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If $\gamma$ is a path from $-i$ to $i$, whose image is contained in $\mathbb C\setminus\mathbb R^-$, find $\int_{\gamma}\frac{dz}z$

Does the integral converge ?, because the path $-i+2it, 0\le t\le1$ is also in $\mathbb C\setminus\mathbb R^-$ and for $t=1/2$ it is $0$

or can I use the fact that $f(z):=\frac1z$ is meromorphic with simple pole at $0$ and define

$\alpha_{\rho}(t):=0+\rho e^{it}, t\in[0,\pi]$ and then $\displaystyle\int_{\alpha_{\rho}}f(z)dz=\text{res}_0(f)\cdot \pi i$

with $\text{res}_0(f)=a_{-1}$ where $a_{-1}$ is the Laurent coefficient of $f$

Is the result $1$ ?

inequal
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    The author(s) probably denote $\mathbb{R}^- = { x\in \mathbb{R} : x \leqslant 0}$. Otherwise, as you note, there is a problem. – Daniel Fischer May 19 '15 at 15:14

2 Answers2

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$\gamma$ is a path from $i$ to $-i$, not necessarily a straight line which passes through $0$.

Since the image of $\gamma$ does not touch $\mathbb R^-$, it means that $\gamma$ cannot "turn around" $0$, so you can't really use the residue theorem because there are no residue in the inside of $\gamma$ (well $\gamma$ is not even a closed curve. But anyhow you can't close it such that $0$ is inside)

So any curve between those two points will result in the same integral.

How to do that? Well since we are on $\mathbb C - \mathbb R^-$, we have that $\log z$ is a holomorphic function (it's not multivalued anymore) and it's derivative is $\frac 1z$, we can use the usual formula;

The result depends only on the endpoints, and we get

$$\int_\gamma \frac 1z dz = \log(i) - \log(-i) =i \frac \pi2 - (-i\frac \pi2) = i\pi$$

Mark Viola
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Ant
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  • You're assuming here, of course, the principal values of the corresponding arguments in $;\log z;$ . – Timbuc May 19 '15 at 16:25
  • @Ant I took the liberty of adding the missing "$i$'s" from the last expression. I hope you don't mind my doing so. – Mark Viola May 19 '15 at 19:03
  • @Dr.MV Of course! Thank you for the correction :-) – Ant May 19 '15 at 19:43
  • @Ant You're welcome. My pleasure. I cannot tell you how many typos I've made on this site. And some users are fast to give a down vote rather than recognize sincere effort along with a typographical error. Oh, well. +1 for you! – Mark Viola May 19 '15 at 19:48
  • @Ant, I am confused. Don't you think the integral is gonna pick a half residue on the way ? – user91411 Jul 15 '20 at 10:42
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Inasmuch as @Ant has already provided an efficient approach, it might be instructive to see a more "brute force" approach.

To that end we parameterize $z$ on $\gamma$ and let $z=x(t)+iy(t)$, $0\le t\le 1$, with $x(0)=x(1)=0$, $y(0)=-1$, and $y(1)=1$.

Thus, $dz=x'(t)dt+iy'(t)dt$ and the integral of interest becomes

$$\begin{align} \int_{\gamma}\frac{dz}{z}&=\int_{0}^1\frac{x(t)x'(t)+y(t)y'(t)}{x^2(t)+y^2(t)}dt+i\int_{0}^1\frac{x(t)y'(t)-y(t)x'(t)}{x(t)^2+y(t)^2}dt\\\\ &=\left. \frac12 \log(x^2(t)+y^2(t))\right|_{t=0}^{t=1}+i\,\left.\arctan\left(\frac{y(t)}{x(t)}\right)\right|_{t=0}^{t=1}\\\\ &=(0-0)+i(\pi/2-(-\pi/2))\\\\ &=i\pi \end{align}$$

as expected!

NOTE: For a more general development, see This Answer.

Mark Viola
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  • Thanks. but you assume $\frac10=\infty$ in the 3rd last line, actually clear but is it allowed ? – inequal May 19 '15 at 19:47
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    @inequal Technically, one ought to use a limit as $t \to 0$ and $t\to 1$. But, the answer prevails. Another way is to redefine the anti-derivative to be $\arctan(y/x)$ for $x\ne 0$ and $\pm \pi/2$ when $x=0$. Thank you for your comment! – Mark Viola May 19 '15 at 19:51