My Attempt: $$\lim _{z \to 0}\frac{1}{\sin \left ( e^{\frac{-1}{z}}\right)} \; does \; not \; exist \;$$
$\Rightarrow z=0 $ is Essential Singularity.
But How to compute the value of integral?
2026-03-26 11:18:21.1774523901
The value of the integral $\oint_{|z|=1}\frac{1}{\sin \left ( e^{\frac{-1}{z}}\right)}$
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Let $f(z)=\dfrac{1}{\sin e^{-\frac{1}{z}}}$, then $\text{res}_\infty f(z)=\lim\limits_{z\to\infty}z(f(\infty)-f(z))$, where $f(\infty)=\lim\limits_{z\to\infty}f(z)=\dfrac{1}{\sin 1}$ (therefore infinity is a removable singular point). So $$\text{res}_\infty f(z)=\lim\limits_{z\to\infty}z\left(\dfrac{1}{\sin 1}-\dfrac{1}{\sin e^{-\frac{1}{z}}}\right)=\dfrac{1}{\sin^21}\lim\limits_{z\to\infty}z(\sin e^{-\frac{1}{z}}-\sin1)=$$$$=\dfrac{1}{\sin^21}\lim\limits_{w\to0}\dfrac{\sin e^{-w}-\sin1}{w}=\dfrac{2}{\sin^21}\lim\limits_{w\to0}\dfrac{\sin\dfrac{e^{-w}-1}{2}\cos\dfrac{e^{-w}+1}{2}}{w}=$$$$=\dfrac{2\cos1}{\sin^21}\lim\limits_{w\to0}\dfrac{\dfrac{e^{-w}-1}{2}}{w}=-\dfrac{\cos1}{\sin^21}.$$ And integral equals $2\pi i\cdot\dfrac{\cos1}{\sin^21}$.