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$\gamma$ is the (triangle) contour $i\longrightarrow-i\longrightarrow1\longrightarrow i$. $\def\rmd{\mathop{}\!\mathrm{d}}$

Using Mathematica to evaluate the directional limit at $0$ on $\gamma$ Limit[1/Sin[1/z], z -> 0, Direction -> -I] I get $0$. [Proof: $z=iy,\lim_{y\to0}\sinh\frac1y=\infty\Rightarrow\lim_{y\to\infty}\frac1{\sinh\frac1y}=0$] So the integrand is bounded and continuous on $\gamma$, so the integral is finite.

Using Mathematica to evaluate the integral NIntegrate[1/Sin[1/z], {z, I, -I, 1, I}]

I get $$\int_\gamma\frac{1}{\sin \left(\frac{1}{z}\right)}=\frac{i\pi }{6}$$ I try to prove this.

Sum of residues at $z=\frac1{\pi n},n=1,2,\dots$ $$\sum _{n=1}^{\infty }\frac1{\frac d{dz}\sin(1/z)|_{z=1/\pi n}}=\sum _{n=1}^{\infty }\frac1{-\frac{\cos(1/z)}{z^2}|_{z=1/\pi n}}=\sum _{n=1}^{\infty } \frac{(-1)^{n+1}}{\pi ^2 n^2}=\frac{1}{12}$$ Then $$\int_\gamma\frac{1}{\sin \left(\frac{1}{z}\right)}\rmd z=2\pi i\cdot\frac{1}{12}=\frac{i\pi }{6}$$ Is this proof correct? I am worried since $f(z)$ has a non-isolated singularity at $z=0$.

hbghlyj
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    The point $z=0$ is not an isolated singularity of $\frac{1}{\sin{(1/z)}}$. It is a limit point of poles because $\frac{1}{n\pi}$ is a pole for each integer $n \neq 0$. Every deleted $\epsilon$ neighborhood of $z=0$ has at least one point of the form $z = \frac{1}{n\pi}$. You cannot express a Laurent series of $\frac{1}{\sin{(1/z)}}$ near $z = 0$. – Accelerator Jun 05 '23 at 20:58
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    @Accelerator Yes. In the last sentence I wrote that "$f(z)$ has a non-isolated singularity at $z=0$", so I am worried about the proof. But in the first part I showed that the integral on contour $\gamma$ is finite. – hbghlyj Jun 05 '23 at 21:00
  • Oh, my bad. For some reason, I read your last sentence as "I am worried if $f(z)$ has a non-isolated singularity at $z=0$." – Accelerator Jun 05 '23 at 21:02
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    @Accelerator Ok. I will fix the last sentence. – hbghlyj Jun 05 '23 at 21:06
  • This is not correct because there is an essential singularity with no local Laurent series expansion at zero, and to skirt around zero... you have to evaluate an error term which you haven't done – FShrike Jun 05 '23 at 21:21
  • Use residue calculus with an indented contour? – hbghlyj Jun 05 '23 at 22:27
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    The contour integral can be evaluated. But the value of the integral has no connection with residue at 0 for the simple reason, residue is defined only for isolated singularities. – Lawrence Mano Jun 06 '23 at 00:29
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    @Accelerator Having the right answer doesn’t mean you had a right method. There is a very genuine problem at zero which needs addressing. Also I invite you to prove the clockwise indent vanishes: $z\csc z^{-1}$ is not convergent as $z\to0$. – FShrike Jun 06 '23 at 08:41
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    Though, if the answer really is $\pi i/6$, then this would imply (and be equivalent to): $$\lim_{n\to\infty}\frac{1}{n}\int_{-\pi/2}^{\pi/2}\csc\left(\frac{\pi}{2}(2n+1)\cdot e^{-i\phi}\right)\cdot e^{i\phi},,\mathrm{d}\phi=0$$Which is not obvious to me. – FShrike Jun 07 '23 at 11:37

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