$\displaystyle \frac{1}{2\pi i}\int \limits _\gamma \sin ^2\frac{1}{\xi}\,d\xi$
$\gamma (t)=Re^{it}$,
$R>0$,
$0\leq t\leq 2\pi$
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.
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:- $\displaystyle\cos(z)=1-\frac{z^{2}}{2!}+\frac{z^{4}}{4!}-...=\sum_{r=0}^{\infty}\frac{(-1)^{r}z^{2r}}{(2r)!}$