# Complex integral on a smooth curve contained in an open set

Thm 5.19 (exactly) says: Let $\gamma\colon[a,b]\rightarrow \mathbb{C}$ be piecewise smooth. Let $F$ be a complex function defined on an open set containing $\gamma^*$, and suppose that $F'(z)$ exists and is continuous at each point of $\gamma^*$. Then $$\int_\gamma F'(z)dz=F(\gamma(b))-F(\gamma(a)).$$

The homework asks to determine all possible values of $$\int_C \frac 1{z^2+1}dz,$$ where $C$ is piecewise smooth from $z=-1$ to $z=1$, but does not include $z=i$ or $z=-i$.

Part of our answer is that if $C_1$ is straight from $-1$ to $1$ (or deformable into such a line), then the integral must equal $\arctan(1)-\arctan(-1)=\frac \pi 2$. Note that $C_1$ passes between the two singularities.

Why doesn't thm 5.19 apply to a curve which passes over (or under), both singularities (and doesn't loop either)?

If $C_2$ passes over both singularities, then it is piecewise smooth. $F$ is still defined on an open set containing $C_2$ (it's not an open disc, and it would have to be drawn carefully so as not to include either singularity, but it is open). And $F'(z)$ still exists and is continuous on the curve.

If my friend (and the perfect grade he got on his homework!), is correct, then if a curve passes over both singularities, it is equal to $\frac \pi2$.

P.S. No need to inform me about loops of the singularities. We've handled that situation using Cauchy's Integral Formula.

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You write that you already know how to deal with loops around the singularities. But that is actually all you need to know. For example if you have a path that passes above both singularities from $-1$ to $1$, you can extend it with one that goes back along the real line to $-1$ and then doubles back (forward) to $1$:

The two parts of the extensions will cancel out each other, so they don't change the integral around the entire path. On the other hand, the extended path can now also be viewed as a sum of a loop around $i$ and the straight path you have already computed.

In this way, any path however wild from $-1$ to $1$ can be decomposed into some number of loops around the singularities (possibly in various directions), plus a single copy of the straight baseline path.

The theorem you quote does apply to the original (unextended) $\gamma$, but you cannot use the same antiderivative as you used to find the integral of the straight path. That's because this $\gamma$ crosses a branch cut you needed to make to define that antiderivative, and you can't extend the domain of the antiderivative all the way to the branch cut from both sides and still have it be continuous on its domain.

On the other hand, the theorem doesn't apply to the extended path in my diagram, because there is no antiderivative that can cover the entire loop around $i$ without any discontinuities.

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You know what the integral is around the loop given by traversing $C_2$ and then traversing $C_1$ backwards (by the Cauchy integral formula, on which you're sold) and you know what the integral is along $C_1$, so you can deduce the integral for $C_2$ by subtracting, even if you're squeamish about the branch cut used in defining the arc tangent.

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There is a way to make your theorem 5.19 applicable to arbitrary paths $\gamma$ connecting $z=-1$ with $z=+1$ that are disjoint from the segment $\sigma:=[-i,i]$, in particular paths that "pass over the singularity at $z=i$":

Let $\Omega:={\mathbb C}\setminus\sigma$. The Möbius transformation $z\mapsto{z-i\over z+i}$ maps $\sigma$ onto ${\mathbb R}_{\leq 0}\subset{\mathbb C}$, therefore $${z-i\over z+i}\notin {\mathbb R}_{\leq 0}\qquad(z\in\Omega)\ .$$ It follows that the function $$F(z):={1\over 2i}{\rm Log}{z-i\over z+i}$$ is analytic in $\Omega$, and one easily computes $F'(z)={1\over z^2+1}$. Therefore by theorem 5.19 for any $\gamma$ of the described kind one has $$\int_\gamma{dz\over z^2+1}=F(1)-F(-1)={1\over 2i}\Bigl({\rm Log}{1-i\over 1+i}-{\rm Log}{-1-i\over -1+i}\Bigr)=-{\pi\over2}\ .$$

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As countinghaus points out, there's a branch cut so $F$ is not holomorphic for the theorem to apply, hence the need to subtract an overhead contour out of a loop to compute that particular integral.

$\hskip 2.3in$

This is a picture of the complex $\arctan(z)$ function from Wikipedia.

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