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I stumbled upon this problem that requires the residue theorem, but there's something I'm not sure I understand. Let $x\in R /(0)$, I want to use the residue theorem to calculate: $$f(x):=\int_R{ 1\over [(x-t)^2+1](t^2+1)}dt$$

Now, the way I would do this is finding and classifying the singularities, then taking an arbitrary circumference containing them and using the theorem as I normally would. However, in this case the solution claims that the "condition of decay" is met, and without adding anything else it only applies the residue theorem to the singularities in the upper half-plane. There is no trace of this condition in my notes, so I'm not exactly sure what that means. The only decay I can see is that for $t$ large the function inside the integral goes to zero, but I still can't grasp the logic behind it. When am I supposed to only consider the positive (in a complex sense) singularities, as opposed to all of them?

Thank you.

Silence
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  • It all depends on the contour that you use. In this case you'll use a semi-circle oriented the first two quadrants. From Cauchy's second integral theorem you only consider the residues INSIDE the contour, so that is why you only consider the positive residues for this contour ($t=i$ and $t=x+i$). If you need further help let me know. – Ty. May 29 '20 at 17:44
  • First of all, thank you for taking the time. I know about contours, what I don't understand is why am I supposed to use an upper semi-circle in this instance, as opposed to a full circle? The way I solved this (with a full circle) the result was $0$, but the solution simply says I'm supposed to consider the upper semi-circle. Why is that? – Silence May 29 '20 at 17:48
  • This may help you: https://math.stackexchange.com/questions/580110/what-are-the-reasons-for-using-a-semi-circle-in-upper-half-plane-of-mathbbc. – Ty. May 29 '20 at 17:50
  • Alright yeah that's definitely what I was looking for. However, the way I understand it, is that the contour choice is sort of arbitrary, except for the fact that this function, as I mentioned, goes to zero for large $t$, which I assume is the infamous condition of decay. Let me ask you this then: if the condition weren't true, choosing a full circle would've been perfectly ok, right? And how does a decaying function imply the necessity of a semi-circle? I apologise for the barrage of questions but I really want to understand this subject fully. Thank you again for your patience. – Silence May 29 '20 at 18:03
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    I haven't taken a complex analysis course yet so I'm not completely sure, but I think that the advantage of using a semi-circle is that the line integral of the segment along the real axis is equal to the original integral that you're solving for (at least for this case), and the line integral along the arc of the semi-circle cancels out due to the condition of decay/Jordan's lemma. Therefore, using a semi-circle is quite advantageous, and a circle doesn't have the line segment along the real axis that a semi-circle has. – Ty. May 29 '20 at 18:10
  • Makes sense, as the integral along the full circle gives in fact zero. Alright well I may not have a full grasp on the hows and whys, but at least I got a procedural answer that I can understand. Thank you very much. – Silence May 29 '20 at 18:20

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