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How do I calculate this complex integral?

$$\displaystyle\oint_{|z|=1}\sin\left ({\displaystyle\frac{1}{z}}\right ) dz$$

I made the Taylor series for this: $$\displaystyle\sum_{n=0}^\infty \displaystyle\frac{(-1)^n*(1/z)^{2n+1}}{(2n+1)!}$$

But now I don't know what to do.

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    $\begingroup$ Do you know how to integrate any of the terms in that series on the circle? $\endgroup$ – Jonas Meyer Nov 1 '13 at 3:32
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You can either apply the Residue Theorem or use the power series of $\sin$. Using the former we have \begin{eqnarray} \int_{|z|=1}\sin\left(\frac{1}{z}\right)\,dz&=&\int_{|z|=1}\frac{1}{z}\,dz+\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)!}\int_{|z|=1}\frac{1}{z^{2n+1}}\,dz\\ &=&\int_{|z|=1}\frac{1}{z}\,dz=\int_0^{2\pi}\frac{ie^{it}}{e^{it}}\,dt=2i\pi. \end{eqnarray}

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Here is a trick. Let $z'=\frac{1}{z}$. Then $$dz=d(\frac{1}{z'})=\frac{-dz'}{z'^{2}}$$

So all you need is to integrate $$-\int_{|z'|=1}\frac{\sin[z']}{z'^{2}}dz'$$

But you can evaluate it by Cauchy's integral formula by noticing this is $\sin[z']'$ at $z'=0$ times some constant.

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    $\begingroup$ Mixing the two meanings of $\,'$ here is a little confusing. Using something like $w=1/z$ would be preferable. $\endgroup$ – Antonio Vargas Nov 1 '13 at 5:12

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