My question is:
$$\oint_{|z| = 1} \frac{sin(\frac{1}{z})}{z-\pi} \,dz$$
The question was that weather this integral exists or not ?
If we see it in terms of singularities,we observe that $\sin(1/z)$ has an essential singularity at $z = 0$. And $z = \pi$ lies outside the given region $|z| = 1$.
But If I replace $z\mapsto 1/z $ in the integral. We get $\oint_{|\frac{1}{z}| = 1}\frac{-sin(z)}{(1-\pi z)(z)}dz$.
Am I right in doing this ?
And can I write $|1/z| = 1 $ as $ |z| = 1$?
Thanks
Yes it is correct. You can compute this last integral with the Cauchy integral formula. With $g(z) = \frac{\sin(z)}{z}$ that is analytic on $|z|< 1+\epsilon$ you have $$\int_{|z|=1}\frac{\sin(z)}{(1-\pi z)z}dz = \frac{-1}{\pi}\int_{|z| = 1}\frac{g(z)}{z-1/\pi}dz = \frac{-1}{\pi}2i \pi g(1/\pi) = -2i\pi \sin(1/\pi)$$ The alternative method is computing $Res(\frac{\sin(1/z)}{z-\pi},0)$.
Expand $$\frac{1}{z-\pi}=\frac{-1/\pi}{1-z/\pi} = \frac{-1}{\pi}\sum_{k=0}^\infty \frac{z^k}{\pi^k}, \qquad \sin(1/z) = \sum_{n=0}^\infty (-1)^n z^{-2n-1}/(2n+1)!$$ Thus $\frac{\sin(1/z)}{z-\pi} =\frac{-1}{\pi}\sum_{k=0}^\infty z^k/\pi^k \sum_{n=0}^\infty (-1)^n z^{2n+1}/(2n+1)!$ and $$Res(\frac{\sin(1/z)}{z-\pi},0) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\frac{-1}{\pi}\frac{1}{\pi^{2n}} = -\sin(1/\pi)$$
And since $\frac{\sin(1/z)}{z-\pi}$ is analytic on $0 < |z| < 1+\epsilon$ : $$\int_{|z| =1} \frac{\sin(1/z)}{z-\pi}dz = 2i \pi \, Res(\frac{\sin(1/z)}{z-\pi},0) = -2i \pi \sin(1/\pi)$$