Reversing the direction of a contour integral

Question

If $$\int_{C} f(z) dz$$ is some contour integral over a closed curve $C$, and $-C$ is the contour taken in the opposite direction, can $$ \int_{-C} f(z) dz$$ be treated as a closed curve around the plane not enclosed in $C$, in the sense that it can be evaluated using the residues not inside $C$? I'm assuming that if it could, it would include the residue at infinity; is this assumption correct? Similarly, can $$\int_{c-i\infty}^{c+i\infty} f(z) dz$$ be treated in a similar manner?

Answer 1

Doing my own research, I have found out that if $f$ is holomorphic on some annulus centered at $0$ with infinite outer radius, the residue at infinity of $f$ is defined to be $$Res(\frac{-1}{z^2}f(\frac{1}{z}), 0)$$ which is consistent with the substitution rule of integration. If $C$ is a circle with radius larger than the inner radius of the annulus, the residue is also$$\frac{1}{2\pi i}\int_{-C} f(z)dz$$ I have not yet found anything for the second part of the question, though I assume it to be inaccurate because it does not seem to be true when taking the inverse Mellin transform of the gamma function.