# Showing Existence of Antiderivative for Complex-Valued Function

I am asked to show that for $$z\in \mathbb{C} \setminus \{0,1\}$$, there exists an analytic (single-valued) function, $$F(z)$$ on $$\mathbb{C} \setminus \{0,1\}$$, such that $$F'=f$$, where $$f(z) = \frac{(1-2z)\cos(2\pi z)}{z^2 (1-z)^2}$$ I know that if $$\int_{\gamma} f(z) dz =0$$ for all closed contours, $$\gamma$$, then $$f$$ has an antiderivative. Furthermore, in the case of the given function above, $$f(z)$$, I know that Res$$(f,0)=$$ Res$$(f,1)=0$$, so using the Residue Theorem I know that for any simple closed contour, $$\gamma$$, we have $$\int_{\gamma} f(z) dz =0$$

However, to ensure that $$f$$ has an antiderivative, I need to show that this is true for all closed $$\gamma$$, not just simple closed $$\gamma$$. How can I go about finishing this last step of the proof?

• Are you allowed to use Taylor series expansion of cosine? If so, the result follows immediately by considering the uniform convergence of the series on proper domains. – Ken Hung Aug 4 at 0:17
• The Residue theorem holds for all (rectifiable) closed curves. Which version are you referring to? – Martin R Aug 4 at 0:18
• For general rectifiable closed curve $\gamma$ in $\mathbb{C}\setminus\{0,1\}$, we have $$\int_{\gamma}f(z)\,\mathrm{d}z=2\pi i \sum_{z_0\in\{0,1\}}\operatorname{wind}(f,z_0)\operatorname{Res}(f,z_0),$$ where $$\operatorname{wind}(f,z_0)=\frac{1}{2\pi i}\int_{\gamma}\frac{\mathrm{d}z}{z}\in\mathbb{Z}$$ is the winding number. It can be proved by deforming $\gamma$ into a formal linear combination of simple closed curves. – Sangchul Lee Aug 4 at 0:25
• What I was trying to emphasize is that in showing that the antiderivative of a certain function exists, one may try to construct a series expansion of the function and try to show that it converges uniformly on some domain. So that you can integrate (here I mean the inverse process of differentiation) the series term by term to find the antiderivative explicitly. – Ken Hung Aug 4 at 0:30
• In this case, Taylor series expansion of cosine might be helpful. – Ken Hung Aug 4 at 0:31

Here is an alternative solution. Write

$$g(z) = f(z) - \left( \frac{1}{z^2} - \frac{1}{(z-1)^2} \right).$$

Then $$g$$ has removable singularities at both $$z=0$$ and $$z=1$$, and so, $$g$$ extends to a holomorphic function on $$\mathbb{C}$$. In particular, $$g$$ has an antiderivative, say $$G(z)$$. Then

$$f(z) = g(z) + \frac{1}{z^2} - \frac{1}{(z-1)^2}$$

has an antideriviative

$$G(z) - \frac{1}{z} + \frac{1}{z-1}.$$

• Thanks for the response. I'm trying to follow your reasoning where you conclude that $g$ has an antiderivative. You say $g$ extends to a holomorphic function and so, in particular, $g$ has an antiderivative. However, just because a function is holomorphic does not mean that it has an antiderivative. As Martin pointed to in a comment above, $1/z$ is holomorphic in $\mathbb{C} \setminus \{0\}$ yet does not have a holomorphic antiderivative. – French Toast Crunch Aug 4 at 3:36
• @FrenchToastCrunch, The crucial difference here is that the domain of $g$ is all of $\mathbb{C}$, which is simply connected. In such case, a holomorphic function always admits an antiderivative. (That is why we bother to remove the poles in this solution!) Alternatively, since $g$ is analytic on all of $\mathbb{C}$, it admits Taylor series that converges to $g$ on all of $\mathbb{C}$, and so, its antiderivative can be represented explicitly via termwise integration. – Sangchul Lee Aug 4 at 3:40
• I see, I'm following now. The last bit I need to put the pieces together is understanding why, if $g$ is a function with removable discontinuities, it extends to a holomorphic function on $\mathbb{C}$. My basic knowledge of analytic continuation is that if $f_1$ and $f_2$ are analytic on domains $D_1$ and $D_2$, respectively, and $f_1=f_2$ on $D_1 \cap D_2$, where $D_1 \cap D_2$ is non-empty, then $f_2$ is the analytic continuation of $f_1$ on $D_2$. I'm trying to use this to conclude what you did about the extension of $g$.If you can clear up this last step I'd be happy to accept your answer. – French Toast Crunch Aug 4 at 4:20
• @FrenchToastCrunch, It seems that you have not learned Riemann's theorem on removable singularities yet. The proof is actually short, and is included in the link. – Sangchul Lee Aug 4 at 15:44

The Residue theorem states that for any (rectifiable) closed curve $$\gamma$$ in $$\mathbb{C} \setminus \{0,1\}$$: $$\int _{\gamma }f(z)\,dz=2\pi i \bigl( \operatorname {I} (\gamma ,0)\operatorname {Res} (f,0) + \operatorname {I} (\gamma ,1)\operatorname {Res} (f,1) \bigr)$$ and that is zero for the given function $$f$$ because both residues are zero (as you already calculated).

Alternatively you could use that $$f(z) = \frac{\cos(2 \pi z)}{z^2} - \frac{\cos(2 \pi (z-1))}{(z-1)^2}$$ and show that $$\frac{\cos(2 \pi z)}{z^2}$$ has an antiderivative in $$\mathbb{C} \backslash \{0\}$$.

• What is $I(\gamma,0)$? Is that an indicator function specifying that the curve $\gamma$ encloses the point $0$? – mjw Aug 4 at 1:12
• @mjw: It is the winding number of $\gamma$ with respect to the point $0$. – Martin R Aug 4 at 1:16
• Okay, thank you. – mjw Aug 4 at 1:19
• @MartinR : Thank you for your response, in the link you provided, the claim is that the Residue Theorem holds for any rectifiable closed curve in $U$, where $U$ is a simply connected open subset of $\mathbb{C}$. However, $\mathbb{C} \setminus \{0,1\}$ is not a simply connected set. I like your alternate suggestion but ultimately I'd be faced with the same problem I was running into. Following along with your suggestion, how do you show $\cos(2 \pi z)/z^2$ has an antiderivative in $\mathbb{C} \setminus \{0\}$? – French Toast Crunch Aug 4 at 3:28
• @FrenchToastCrunch: The curve integral is only defined for rectifiable curves. – Martin R Aug 4 at 4:19

The function $$f$$ is holomorphic on $$\mathbb{C} \backslash \{0,1\}$$ because $$f$$ is the quotient of holomorphic functions. Since $$f$$ is holomorphic in this set, it has a primitive $$F$$ there so that $$F^\prime=f$$. Or is the question to find $$F$$?

UPDATE

This answer is not complete. Leaving it here for a while because the comment from @Martin R was helpful.

• A holomorphic function does not necessarily have a (holomorphic) antiderivative. The standard counterexample is $1/z$ in $\mathbb{C} \backslash \{0\}$. – Your argument would only work in a simply-connected domain. – Martin R Aug 4 at 0:59
• Okay, so (in general) there is not necessarily an $F$ that works for the whole set, if not for every closed curve $\gamma$, $\oint_\gamma f(z) \,dz = 0$. I am glad I attempted an answer. I learned something. Thank you! – mjw Aug 4 at 1:09