The value of $I=\int_{|z|=2}z^2\sin\frac{1}{z}dz$?

Question

I want to get the value of $I=\int_{|z|=2}z^2\sin\frac{1}{z}dz$.

My idea is to let $t=\frac{1}{z}$, then $$I=\int_{|z|=2}z^2\sin\frac{1}{z}dz=\int_{|t|=\frac{1}{2}}\frac{\sin t}{t^4}dt=2\pi i \times-\frac{1}{6}=-\frac{\pi i}{3}$$ Is this right?

Answer 1

Your result has the wrong sign because you did not take into account that the substitution $t=-\frac{1}{z}$ reverts the orientation of the circle.

You could also apply the residue theorem directly to $$ z^2 \sin \frac 1z = z^2 \left( \frac 1z - \frac{1}{3!z^3} + \ldots \right) = z -\frac{1}{6z} + \ldots $$ so that $$\operatorname{Res}(z^2 \sin \frac 1z, 0) = -\frac 16$$ and therefore $$ I = 2 \pi i \operatorname{Res}(z^2 \sin \frac 1z, 0) = - \frac{\pi i}{3} \,. $$