Computing $\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz$ using Cauchy integral formula

Question

Let $\alpha(t) = re^{it}$ where $|a|<r<|b|$ and $t \in [0,2\pi]$. I'd like to compute $$\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz \ \ \ \ n, m \in \mathbb{N}.$$ It appears that the answer is $$2\pi i (-1)^m{n + m -2 \choose n-1}\frac{1}{(b-a)^{n+m-1}}.$$ I try to compute it but I'm not sure how that's the final answer. Here is my attempt:

Let $f(z) = \frac{1}{(z-b)^m}$, then by Cauchy integral formula $$\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz = \int_{\alpha}\frac{f(z)}{(z-a)^n}dz = \frac{2\pi i}{(n-1)!}f^{(n-1)}(a).$$ Trying to find an expression of $f^{(n-1)}(z)$ \begin{align*} f^{(1)} &= -m\frac{1}{(z-b)^{m-1}}\\ f^{(2)} &= m(m-1)\frac{1}{(z-b)^{m-2}}\\ f^{(3)} &= -m(m-1)(m-2)\frac{1}{(z-b)^{m-3}}\\ &\vdots\\ f^{(n-1)} &= (-1)^{n-1}m(m-1)(m-2)\dotsc(m-n + 2)\frac{1}{(a-b)^{m-n+1}} \end{align*} I think this should be correct. So, \begin{align*} \int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz &= \frac{2\pi i}{(n-1)!}(-1)^{n-1}m(m-1)(m-2)\dotsc(m-n + 2)\frac{1}{(a-b)^{m-n+1}}\\ &= \frac{2\pi i}{(n-1)!}(-1)^{2n-2-m}m(m-1)(m-2)\dotsc(m-n + 2)\frac{1}{(b-a)^{m-n+1}}\\ &=\frac{2\pi i}{(n-1)!}(-1)^{m}m(m-1)(m-2)\dotsc(m-n + 2)\frac{1}{(b-a)^{m-n+1}}. \end{align*} But the exponent on $(b-a)$ is not the same. Further, I have no clue how I can get the binomial coefficient. Writing out the binomial coefficient in the given answer doesn't seem to help me either. $${n + m -2 \choose n-1} = \frac{(n+m-2)!}{(n-1)!(m-1)!} = \frac{(n+m-2)(n+m-3)\dotsc 2\cdot 1}{(n-1)!(m-1)!} $$

NOTE: I haven't learned residue theorem yet.

Answer 1

Answering my own question, thanks to the comments. Let $f(z) = \frac{1}{(z-b)^m}$, then by Cauchy integral formula $$\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz = \int_{\alpha}\frac{f(z)}{(z-a)^n}dz = \frac{2\pi i}{(n-1)!}f^{(n-1)}(a).$$ Trying to find an expression of $f^{(n-1)}(z)$ \begin{align*} f^{(1)} &= -m\frac{1}{(z-b)^{m+1}}\\ f^{(2)} &= m(m+1)\frac{1}{(z-b)^{m+2}}\\ f^{(3)} &= -m(m+1)(m+2)\frac{1}{(z-b)^{m+3}}\\ &\vdots\\ f^{(n-1)} &= (-1)^{n-1}m(m+1)(m+2)\dotsc(m+n - 2)\frac{1}{(a-b)^{m+n-1}} \end{align*} So, \begin{align*} \int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz &= \frac{2\pi i}{(n-1)!}(-1)^{n-1}m(m+1)(m+2)\dotsc(m+n - 2)\frac{1}{(a-b)^{m+n-1}}\\ &= \frac{2\pi i}{(n-1)!}(-1)^{n-1-m-n+1}m(m+1)(m+2)\dotsc(m+n - 2)\frac{1}{(b-a)^{m+n-1}}\\ &=\frac{2\pi i}{(n-1)!}(-1)^{-m}m(m+1)(m+2)\dotsc(m+n - 2)\frac{1}{(b-a)^{m+n-1}}\\ &=\frac{2\pi i}{(n-1)!}(-1)^{m}m(m+1)(m+2)\dotsc(m+n - 2)\frac{1}{(b-a)^{m+n-1}}. \end{align*} Note that $${n + m -2 \choose n-1} = \frac{(n+m-2)(n+m-3)\dotsc m}{(n-1)!}.$$ Hence, $$\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz = 2\pi i(-1)^m{n + m -2 \choose n-1}\frac{1}{(b-a)^{m+n-1}}.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$

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\begin{align} &\bbox[5px,#ffd]{\left. \oint_{\verts{z}\ =\ r}\,\, {\dd z \over \pars{z - a}^{\,n}\pars{z - b}^{\,m}} \,\right\vert_{\substack{n, m\ \in\ \mathbb{N} \\[1mm] \verts{a}\ <\ r\ <\ \verts{b}}}} \\[5mm] = &\ 2\pi\ic\on{Res}\bracks{{1 \over \pars{z - a}^{\,n} \pars{z - b}^{\,m}}, z = a}\label{1}\tag{1} \end{align}

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\begin{align} \mbox{Note that}\ & {1 \over \pars{z - b}^{m}} = {1 \over \bracks{a - b + \pars{z - a}}^{m}} \\[5mm] & = {1 \over \pars{a - b}^{m}} \bracks{1 + {z - a \over a - b}}^{-m} \\[5mm] & = {1 \over \pars{a - b}^{m}} \sum_{k = 0}^{\infty}{-m \choose k} \pars{z - a \over a - b}^{k} \end{align} (\ref{1}) becomes \begin{align} &\bbox[5px,#ffd]{\left. \oint_{\verts{z}\ =\ r}\,\, {\dd z \over \pars{z - a}^{\,n}\pars{z - b}^{\,m}} \,\right\vert_{\substack{n, m\ \in\ \mathbb{N} \\[1mm] \verts{a}\ <\ r\ <\ \verts{b}}}} \\[5mm] = &\ 2\pi\ic{1 \over \pars{a - b}^{m}} \bracks{{-m \choose n - 1} \pars{1 \over a - b}^{n - 1}} \\[5mm] = &\ 2\pi\ic{1 \over \pars{a - b}^{m + n - 1}} {m + \bracks{n - 1} - 1 \choose n - 1}\pars{-1}^{n - 1} \\[5mm] = &\ \bbx{2\pi\ic\,\pars{-1}^{m} {n + m - 2 \choose n - 1} {1 \over \pars{b - a}^{n + m - 1}}} \\ & \end{align}