Contour integral $\int_{|z|=1}\exp(1/z)\sin(1/z)dz$ Evaluate the contour integral $$\int_{|z|=1}\exp(1/z)\sin(1/z)\,dz$$ along the circle $|z|=1$ counterclockwise once.
The singularities are $\dfrac1{\pi k},k\in\mathbb{Z}$ plus the limit point $0$. So I can't apply the residue theorem. Any other alternative?
 A: Note that $\vert z \vert = 1 \implies \vert 1/z \vert = 1$. Hence, setting $w=1/z$, we get that
\begin{align}
\int_{\vert z \vert = 1} \exp(1/z) \sin(1/z) dz & = \overbrace{\underbrace{\int_{\vert w \vert = 1} \exp(w) \sin(w) \dfrac{dw}{w^2}}_{\text{by change in orientation of integral}}}^{\text{Negative sign gets cancelled}}\\
& = \int_{\vert w \vert = 1} \dfrac{\left(1+ \mathcal{O}(w)\right)(1+\mathcal{O}(w^2))}w dw = 2 \pi i
\end{align}
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\color{#66f}{\large\int_{\verts{z}\ =\ 1}\exp(1/z)\sin(1/z)\,\dd z}
&=\int_{\verts{z}\ =\ 1}{1 \over 2\ic}\bracks{%
\exp\pars{1 + \ic \over z}- \exp\pars{1 - \ic \over z}}\,\dd z
\\[3mm]&=2\pi\ic\,{\pars{1 + \ic} - \pars{1 - \ic} \over 2\ic}
=\color{#66f}{\Large 2\pi\ic}
\end{align}
