Hm, if it's based on this ratio of contour integrals, shouldn't it be possible to do better than this? Like why would it be so hard to find the residues for these poles? Shouldn't that be just a bit of formal Laurent series manipulation? What am I missing here?
If the series don’t cancel out nicely (likely, I would guess), wouldn’t you end up with some infinite sum?
If so, as this example shows (contour integrals in a closed form?) “closed form” is loosely defined, but I think most would say something with an infinite sum wouldn’t be one (but then, https://en.wikipedia.org/wiki/Closed-form_expression#Analyti... says:
“In particular, special functions such as the Bessel functions and the gamma function are usually allowed, and often so are infinite series and continued fractions. On the other hand, limits in general, and integrals in particular, are typically excluded.[citation needed]”
No, you shouldn't end up with an infinite sum; if you only want to know the series finitely far out, you only need to know finitely many terms.
However, it seems the problem is hard for other reasons. I had assumed, without checking, that they'd put the center of the circle at 3pi/8 because it's a pole. Nope! As best I can tell from some graphing tools, there is indeed precisely one pole in the circle, and it's on the real line, but it's close to the right endpoint; I don't know that it has any nice form. So I imagine that getting any sort of exact series expansion around there -- or even just getting exactly the first few terms, i.e. the first few derivatives there, which would be all you'd need -- would be difficult for that reason.
(Although, the higher the order of the zero, the more initial terms you'd need...)
Edit: Actually, I guess it looks like a zero of order 1? Except that doesn't make sense, because then the top contour integral would be zero...
