contour integration by residues still works?
c) Is it possible to define a class of integrals that are solvable by residues, and yet do not satisfy Cauchy-Riemann?
I have checked your example (the one in Appendix ?) and the CR condition holds. The point is that the contour (path) of integration must lie in an annulus where the function in question is holomorphic (analytic); note that residues are given by the coefficient a_{-1} in the Laurent series.
Yesterday, I promissed to check if the assumptions can be weaker. I don't think so because the crucial point is that the integral along the closed path must vanish what is guaranteed by holomorphity. I can imagine a non-holomorphic function with vanishing integral along such path that fulfils the the residue theorem formula but, I think, this would be only a numerical coincidence. But an example would be interesting :-).
Dear Derek,
Sorry for the poor English of my answer. I may use some incorrect English words in this answer.
I can say my opinion with an example. Suppose you want to calculate integral of a complex variable function like “f(z) = f(x+iy) = U(x,y)+iV(x,y)” in the area D by residues method.
1. When you integrate using the method of residues, your function should be analytical expect for countable points in the area of your integral.
2. Beside some other conditions (Ux,Uy,Vx,Vy should be continues), if Cauchy-Riemann conditions are satisfied, f is analytical.
So Cauchy-Riemann is not a necessary condition, it is sufficient (if other conditions will be satisfied too)
Hi Derek!
In the standard formulation of the of the theorem in question the function actually has to be holomorphic and Cauchy-Riemann is a weaker condition! (In, addition the de partial derrivatives have to be continuous.). But as I remember, in the proof one uses the Green theorem and Cauchy-Riemann eqs. I have to check if such a formula can be proven with weaker assumptions. There Looman-Menchoff theorems that give the weaker condition for a continuous function to be holomorphic- if I remember correctly the the CR condition has to hold only almost everywhere.
@ Derek. What do you mean when at pag 152 you state that "because the conjugation on the denominator of the original integral, the Cauchy-Riemann equations are not satisfied."? If with "the original integral" you mean eq (6), I do not see any conjugation on the denominator.
(CR) is necessary condition: Holomorphy => CR.
(CR+Continuous Partial Derivatives) is sufficient condition.
If you are talking about "if and only if-type conditions", then CR is surely one of those
I think that the integrand of eq (6) is holomorphic everywhere except for the singularities at +-s_1 and +-s_2, therefore the Cauchy-Riemann conditions are satisfied, as well as the hypotheses of the Cauchy's residue theorem.
I have checked your example (the one in Appendix ?) and the CR condition holds. The point is that the contour (path) of integration must lie in an annulus where the function in question is holomorphic (analytic); note that residues are given by the coefficient a_{-1} in the Laurent series.
Yesterday, I promissed to check if the assumptions can be weaker. I don't think so because the crucial point is that the integral along the closed path must vanish what is guaranteed by holomorphity. I can imagine a non-holomorphic function with vanishing integral along such path that fulfils the the residue theorem formula but, I think, this would be only a numerical coincidence. But an example would be interesting :-).
Consider the function f(x,y)=u(x,y)+iv(x,y)=y/(x^2+y^2)+i*x/(x^2+y^2) which In z terms is simply f(z)=i/z.
The real functions u(x,y), v(x,y) satisfy the conditions C-R and they are harmonic functions.
The complex function f is not analytic at origin z=0.
The path integral of f over any curve that contains the origin (0,0) is -2*pi.
The Res(f,0)=i, thus Integral=2*pi*i*i=-2*pi.
That is because u, and v have not cintinuous partial derivatives at origin.
let function f(x,y)=u(x,y)+iv(x,y) is differential at z=x+iy and partial derivative (Ux,Uy,Vx,Vy ) exist and continue Ux = Vy and Uy= -Vx which is necessary conditions for f(z) to be analytic alsso sufficient for f(z) to be analytic in domain D.
Dear Derek Abbott,
Cauchy-Riemann Equations is necessary condition but is not sufficient for analyticity. Because,
1. If f=u+iv is analytic (holomorphy) ==> CR is satisfied.
2. If CR is satisfied and ux , uy , vx , vy are exist-continuous ==> f is analytic.
With regards..
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