calculating a complex integral with essential singularity I want to calculate the integral of the complex function $\dfrac 1 {\sin\frac{1}{z}} $ along the curve $|z|=\frac{1}{5}$. I wanted to use cauchy's residues theorem but the function has an essential singularity at $z=0$ which lies inside the curve so I don't know what to do.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\oint_{\verts{z} = 1/5}{\dd z \over \sin\pars{1/z}}
\,\,\,\stackrel{z\ \mapsto\ 1/z}{=}\,\,\,
\oint_{\verts{z} = 5}{\dd z \over z^{2}\sin\pars{z}}
\\[5mm] & =
2\pi\ic\braces{%
\lim_{z \to 0}{1 \over 2!}\totald[2]{}{z}\bracks{z \over \sin\pars{z}} +
\lim_{z \to -\pi}
\bracks{z + \pi \over z^{2}\sin\pars{z}} +
\lim_{z \to \pi}\bracks{z - \pi \over z^{2}\sin\pars{z}}}
\\[5mm] & =
2\pi\ic\pars{{1 \over 6} + {-1\phantom{^{2}} \over \pi^{2}} +
{-1\phantom{^{2}} \over \pi^{2}}} =
\bbx{\pars{{1 \over 3}\,\pi - {4 \over \pi}}\ic}
\end{align}

There are three poles inside the contour $\ds{\braces{z\ \mid\ \verts{z} = 5}}$: single ones at $\ds{\pm\pi}$ and a third order one at $\ds{z = 0}$.

