I have to calculate this integral: $$ \int_{|z|=2} \frac{dz}{z^3(z^{10}-2)}. $$ It is clear that we can make the calculation using the residue theorem, but there are 13 residues inside the curve and it seems unnecessary to do it.
The exercise says that it can be useful to make the following variable change $\zeta = 1/z$, but I have never used variable changes for calculating integrals along curves.
How can I use this variable change to calculate the integral? Thank you very much.
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Let $\zeta=1/z$, then $z=1/\zeta$, so "$dz=-d\zeta/\zeta^2$". Your integral is $$\int_{C}-\frac{d\zeta}{\zeta^{-1}(\zeta^{-10}-2)}$$ where $C$ is the circle centre $0$ and radius $1/2$ but taken in the negative (clockwise) direction. Changing orientation, this equals $$\int_{|\zeta|=1/2}\frac{d\zeta}{\zeta^{-1}(\zeta^{-10}-2)} =\int_{|\zeta|=1/2}\frac{\zeta^{11}d\zeta}{1-2\zeta^{10}}. $$ One can now apply Cauchy's theorem...
Since $$ \int_{|z|=2} \frac{dz}{z^3(z^{10}-2)} = \int_{|z|=R} \frac{dz}{z^3(z^{10}-2)} $$ for all $R > 2$ (why?), you may as well compute the limit as $R \to \infty$, and this can be done by estimating the size of the integrand.