Question: Show that for R > 1 $\int_{|z|=1} \frac{z^{2011}}{2z^{2012}-1} dz = \int_{|z|=R} \frac{z^{2011}}{2z^{2012}-1} dz$
Thoughts thus far: (i) I know that we cannot use Cauchy's integral formula because neither is f analytic (because there are singularities within the unit circle) nor would the reduced form fit the form of Cuachy's integral formula (i.e. z0 would not be fixed). (ii) The above integral may be evaluated to ln(2z^2012-1)/4024 using u substitution where u=2z^2012, but I am not certain that this evaluation is even correct. (iii) I have also considered reducing it to polar form and subtracting the two integrals to show that the difference is zero, but I get stuck with $\int_{|z|=1} \frac{z'(\theta)z(\theta)^{2011}}{2z(\theta)^{2012}-1}dz - \int_{|z|=R} \frac{z'(\theta)z(\theta)^{2011}}{2z(\theta)^{2012}-1} d\theta$ $\int_{unit\space circle} \frac{i\theta e^{i\theta}e^{2011i\theta}}{2e^{2012i\theta}-1} - \frac{i\theta Re^{i\theta}R^{2011}e^{2011i\theta}}{2R^{2012}e^{2012i\theta}-1}\ d\theta$ I am unable to deal with the R terms, which do not appear to disappear easily. I have also considered multiplying by the conjugate of z, but again we get stuck with R terms on the RHS. (iv) I am led to believe that, rather than simple ignorance of calculating the above integral, there is something conceptual or extremely fundamental that I am missing that allows us to disregard the value of R (for R > 1).
Any help would be greatly appreciated. Thank you in advance.