Show the value of a complex integral is independent of R for R > 1 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.
 A: This seems to be a straight forward application of Cauchy's Residue Theorem. It tells us that the integral of a function around a contour is $2\pi i$ times the sum of the residues at the poles inside the contour of the function. 
Your function has a pole when $2z^{2012}-1=0$. There are 2,012 simple poles and they all lie on the circle $|z| = \sqrt[2012]{1/2} \approx 0.9997.$ Any circular contour $|z| = R$ will contain all of these poles in its interior provided $R > \sqrt[2012]{1/2}$. It follows that the integral around any circular contour $|z| = R$ will be the same provided $R > \sqrt[2012]{1/2}$. In your case $|z|=1$ and $|z|=R$ with $R>1$ are two such contours. 
(If $R < \sqrt[2012]{1/2}$ then the integral will be zero because the integrand is holomorphic on and inside the contour $|z| = R$.)
A: Hint: Draw a segment between the 2 circles, and perform 2 integrals over the given paths.
The evaluation with $\log$ is also correct. But for that you should now that $\log$ is not unique (since $\exp$ is periodic with $2\pi i$ period!), but unique up to $+2k\pi i$ for $k\in\mathbb Z$. If the origo (the pole) is surrounded only once, you get 
$$\int_{|z|=e^r}\frac 1z = [\log z]_{z=e^r}^{e^{r+2\pi i}} = (r+2\pi i)-r = 2\pi i$$
A: This is not an answer to your original question and I think that has been already taken care of. But there is an alternative approach to evaluating the integral instead of computing all the residues. 
Note that when $z \in \{|z|=R\}$ with $R>>1$, we can write the integrand as
$\dfrac{z^{2011}}{2z^{2012}}(1-\dfrac{1}{2z^{2012}})^{-1} = \dfrac{1}{2z}\sum_{n=0}^{\infty}(\dfrac{1}{2z^{2012}})^n$.
When you integrate this function over the unit circle only the term involving $\frac{1}{z}$ will produce a non-zero residue and which will happen only for $n=0$ in the power series above. Hence the integral will be equal to $\pi i$.
