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My two questions are bolded below.

Hypothesis: Let $\gamma$ denote the circle about the origin of radius $2$.

Goal: Compute

$$ \int_{\gamma} z^n(1 - z)^m\ dz $$

Attempt:

  1. We have that $$ \int_{\gamma} z^n(1 - z)^m\ dz = \int_{\gamma} z^n(-1)^m(z-1)^m\ dz $$

  2. Take the integral of the inverse of the integrand. Once we figure out an answer to this question, we can inverse that answer to find the integral of our original integrand. Is this correct reasoning?

  3. Assuming this is correct reasoning, we have that

    $$ \int_\gamma {1 \over z^n(-1)^m(z-1)^m}\ dz = \int_\gamma {{1 \over z^n}(-1)^{(m-1)+1} \over (z-1)^{(m-1)+1}}\ dz = {2 \pi i \over n!} f^{(m-1)}(1) \text{ s.t. } f(z) = ?? $$

    Here is $f(z) = {(-1)^{m} \over z^n}$? I'm trying to make heavy use of Cauchy's integral formula but think I've computationally confused myself in that pursuit. How does one finish this computation?

user1770201
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    I don't think your reasoning in 2) is correct, if I understand you correctly. $\int_C{f(x)dx}= \left( \int_C{\frac{1}{f(x)}dx}\right)^{-1}$ is not true in most cases. – Hayden Apr 21 '14 at 00:05
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    Also, assuming $n$ and $m$ are positive integers, then the answer is simply zero by Cauchy's Theorem. Otherwise, use the Residue Theorem... – Hayden Apr 21 '14 at 00:06
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    The integrand is a polynomial, therefore it has antiderivative on the simply connected set $\mathbb C$. Since $\gamma$ is closed, the integral is zero. See this. – Git Gud Apr 21 '14 at 00:06
  • Hayden, is this the argument you had in mind: If $n$ and $m$ are positive integers, then we have that $z^n(1-z)^m \in \mathbb{C}[x] \implies z^n(1-z)^m$ is analytic $\implies$ $\int_{|z|=2} z^n(1-z)^m\ dz = 0$ via Cauchy's Theorem. – user1770201 Apr 21 '14 at 00:42

3 Answers3

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As others have said, if both $m$ and $n$ are non-negative integers, then the function $f(z)=z^n(1-z)^m$ is entire,and the integral is zero.

If $m\ge 0$ and $n<0$, then $$z^n(1-z)^m = \sum_{k=0}^m \binom{m}{k}(-1)^kz^{k+n}$$ so that $$\int_\gamma f(z) \, dz =2\pi i \, \text{res}_{z=0} f(z) = 2\pi i \, \binom{m}{-n-1}(-1)^{-n-1}$$

If $m<0$ and $n\ge 0$, then let $p=z-1$, and

$$z^n(1-m)^m = (-1)^m (p+1)^n p^m =(-1)^m \sum_{k=0}^n \binom{n}{k} p^{m+k}$$

and

$$\int_\gamma f(z)\, dz= \int_{\gamma^\star} f(p) \, dp =2\pi i \, \text{res}_{p=0} f(p) = -2\pi i \, \binom{n}{-m-1}$$

It can be shown that if both $m$ and $n$ are negative, then the residues at $z=1$ and at $z=0$ add to zero, so again the integral is zero.

mjw
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  • Best answer. Though the use of the binomial coefficient is a bit deceiving - I would prefer to use Gamma/Beta functions. – K.defaoite Jul 28 '21 at 02:41
  • I do like your suggestion! Your answer (or your expression) may very well be more elegant. Since we are dealing with a binomial, this seemed the most natural way of writing it. – mjw Jul 28 '21 at 02:47
  • @K.defaoite, is this how you would write it?: $\displaystyle \binom{m}{-n-1} = \frac{B(m+1,n+1)}{\Gamma(-n)\Gamma(n+1)}.$ If so, why is this less deceiving/more intuitive? – mjw Aug 02 '21 at 22:03
  • I tend to reserve combinatorial notation such as binomials and factorials for natural valued arguments. This is just personal preference. – K.defaoite Aug 03 '21 at 15:04
  • Here, $-n-1$ is non-negative, so it's a $\textit{natural}$ number. – mjw Aug 03 '21 at 16:01
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The integral is zero if $n,m\in\Bbb Z^+$, by Cauchy's theorem (you are integrating a holomorphic (analytic) function over a closed curve).

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If $n,m$ are positive integers the integrand is holomorphic on the whole $\mathbb C$ hence it has NO singularities and so NO residues. Then you can conclude by Residue Thm that the intergral is zero.

Joe
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