# Is a function analytic iff it has antiderivative?

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?

• But if by primitive of $f$ you mean there is an $F$ with$F'=f$, then $F$ itself is differentiable and therefore analytic -- and then so is $f$.
– gary
Jul 26, 2017 at 17:40
• Are you familiar with Morera's theorem? Jul 26, 2017 at 17:42
• @gary Thank you for the answer. But what about the reverse? If f is analytic do we always have antiderivative such that $F'=f$? Jul 26, 2017 at 17:45
• @Daniel Li. Yes, if the domain is simply-connected.
– gary
Jul 26, 2017 at 17:49
• @DanielLi: Please see the "Sufficiency" part of en.wikipedia.org/wiki/Antiderivative_(complex_analysis) , near the bottom, and let me know if you still have questions.
– gary
Jul 26, 2017 at 18:04

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
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.