A residue formula
=================


I have strong evindence to believe that the following identity holds:
$$
\frac{n!}{2\pi i}\oint\_{|z-1|=\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}} = d^{-n-1}\prod\_{j=1}^{n}\left(a-jd\right),\quad a,d,n+1 \in \mathbb{Z}\_{\ge 1}.
$$



> 
> Have you seen this formula before? Any ideas on how to prove it?
> 
> 
> 


Any suggestions are welcome. I'll describe my partial success in the next section. I'm including the "combinatorics" tag because both of my methods got stuck in a combinatorial statement.


Verification for a finite sample and proof for small $n$
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I verified this for thousands of particular values of $(a,d,n)$ and proved it for all $(a,d)$ with small $n$.


To get a proof for small $n$ one can write out the power series expansion in order to find the coefficient of $(z-1)^{-1}$. However, this gets increasingly complicated for higher $n$ and I don't think a proof for all $n$ will emerge from this line of attack unless you are a brilliant combinatorialist.


In order to check this equality for particular values of $(a,d,n)$, it makes a huge difference in computation time to combine [Cauchy's integral formula](https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula) with [Faà di Bruno's formula](https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula) for expanding higher derivatives of compositions. Despite the advantage of quick evaluation, these formulas seem too unwieldy to get a proof for arbitrary $(a,d,n)$.



A natural residue formula
Let us consider the function $w=z^d$ around $z=1$. It is holomorphic and locally invertible with a holomorphic inverse. So we can think of $z$ as $w^{1/d}$ where the latter symbol means a branch of the $d$-th root function taking $1$ at $1$. With this notation, the integral can be written as
$$\frac{n!}{2\pi i}\oint\_{|z-1|=\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}}=\frac{n!}{2\pi i}\oint\frac{1}{d}\frac{w^{(a-d)/d} \mathrm{d}w}{(w-1)^{n+1}}.$$
Here, we understand $w^{(a-d)/d}$ as the $(a-d)$-th power of $z=w^{1/d}$, and the integration is on the closed curve $\{w=(1+\epsilon\cdot e^{i t})^d:\ 0\leq t\leq2\pi\}$ with winding number $1$ around $w=1$. Hence, by Cauchy's integral formula, the right hand side equals $1/d$ times the $n$-th derivative of $w^{(a-d)/d}$ at $w=1$:
$$\frac{n!}{2\pi i}\oint\_{|z-1|=\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}}=\frac{1}{d}\prod\_{j=1}^n\left(\frac{a-d}{d}+1-j\right)=d^{-n-1}\prod\_{j=1}^{n}\left(a-jd\right).$$


**P.S.** I have not seen this formula before.


