An elegant Multi-Integral that implies an even more elegant determinant identity of Dougherty and McCammond

Tewodros AMDEBERHAN and Doron ZEILBERGER

The first purpose of this note is to prove the following elegant identity.

Theorem A: Let z_{1}, … z_{n} be commuting indeterminates, let n be a positive integer, and let a_{1}, …, a_{n}, and b be non-negative integers. Let ā := ∑_{i=1}^{⌈n/2⌉} a_{2i−1}. Then

∫_{0}^{z_{1}} … ∫_{0}^{z_{n}} ∏_{i=1}^{n} x_{i}^{b} ∏_{1≤j,k≤n}(x_{j} − z_{k})^{a_{k}} ∏_{1≤i<j≤n}(x_{j} − x_{i}) dx_{n} … dx_{1}

= (−1)ā ∏_{1≤i<j≤n}(z_{j} − z_{i})^{a_{i}+a_{j}+1} · ∏_{i=1}^{n} z_{i}^{a_{i}+b+1} · \frac{b! ∏_{i=1}^{n} a_{i}!}{(n + b + ∑_{i=1}^{n} a_{i})!}.  (1)

The second purpose is to deduce from it (and thereby give a shorter proof) of the following even more elegant identity, discovered, and first proved in [1].

Theorem B (Dougherty and McCammond): Let

p(Z) := ∫_{0}^{Z} ∏_{i=1}^{n}(w − z_{i})^{a_{i}} dw,

and let J(z_{1}, …, z_{n}) be the n × n matrix whose (i, j) entry is J(z_{1}, …, z_{n})_{i,j} := \frac{∂}{∂z_{i}} p(z_{j}), then

det J(z_{1}, …, z_{n}) = \frac{∏_{i=1}^{n} a_{i}!}{(∑_{i=1}^{n} a_{i})!} · ∏_{i=1}^{n}(−z_{i})^{a_{i}} · ∏_{{1≤i,j≤n \atop i≠j}}(z_{i} − z_{j})^{a_{j}}.  (2)

Proof that A ⇒ B: Let’s rewrite the determinant and apply Cauchy’s alternant formula so that

det(J) = det (−a_{i} ∫_{0}^{z_{j}} ∏_{k=1}^{n}(w − z_{k})^{a_{k}} \frac{dw}{w − z_{i}} ) = det (−a_{i} ∫_{0}^{z_{j}} ∏_{k=1}^{n}(x_{j} − z_{k})^{a_{k}} \frac{dx_{j}}{x_{j} − z_{i}} )

= ∏_{i=1}^{n}(−a_{i}) ∫_{0}^{z_{1}} … ∫_{0}^{z_{n}} ∏_{1≤j,k≤n}(x_{j} − z_{k})^{a_{k}} · det (\frac{1}{x_{j} − z_{i}} )_{i,j}^{1,n} dx_{n} ⋯ dx_{1}

= ∏_{i=1}^{n}(−a_{i}) ∫_{0}^{z_{1}} … ∫_{0}^{z_{n}} ∏_{1≤j,k≤n}(x_{j} − z_{k})^{a_{k}} · [\frac{∏_{1≤i<j≤n}(z_{i} − z_{j})(x_{j} − x_{i})}{∏_{1≤i,j≤n}(x_{j} − z_{i})}] dx_{n} ⋯ dx_{1}

= ∏_{i=1}^{n}(−a_{i})· ∏_{1≤i<j≤n}(z_{i} −z_{j})·∫_{0}^{z_{1}} … ∫_{0}^{z_{n}} ∏_{1≤j,k≤n}(x_{j} −z_{k})^{a_{k}−1} · ∏_{1≤i<j≤n}(x_{j} −x_{i}) dx_{n} ⋯ dx_{1}.

But by Theorem A, with b = 0 and (a_{1}, …, a_{n}) replaced by (a_{1} − 1, … a_{n} − 1), this equals

= (−1)^{ā−⌈n/2⌉} ∏_{i=1}^{n}(−a_{i}) ∏_{1≤i<j≤n}(z_{i} − z_{j}) ∏_{1≤i<j≤n}(z_{j} − z_{i})^{a_{i}+a_{j}−1} ∏_{i=1}^{n} z_{i}^{a_{i}} \frac{∏_{i=1}^{n}(a_{i} − 1)!}{(∑_{i=1}^{n} a_{i})!}

= \frac{∏_{i=1}^{n} a_{i}!}{(∑_{i=1}^{n} a_{i})!} ∏_{i=1}^{n}(−z_{i})^{a_{i}} ∏_{{1≤i,j≤n \atop i≠j}}(z_{i} − z_{j})^{a_{j}}.

