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I am studying for an exam and I was doing practice questions, but I am stuck on this one. I'm not sure how to evaluate it using a complex integral, which is what we're supposed to do.

I was thinking that I can replace $2\sin(\theta)$ by $\frac{1}{i}(z-\frac{1}{z})$, and then evaluate it as $$\int e^{-i(z-\frac{1}{z})}dz$$ on $|z|=1$, but I am stuck on how to actually proceed, since there is a singularity $z=0$ for the $1/z$ part. Can anyone help explain how to do this?

Deirdre
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  • This is related to the modified Bessel function of the first kind. – Sangchul Lee Dec 20 '21 at 17:56
  • Ahh okay, I did not know that lol. Is there no other, more simple way to solve this integral though? – Deirdre Dec 20 '21 at 18:02
  • I am pretty sure the Bessel functions are non-elementary, so there is no hope of obtaining an elementary closed form for your integral. Is this really what the original question is asking? (By the way, it is possible to give a series representation of your integral.) – Sangchul Lee Dec 20 '21 at 18:05
  • You can get the result as a series that converges fast by using Taylor and Laurent series – Conrad Dec 20 '21 at 18:08
  • @Conrad Ohhh yeah okay. So I can get the Taylor series for e, and work ahead from there? Or the Taylor/Laurent series for the power part? – Deirdre Dec 20 '21 at 18:12
  • @SangchulLee ahh yeah, so I'm trying to evaluate it using complex integrals or the residue theorem, so I was asking in that sense. I'll edit my question to clarify it – Deirdre Dec 20 '21 at 18:12
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    Hint: $e^{2\sin\theta}=e^{z/i}e^{i/z}=\sum_{j,,k\ge0}\frac{i^{k-j}z^{j-k}}{j!k!}$ has $z^{-1}$ coefficient $i\sum_{j\ge0}\frac{1}{j!(j+1)!}$. This can be written nicely, but only with Bessel functions. – J.G. Dec 20 '21 at 18:24

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