How to calculate this integral using the residue theorem? $\displaystyle \frac{1}{2\pi i}\int \limits _\gamma \sin ^2\frac{1}{\xi}\,d\xi$
$\gamma (t)=Re^{it}$,
$R>0$,
$0\leq t\leq 2\pi$
Could you help me solve this equation? I did not understand this topic very well, I managed to get the result but using a different method.
The value of the integral is 0, what is asked of me is to justify why it is 0.
Thank you in advance.
 A: Since you want to use Residue Theorem, why not try and calculate the residue at $z=0$ (coefficient of $\frac{1}{z}$ in the Laurent Series expansion)?
Notice that the function has only one singularity at $z=0$ and is holomorphic inside the punctured disc $0<|z|\leq R \,,R>0$ . That is it is holomorphic inside the disc at all points but at $z=0$. Hence by Cauchy's Residue Theorem we have $\displaystyle\int_{\gamma}\sin^{2}(\frac{1}{z})\,dz =2\pi i\cdot\text{Res}_{z=0}$
$\sin^{2}(\frac{1}{z})$.
But $\sin^{2}(\frac{1}{z})=\frac{1}{2}\cdot\bigg(1-\cos(\frac{2}{z})\bigg)$
The Laurent Series expansion is then $\frac{1}{2}(1-(1-\frac{1}{2!}(\frac{2}{z})^{2}+...))$ (By using the Taylor expansion for $\cos(z)$) which should directly tell you that the residue at $z=0$ (coefficient of $\frac{1}{z}$ in the expression above) is $0$.
Hence the integral $\displaystyle\int_{\gamma}\sin^{2}(\frac{1}{z})\,dz =2\pi i\cdot\text{Res}_{z=0}= 0 $ by the Residue Theorem.
Note:- $\displaystyle\cos(z)=1-\frac{z^{2}}{2!}+\frac{z^{4}}{4!}-...=\sum_{r=0}^{\infty}\frac{(-1)^{r}z^{2r}}{(2r)!}$