Proof of Theorem A: The proof is by induction on n and b. When n = 1 and b = 0 this is saying that ∫_{0}^{z_{1}}(x_{1} − z_{1})^{a_{1}} dx_{1} = (−1)^{a_{1}} \frac{(z_{1})^{a_{1}+1}}{a_{1}+1} , whose proof is left to the reader’s five-year-old.

Let’s denote the statement of theorem A by A(n, b).

Proof that A(n, b) ⇒ A(n, b + 1)

We claim that both sides of Eq. (1), let’s call them L(a_{1}, …, a_{n}; b) and R(a_{1}, …, a_{n}; b) respectively, satisfy the recurrence

X(a_{1}, …, a_{n}; b+1) = ∑_{i=1}^{n} (∏_{{j=1 \atop j≠i}}^{n} \frac{z_{j}}{z_{j} − z_{i}} )·X(a_{1}, ….a_{i}+1, …, a_{n}; b)+(∏_{j=1}^{n} z_{j})·X(a_{1}, …, a_{n}; b).

(3)

In other words, if you replace X by either L or R you get a true statement. Regarding the left side of (1), in fact, this identity is already true if you replace X by the integrand of the left side of (1), since there are no x_{i}’s in sight, it is still true when you integrate with respect to x_{1}, …, x_{n}. We leave both checks as pleasant exercises for the reader.

Proof that A(n − 1, b) for all b implies A(n, 0)

Fix a_{1}, …, a_{n}. Let V (x_{1}, …, x_{n}) := ∏_{1≤i<j≤n}(x_{j} − x_{i}) and

F (x_{1}, … x_{n}; z_{1}, …, z_{n}) := ∏_{1≤j,k≤n}(x_{j} − z_{k})^{a_{k}}.

We claim (check!) that

(n + ∑_{i=1}^{n} a_{i}) F (x_{1}, … x_{n}; z_{1}, …, z_{n})V (x_{1}, …, x_{n})

= ∑_{i=1}^{n}(−1)^{i}V (x_{1}, …, x_{i−1}, x_{i+1}, …, x_{n}) · \frac{∂}{∂x_{i}} [∏_{j=1}^{n}(x_{i} − z_{j}) · F (x_{1}, …, x_{n}; z_{1}, …, z_{n})] .  (4)

Applying ∫_{0}^{z_{1}} … ∫_{0}^{z_{n}} dx_{n} ⋯ dx_{1}, we get

(n + ∑_{i=1}^{n} a_{i}) ∫_{0}^{z_{1}} … ∫_{0}^{z_{n}} F (x_{1}, … x_{n}; z_{1}, …, z_{n})V (x_{1}, …, x_{n})dx_{n} ⋯ dx_{1}

= ∑_{i=1}^{n}(−1)^{i} ∫_{0}^{z_{1}} … ∫_{0}^{z_{n}} V (x_{1}, …, x_{i−1}, x_{i+1}, …, x_{n})·\frac{∂}{∂x_{i}} [∏_{j=1}^{n}(x_{i} − z_{j}) · F (x_{1}, …, x_{n}; z_{1}, …, z_{n})] dx_{n} ⋯ dx_{1}

= ∑_{i=1}^{n}(−1)^{i} ∫_{0}^{z_{1}} … ∫_{0}^{z_{i−1}} ∫_{0}^{z_{i+1}} … ∫_{0}^{z_{n}} V (x_{1}, …, x_{i−1}, x_{i+1}, …, x_{n})dx_{n} ⋯ dx_{i+1}dx_{i−1} ⋯ dx_{1}

× ∫_{0}^{z_{i}} \frac{∂}{∂x_{i}} [∏_{j=1}^{n}(x_{i} − z_{j}) · F (x_{1}, …, x_{n}; z_{1}, …, z_{n})] dx_{i}.

By the Fundamental Theorem of Calculus, we have

∫_{0}^{z_{i}} \frac{∂}{∂x_{i}} [∏_{j=1}^{n}(x_{i} − z_{j}) · F (x_{1}, …, x_{n}; z_{1}, …, z_{n})] dx_{i} = ∏_{j=1}^{n}(x_{i} −z_{j})·F (x_{1}, …, x_{n}; z_{1}, …, z_{n})|_{x_{i}=0}^{x_{i}=z_{i}}

= ∏_{j=1}^{n}(x_{i} − z_{j})^{a_{j}+1} ∏_{j=1}^{n} ∏_{{1≤i′≤n \atop i′≠i}}(x′_{i} − z_{j})^{a_{j}} |_{x_{i}=0}^{x_{i}=z_{i}} = − ∏_{j=1}^{n}(−z_{j})^{a_{j}+1} ∏_{j=1}^{n} ∏_{{1≤i′≤n \atop i′≠i}}(x′_{i} − z_{j})^{a_{j}}.

