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I was reading this Wikipedia article’s example of analytic continuation using the function: $$f(z)=\sum_{k=0}^{\infty}(-1)^k(z-1)^k$$ I was able to understand the process until this step: $$\sum_{n=0}^{\infty}(-1)^n\int_{0}^{2\pi}\frac{\sum_{m=0}^{n}{n \choose m}(a-1)^{n-m}(re^{i\theta})^m}{(re^{i\theta})^k}d\theta=(-1)^ka^{-k-1}$$

Can someone explain how these two expressions are equivalent? Note that k will always be an integer. I believe this is all the needed information to explain this equivalence but if it is not the rest should be in the above Wikipedia article. Thanks in advance.

Snacc
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Let $p \ne 0$. Note that

$$ \int_0^{2\pi} (e^{i\theta})^p d\theta= \int_0^{2\pi} e^{pi\theta}d\theta = \frac{e^{pi\theta}}{pi}|_{\theta = 0}^{2\pi} = 0. $$

So in the integral you're having trouble with, the only summand the value of which is non-zero is precisely when the numerator $(re^{i\theta})^n$ cancels out with the denominator $(re^{i\theta})^k$. You should be able to work out the rest.

xyzzyz
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  • I was able to get to this with your method: $\sum_{n=k}^{\infty}(-1)^n{n \choose k}(a-1)^{n-k}$ but I cannot figure out what to do next. How would you show that this power series equals $(-1)^ka^{-k-1}$ – Snacc Oct 16 '19 at 00:17
  • There is an answer to the question here. – jvkloc Nov 18 '22 at 09:53