Here is my worksheet for the simple example 1 / sqrt(z)

MP_contour_bySqr.mws

MP_contour_bySqr.pdf

It is as follows, I think, f:= t-> g(c(t))*D(c)(t), c = unit circle, t in 0 .. 1:

It is defined, but because of branch cut for sqrt it is not continous. 

It jumps, where the input for sqrt has imaginary part = 0 and negative real
part, i.e. where 1/g(c(t))^2 does that.

This is exactly the case, where t=1/3 or t=2/3:

'g(z)': '%'=%;
denom(g(c(t))^2);
plot([Re(%), Im(%)], t=0 .. 1, color=[red,blue]);


Now manually split into continous parts and let Maple do that work then
(which would not be needed in Math):

'Int(f(t), t=0/3 .. 1/3)'; v1:= combine(value(%));
'Int(f(t), t=1/3 .. 2/3)'; v2:= value(%);
'Int(f(t), t=2/3 .. 3/3)'; v3:= combine(value(%));


                                            1/2
                                        3 11    + 10
                v1 := 1/4 I Pi - 1/4 ln(------------)
                                               1/2
                                          2 + 3

                           v2 := -1/2 I Pi

                                               1/2
                                          2 + 3
                v3 := 1/4 I Pi - 1/4 ln(------------)
                                            1/2
                                        3 11    + 10


'v1+v2+v3': '%'=simplify(%);

                           v1 + v2 + v3 = 0



I will do the more simple case 1/sqrt(z) (and the other one) separately.

It is as follows, I think, f:= t-> g(c(t))*D(c)(t), c = unit circle, t in 0 .. 1:

It is defined, but because of branch cut for sqrt it is not continous. 

It jumps, where the input for sqrt has imaginary part = 0 and negative real
part, i.e. where 1/g(c(t))^2 does that.

This is exactly the case, where t=1/3 or t=2/3:

'g(z)': '%'=%;
denom(g(c(t))^2);
plot([Re(%), Im(%)], t=0 .. 1, color=[red,blue]);


Now manually split into continous parts and let Maple do that work then
(which would not be needed in Math):

'Int(f(t), t=0/3 .. 1/3)'; v1:= combine(value(%));
'Int(f(t), t=1/3 .. 2/3)'; v2:= value(%);
'Int(f(t), t=2/3 .. 3/3)'; v3:= combine(value(%));


                                            1/2
                                        3 11    + 10
                v1 := 1/4 I Pi - 1/4 ln(------------)
                                               1/2
                                          2 + 3

                           v2 := -1/2 I Pi

                                               1/2
                                          2 + 3
                v3 := 1/4 I Pi - 1/4 ln(------------)
                                            1/2
                                        3 11    + 10


'v1+v2+v3': '%'=simplify(%);

                           v1 + v2 + v3 = 0



I will do the more simple case 1/sqrt(z) (and the other one) separately.

@Markiyan Hirnyk 

Missing homotopy invariance

@Markiyan Hirnyk 

Missing homotopy invariance

Hm ... I did it once as it is sketched in Cartan (IIRC). What is your way to
define the path integral? Let's do infinity later.

The point is: sqrt(z) or 1/sqrt(z) do not allow Cauchy's theorem, there is
no Laurent series in z=0 (and ignore the branch cut for now).

Try it with the simple case h:= z -> 1/sqrt(z) and unit path around zero.

Hm ... I did it once as it is sketched in Cartan (IIRC). What is your way to
define the path integral? Let's do infinity later.

The point is: sqrt(z) or 1/sqrt(z) do not allow Cauchy's theorem, there is
no Laurent series in z=0 (and ignore the branch cut for now).

Try it with the simple case h:= z -> 1/sqrt(z) and unit path around zero.

I think it should be zero as well: 
plotting Re and Im suggests to split integration in 1/3 and in 2/3

h:= z-> 1/sqrt(4*z^2+4*z+1/2);

f:= t-> h(c(t))*D(c)(t);

my:= 0 .. 1/3;
Int(Re(f(t)), t=my, method = _d01akc) + Int(Im(f(t)), t=my, method = _d01akc)*I:
v1:=evalf(%);

my:= 1/3 .. 2/3;
Int(Re(f(t)), t=my, method = _d01akc, epsilon=1e-10) + Int(Im(f(t)), t=my, method = _d01akc)*I:
v2:=evalf(%);

my:= 2/3 .. 3/3;
Int(Re(f(t)), t=my, method = _d01akc) + Int(Im(f(t)), t=my, method = _d01akc)*I:
v3:=evalf(%);

'v1+v2+v3': '%'=%;
                      f := t -> h(c(t)) D(c)(t)

                                     -14         -14
                v1 + v2 + v3 = 0.4 10    - 0.1 10    I

Edited formal solutions

'Int(f(t), t=0/3 .. 1/3)'; v1:= combine(value(%));
'Int(f(t), t=1/3 .. 2/3)'; v2:= value(%);
'Int(f(t), t=2/3 .. 3/3)'; v3:= combine(value(%));

