If $\displaystyle \int_\gamma f(z) \, dz = i$ and $\displaystyle\int_\gamma g(z)\,dz = 3-i,$ find $\displaystyle\int_{-\gamma} (2f(z) + ig(z))\,dz$?

Question

Can't figure our this reverse path integral at all, can anyone help?

Answer 1

We know that the integral along a directed smooth curve $\gamma$ is given by $$ \int_\gamma f(z)\,dz = \int_a^b f(\gamma(t)) \gamma'(t)\,dt, $$ where we have a parametrization of the curve $\gamma$ consistent with its direction.

Suposing that what is meant by "$-\gamma$" is inverting the path direction, then $$ \int_{-\gamma} f(z)\,dz = \int_{-b}^{-a} f(\gamma(-t)) (-\gamma'(-t))\,dt = - \int_{a}^{b} f(\gamma(u))\gamma'(u)\,du = -\int_\gamma f(z)\,dz, $$ where we have performed the change of variables $u=-t$.

More details can be found in Contour integration and reversing orientation of parametrisation.

Using the above result along with the integral being a linear operator, we have that \begin{align*} \int_{-\gamma} (2 f(z) + i g(z))\,dz &= -\int_{\gamma} (2 f(z) + i g(z))\,dz\\ &= -2\int_{\gamma} f(z) \,dz - i\int_{\gamma} g(z)\,dz \\ &= -2 i - i (3-i) \\ &= \boxed{-1 - 5i}. \end{align*}