Are the integrals of the function:
$$f(z)=\frac{1}{z+1}+\frac{1}{(z+1)^2}+e^{\frac{1}{z}}$$
path independent in the following domain:
$$D= \{Re z >0\}\setminus\{1\}$$
My thoughts on the problem:
The domain is the real parts of z greater than zero with the exception of 1 because if the value of z is 1 then the function fails to be analytic at that point.
I also know that integrals of f(z) are path dependent in D, if there exists at least one closed contour such that the integral over that contour does not equal zero. To verify path dependence it is enough to find a single closed contour with non-zero integrals.
I struggle the most with deciding if they are path INdependent or not. I know that there are three equivalent forms of path independence:
1) If the domain is simply connected then if f is analytic in the domain, then the integrals are path independent.
2) If there exists a function f defined in the domain D which is the anti-derivative of the original function then the integrals are path independent.
3) for every two points and for any two contours lying in D, the contour completely lies in D with the same initial and terminal points (the definition)
Can anyone help me write up a formal definition for this problem?
Best Answer
I think that after you mention that your function has poles in:
everything gets easier :)
The point $1$ is a removable singularity, so you can continue your function onto the whole right half-plane (there is a theorem about it). Your domain doesn't contain $0$ and $-1$, so your function is holomorphic in it. Hence, its integral is path-independent.
If you are allowed to use the Cauchy theorem to prove it, you can check it this way: take two different paths between points $z_1$ and $z_2$, and join them together. They form a closed contour. By Cauchy theorem, the integral of a holomorphic function over a closed contour is zero in a simply-connected domain (which yours is, after you removed the singularity in $1$). So your integral over one path is equal to minus integral over other path. Now, change the direction to get rid of the minus, and you're done.