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Consider the complex function $f(z) = \frac{1}{z^{2}}$. How come the closed integral about the centre vanishes, i.e. $\int_{\partial B_{r}(0)} f(z) dz = 0$, where $r>0$. The function clearly has a pole of second order at $z=0$ and therefore isn't holomorphic on a disc centred at zero.

playdis
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    This thread may also help you, https://math.stackexchange.com/questions/1165813/why-doesnt-frac-1-z-have-an-antiderivative-in-mathbbc-setminus-0 – Someone86 Jan 16 '19 at 11:49
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    Cauchy's theorem says that every holomorphic function has a vanishing integral over a loop, but no one said that if the integral over a loop is vanishing, then the function is holomorphic. – Vasily Mitch Jan 16 '19 at 11:50

2 Answers2

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Three answers:

  1. The integral evaluates to the residue, which is the coefficient of $z^{-1}$ in the Laurent expansion at $0$ of the function. But this is clearly zero, as the Laurent expansion of $z^{-2}$ around $z_0=0$ is again $z^{-2}$, so the coefficient of $z^{-1}$ is $0$.

  2. The function has an antiderivative: $-z^{-1}$ which is continuous on the path you are integrating over.

  3. Evaluating the integral directly via polar coordinates $z=re^{i\varphi}$ gives zero.

b00n heT
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The fact that integrals of holomorphic functions along ["nice"] closed curves are $0$ does not mean those non-holomorphic functions cannot be integrated to be $0$.

To calculate this, just use the parametrization $z = r \mathrm e^{\mathrm i t}, t \in [0, 2\pi)$.

Alternatively, we consider the entire function $f(z) \equiv 1$. Then by the Cauchy integral formula and the derivative formula, $$ f'(z) = \frac {1!}{\mathrm i2\pi}\int_{\partial B_r (z)} \frac {f(\zeta)}{(\zeta - z)^2} \,\mathrm d\zeta, $$ then your integral is just $f'(0) = 0$.

xbh
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