Going back we have that the left side of Eq. (1), when b = 0, is

\frac{−1}{n + ∑_{i=1}^{n} a_{i}} · ∏_{j=1}^{n}(−z_{j})^{a_{j}+1} · ∑_{i=1}^{n}(−1)^{i} ∫_{0}^{z_{1}} … ∫_{0}^{z_{i−1}} ∫_{0}^{z_{i+1}} … ∫_{0}^{z_{n}} V (x_{1}, …, x_{i−1}, x_{i+1}, …, x_{n})

× ∏_{j=1}^{n} ∏_{{1≤i′≤n \atop i′≠i}}(x′_{i} − z_{j})^{a_{j}}dx_{n} ⋯ dx_{i+1}dx_{i−1} ⋯ dx_{1}

= \frac{−1}{n + ∑_{i=1}^{n} a_{i}} · ∏_{j=1}^{n}(−z_{j})^{a_{j}+1} · ∑_{i=1}^{n}(−1)^{i} ∫_{0}^{z_{1}} … ∫_{0}^{z_{i−1}} ∫_{0}^{z_{i+1}} … ∫_{0}^{z_{n}} V (y_{1}, …, y_{n−1})

× ∏_{j=1}^{n} ∏_{i=1}^{n−1}(y_{i} − z_{j})^{a_{j}}dy_{n−1} ⋯ dy_{1}.

We now claim that

∑_{i=1}^{n}(−1)^{i−1} ∫_{0}^{z_{1}} … ∫_{0}^{z_{i−1}} ∫_{0}^{z_{i+1}} … ∫_{0}^{z_{n}} V (y_{1}, …, y_{n−1}) · ∏_{j=1}^{n} ∏_{i=1}^{n−1}(y_{i} − z_{j})^{a_{j}}dy_{n−1} ⋯ dy_{1}

= ∫_{z_{1}}^{z_{2}} … ∫_{z_{1}}^{z_{n}} V (y_{1}, …, y_{n−1}) · ∏_{j=1}^{n} ∏_{i=1}^{n−1}(y_{i} − z_{j})^{a_{j}}dy_{n−1} ⋯ dy_{1}.  (5)

In order to prove this, notice that each of the integrands on the left, and the integrand on the right, are anti-symmetric in their arguments. Hence, for any given permutation of the integration variables, the effect is to multiple it by the sign of that permutation. Calling the common integrand f(y_{1}, …, y_{n−1}) and denoting A_{n}(i) = P er(1, …, i − 1, i + 1, … n), we claim that

∑_{i=1}^{n}(−1)^{i−1} ∑_{π∈A_{n}(i)} sgn(π) ∫_{0}^{z_{π(1)}} … ∫_{0}^{z_{π(i−1)}} ∫_{0}^{z_{π(i+1)}} ⋯ ∫_{0}^{z_{π(n)}} f(y_{1}, …, y_{n−1})dy_{n−1} ⋯ dy_{1}

= ∑_{π∈A_{n}(1)} sgn(π) ∫_{z_{1}}^{z_{π(2)}} … ∫_{z_{1}}^{z_{π(n)}} f(y_{1}, …, y_{n−1})dy_{n−1} ⋯ dy_{1}.  (6)

Since both sides of Eq. (6) are (n − 1)! times the respective sides of Eq. (5), if we can prove (6), then (5) would follow.

But surprise!, Eq. (6) is valid for any integrand! It is just a relation between regions in R^{n−1} that is equivalent to an easy symmetric function identity, that we also leave as a pleasant exercise to the reader. Now make the change of variables (y_{1}, …, y_{n−1}) → (y_{1} − z_{1}, …, y_{n−1} − z_{1}), thereby making it a case of A(n − 1, b) with b = a_{1}; and a_{1}, …, a_{n−1} replaced by a_{2}, …, a_{n}, respectively; and z_{1}, …, z_{n−1} replaced by z_{2} − z_{1}, …, z_{n} − z_{1}, respectively. Plugging it in and simplifying, completes the induction.

Comment: Readers that prefer not do the ‘exercises’ can convince themselves of all the claims, empirically, by playing with the Maple package CritVal.txt available from the front of this paper

https://sites.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/crit.html .

Reference

[1] M. Dougherty, J. M. McCammond, Critical Points, critical values, and a determinant identity for complex polynomials, Proc. Amer. Math. Soc. 148 (2020), 5722-5289.

https://arxiv.org/abs/1908.10477 .

Tewodros Amdeberhan, Department of Mathematics, Tulane University, New Orleans, LA 70118

Email: tamdeber@tulane.edu

Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway, NJ 08854-8019, USA.

Email: DoronZeil@gmail.com

Jan. 7, 2022