I am trying to learn complex analysis, and want to understand how to evaluate contour integrals without using the residue theorem. But I don't understand how an integral over a closed contour can be non-zero. Here is a “proof'' that every such integral is zero:
Consider the integral $\oint_\gamma f(z)\ dz$ for $\gamma$ a simple closed contour, and $f$ meromorphic in the region bounded by $\gamma$. Parametrize $\gamma$ by $[0,1]\to\mathbb{C}$. Then by the definition of the contour integral, we have $\oint_\gamma f(z) dz = \int_0^1 f(\gamma(t))\gamma\prime(t)\ dt$. Now let $w = \gamma(t)$, so $dw = \gamma\prime(t)\ dt$. Thus, $\oint_\gamma f(z)\ dz = \int_0^1 f(w)\ dw$. Now let $F(z)$ be the antiderivative of $f$, so $dF/dZ = f$. Then we have $\oint_\gamma f(z) dz = \int_0^1 f(w)\ dw = f(w)\vert_1 – f(w)\vert_0 = f(\gamma(1)) – f(\gamma(0)) = 0$, where the last step follows from noting that $\gamma$ is closed and therefore the image of the endpoints of the interval $[0, 1]$ under $\gamma$ must be equal. QED
Obviously, this is incorrect. I know that the contour integral is not always zero, but I don't understand where the error is. Perhaps I have the wrong idea of how to evaluate a complex contour integral by hand? Any assistance is greatly appreciated!
Best Answer
Your fake proof assumes that every meromorphic function has an antiderivative. That is false. For instance, $z\mapsto\frac1z$ has no antiderivative. And the simplest way of proving that consists in showing that$$\oint_{|z|=1}\frac{\mathrm dz}z=2\pi i\ne0.$$