                                            1/2
                                        6 34    + 35
                v1 := 1/4 I Pi - 1/4 ln(------------)
                                            1/2
                                        2 42    + 13


                           v2 := -1/2 I Pi

                                            1/2
                                        2 42    + 13
                v3 := 1/4 I Pi - 1/4 ln(------------)
                                            1/2
                                        6 34    + 35

'v1+v2+v3': '%'=simplify(%);

                           v1 + v2 + v3 = 0

I think it should be zero as well: 
plotting Re and Im suggests to split integration in 1/3 and in 2/3

h:= z-> 1/sqrt(4*z^2+4*z+1/2);

f:= t-> h(c(t))*D(c)(t);

my:= 0 .. 1/3;
Int(Re(f(t)), t=my, method = _d01akc) + Int(Im(f(t)), t=my, method = _d01akc)*I:
v1:=evalf(%);

my:= 1/3 .. 2/3;
Int(Re(f(t)), t=my, method = _d01akc, epsilon=1e-10) + Int(Im(f(t)), t=my, method = _d01akc)*I:
v2:=evalf(%);

my:= 2/3 .. 3/3;
Int(Re(f(t)), t=my, method = _d01akc) + Int(Im(f(t)), t=my, method = _d01akc)*I:
v3:=evalf(%);

'v1+v2+v3': '%'=%;
                      f := t -> h(c(t)) D(c)(t)

                                     -14         -14
                v1 + v2 + v3 = 0.4 10    - 0.1 10    I

Edited formal solutions

'Int(f(t), t=0/3 .. 1/3)'; v1:= combine(value(%));
'Int(f(t), t=1/3 .. 2/3)'; v2:= value(%);
'Int(f(t), t=2/3 .. 3/3)'; v3:= combine(value(%));

                                            1/2
                                        6 34    + 35
                v1 := 1/4 I Pi - 1/4 ln(------------)
                                            1/2
                                        2 42    + 13


                           v2 := -1/2 I Pi

                                            1/2
                                        2 42    + 13
                v3 := 1/4 I Pi - 1/4 ln(------------)
                                            1/2
                                        6 34    + 35

'v1+v2+v3': '%'=simplify(%);

                           v1 + v2 + v3 = 0

because of sqrt there is no Laurent series (integer exponents)

because of sqrt there is no Laurent series (integer exponents)

Maple has a very generic way by 'RootOf' ...

For numerics I am only aware of inverting 'erf' over the Reals in double precision, but Maple allows arbitrary precision. So a starting value can be feed to fsolve.

It is not quite clear to me what you want as procedure.

Edited: Here is a simple & brute one (search for Betten, Inverse of the Error Function)
through erf(xi) ~ tanh(1.201270935*xi) which gives a guess.

There are better ones, the following is due to Koopman (years ago in a discussion group on the usenet)

approxK_invErf := x -> 1/2*signum(x)*(-4*ln(1-abs(x))-2*ln(1-2*ln(1-abs(x))+4*(-.3714448e-1*ln(1-abs(x))+.1073009183012759)*ln(1-abs(x))^2/(-2*(-.6907678e-1*ln(1-abs(x))+.44739441)*ln(1-abs(x))+1)))^(1/2);

Maple has a very generic way by 'RootOf' ...

For numerics I am only aware of inverting 'erf' over the Reals in double precision, but Maple allows arbitrary precision. So a starting value can be feed to fsolve.

It is not quite clear to me what you want as procedure.

Edited: Here is a simple & brute one (search for Betten, Inverse of the Error Function)
through erf(xi) ~ tanh(1.201270935*xi) which gives a guess.

There are better ones, the following is due to Koopman (years ago in a discussion group on the usenet)

approxK_invErf := x -> 1/2*signum(x)*(-4*ln(1-abs(x))-2*ln(1-2*ln(1-abs(x))+4*(-.3714448e-1*ln(1-abs(x))+.1073009183012759)*ln(1-abs(x))^2/(-2*(-.6907678e-1*ln(1-abs(x))+.44739441)*ln(1-abs(x))+1)))^(1/2);

Maple finds a simple solution if writing it as integral

  with(inttrans):

  g := (p, q) -> 1/(x+p)/(x+(p*q-1)/q):
  G:=unapply( g(3, 2) , x);

  mellin(f(x),x,s); convert(%, Int);
  eval(%, f=G);
  value(%): simplify(%);


                            (-s + 1)  s      (s - 1)
                      Pi (-2         5  + 5 3       )
                 -2/5 -------------------------------
                                 sin(Pi s)


PS. sometimes it may be usefull to write as Fourier transform:

  mellin(f(x),x,s) = fourier(f(exp(w)),w,I*s); convert(%, Int);
  IntegrationTools:-Change(lhs(%), x=exp(w), w) = rhs(%); 
  #lhs(%%) =Change(rhs(%%), w=ln(x), x);
  is(%); 

           mellin(f(x), x, s) = fourier(f(exp(w)), w, s I)

                                 true

you mean in h=0, no?