Fundamental theorem of line integral states that for any function $f$ that has an antiderivative $F$, integrating $f$ from point $a$ to point $b$ yields $F(b) - F(a)$, which would imply integration over a closed path yields $0$; However, Cauchy theorem requires the function to be analytic to guarantee $0$ on closed path integration. So does this mean any function that has primitive function $F$ will automatically be analytic and vice versa?
2026-03-29 22:13:38.1774822418
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Is a function analytic iff it has antiderivative?
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Because the derivative of a holomorphic function is itself holomorphic, any function that admits a primitive is holomorphic. However, the function $z\in \mathbb C\smallsetminus 0 \longmapsto z^{-1}\in \mathbb C$ admits no primitive on all its domain, because it winds once around the origin, and any function that admits a primitive must wind zero times around it.
Yes, if $f$ is analytic and defined in a simply-connected domain $D$, it will have an analytic antiderivative given by $ F(z)=\int_{\gamma_z}f(z)dz$ , where $\gamma$ is any curve living in $D$, i.e., $\frac {d}{dz}F(z):=\int_{\gamma_z}f(t)dt=f(z)$. See, e.g.:http://planetmath.org/antiderivativeofcomplexfunction