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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 $n$

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 with Faà di Bruno's formula for expanding higher derivatives of compositions. Despite the advantage of quick evaluation, these formulas seem too unwieldy to get a proof for arbitrary $(a,d,n)$.

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    $\begingroup$ Do you mean $|z-1|=\epsilon$? $\endgroup$
    – Brendan McKay
    Commented Feb 5, 2018 at 12:32
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    $\begingroup$ Is it not just a matter of changing variables $w=z^d$ and using straightforward Cauchy? $\endgroup$
    – thedude
    Commented Feb 5, 2018 at 12:49
  • $\begingroup$ There is the following formula $$\sum_{n=-\infty}^\infty \left({a\atop b+\alpha n}\right)v^{b+\alpha n}=\frac{1}{\alpha}(1+v)^a,\quad |v|=1,~ |\arg v|<\pi,~ 0<\alpha\le 1$$ proved in the article T.J. Osler, Taylor's series generalized for fractional derivatives and applications, SIAM J. Math. Anal. 2, 37-48 (1971) that formally can be written as a fractional residue formula. $\endgroup$
    – Nemo
    Commented Feb 8, 2018 at 17:21

1 Answer 1

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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.

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    $\begingroup$ You answered as I was writing my comment $\endgroup$
    – thedude
    Commented Feb 5, 2018 at 12:51
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    $\begingroup$ @thedude: Yes, I noticed this. So we were thinking the same! $\endgroup$
    – GH from MO
    Commented Feb 5, 2018 at 12:51
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    $\begingroup$ Thank you for your answer. It saved me from a tremendous amount of effort, as the power series expansions were just terrible. $\endgroup$
    – Emre
    Commented Feb 5, 2018 at 13:42

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