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I'm struggling to interpret the notation for this integral and can't find a direct definition for it. $\int_{(1/\mathcal{L})}f(s)ds$. Here, $\mathcal{L}:=log(x)$ for a large $x\in\mathbb{N}$. What does this path mean? Is it the parametrization $t\rightarrow 1/\mathcal{L}+it$ for $t\in\mathbb{R}$? The complex circle centre zero radius $1/\mathcal{L}$? (The paper is Zhang's Bounded Gaps Between Primes, lemma 2, if that's relevant).

jshpmm
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  • The integral there rather reads $\int_{|s|=1/\mathcal{L}}\cdots ds $, so it’s the complex integral along the origin-centred circle of radius $1/\mathcal{L}$. – Jack LeGrüß Oct 15 '20 at 19:08
  • @JackLeGrüß I'm talking about the integral above where it explicitly says $|s|=1/\mathcal{L}$, where there's no error term on the integral and the path is only denoted by $(1/\mathcal{L})$. Do they denote the same thing? I wouldn't have thought so since there is then no reason for an error to appear. – jshpmm Oct 15 '20 at 19:21
  • From what I am understanding, you’re referring to the proof of Lemma 2 and not Lemma 3 (as stated in your post)—regarding $\mathcal{A}_1(d)$ and $\mathcal{A}_2(d)$. – Jack LeGrüß Oct 15 '20 at 19:35
  • I am, actually, I've edited the post to reflect this. My mistake. – jshpmm Oct 15 '20 at 19:36
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    Owing to the context, I believe it is the vertical line at $\Re(s)=1/\mathcal{L}$, that is $$\int_{(1/\mathcal{L})}\cdots ds = \int_{1/\mathcal{L}-i\infty }^{1/\mathcal{L}+i\infty} \cdots ds ,.$$ (Though, I must admit that I couldn’t be so sure; I can’t locate any remark on notation on a quick skimming through, and I’m trying to understand that based on the definition and the way he attempts to move the region and bounds them). Hopefully, someone who’s checked it thoroughly might be able to confirm or explain it accurately) – Jack LeGrüß Oct 15 '20 at 19:40

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