Palestine Journal of Mathematics

Vol. 11 (Special Issue )(2022) , 1–6

© Palestine Polytechnic University-PPU 2022

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

Tewodros Amdeberhan and Doron Zeilberger

Communicated by Ayman Badawi

MSC 2010 Classifications: Primary 05A15, 05A05; Secondary 30C15.

Keywords and phrases: Cauchy’s alternant formula, Vandermonde and Jacobian matrices and determinants, critical values, multi-integrals, anti-symmetric functions.

Abstract We state and prove an explicit evaluation of a certain multi-variate integral and use it to furnish a new, and shorter, proof of an elegant determinant identity of Michael Dougherty and Jon McCammond that came up in their study of critical values of a complex polynomial.

1 Introduction

Beardon-Carne-Ng [1] investigated invertibility of the Jacobian of the polynomials

p(z) = ∫_{0}^{z}(w − z_{1}) ⋯ (w − z_{n}) dw

given by the n-by-n matrix

J J J(z) = (\frac{∂}{∂z_{i}} p(z_{j}))_{i,j}^{1,n} .

Using Topological arguments and techniques from Several Complex variables, the authors [1] show that every n-tuple of complex numbers arises as the critical values of some polynomial by proving that the determinant det(J J J) ≠ 0, as along as z_{1}, …, z_{n} are non-zero and distinct. More recently, Dougherty and McCammond [2] computed det(J J J), explicitly, and reproved the above-mentioned result of [1] with a different method. Some nomenclature adopted in the sequel: bold-face lower case letters are reserved for vectors, such as zzz = (z_{1}, …, z_{n}) and ẑ ẑ ẑ^{i} = (z_{1}, …, z_{i−1}, z_{i+1} , …, z_{n}). Given a a a = (a_{1}, …, a_{n}) ∈ ℤ_{≥0}^{n}, denote ā ā ā = ∑_{i=1}^{⌈n/2⌉} a_{2i−1} where ⌈·⌉ is the ceiling function while ⌊·⌋ is the floor function. Let V_{n}(x x x) = ∏_{1≤i<j≤n}(x_{j} − x_{i}) be the Vandermonde determinant of the Vandermonde matrix M M M = (x_{j}^{i−1})_{i,j}^{1,n}. We write dx dx dx = dx_{1} ⋯ dx_{n}, the dimension n being suppressed when it is clear from the context.

2 The Main Theorem

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

Theorem 2.1. 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. Then

∫_{0}^{z_{n}} … ∫_{0}^{z_{1}} ∏_{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_{1} ⋯ dx_{n}

= (−1)^{ā ā 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)

2  Tewodros Amdeberhan and Doron Zeilberger

3 An Example

We offer an illustrative example for Theorem 2.1. Choose n = 2, b = 2022, a_{1} = 1, a_{2} = 2. Then Eq. (1) reads

∫_{0}^{z_{2}} ∫_{0}^{z_{1}}(x_{1}x_{2})^{2022}(x_{1}−z_{1})(x_{2}−z_{1})(x_{1}−z_{2})^{2}(x_{2}−z_{2})^{2}(x_{2}−x_{1}) dx_{1}dx_{2} = \frac{− (z_{2} − z_{1})^{4}z_{1}^{2024}z_{2}^{2025}}{17025293698597800}.

4 An interesting consequence

The second purpose our work here is to deduce from Theorem 2.1 (and thereby give a shorter proof) of the following even more elegant identity, discovered, and first proved in [2]. Theorem 4.1. (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

detJ(z_{1}, …, z_{n}) = \frac{∏_{i=1}^{n} a_{i}!}{(∑_{i=1}^{n} a_{i})!} · ∏_{i=1}^{n}(−z_{j})^{a_{j}} · ∏_{\begin{matrix}1≤i,j≤n \\ i≠j \end{matrix}}(z_{i} − z_{j})^{a_{j}}.  (2)

Proof that Theorem 2.1 ⟹ Theorem 4.1: 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_{n}} … ∫_{0}^{z_{1}} ∏_{1≤j,k≤n}(x_{j} − z_{k})^{a_{k}} · det(\frac{1}{x_{j} − z_{i}} )_{i,j}^{1,n} dx dx dx

= ∏_{i=1}^{n}(−a_{i}) ∫_{0}^{z_{n}} … ∫_{0}^{z_{1}} ∏_{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 dx dx

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

But by Theorem 2.1, with b = 0 and (a_{1}, …, a_{n}) replaced by (a_{1} − 1, … a_{n} − 1), this equals

det(J) = (−1)^{ā ā 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_{j})^{a_{j}} ∏_{\begin{matrix}1≤i,j≤n \\ i≠j \end{matrix}}(z_{i} − z_{j})^{a_{j}}. □

5 Proof of Theorem 2.1:

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

Let’s denote the statement of Theorem 2.1 by A(n, b).

Proof that A(n, b) ⟹ A(n, b + 1):

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

X(a a a; b + 1) = ∑_{i=1}^{n} (∏_{\begin{matrix}j=1 \\ j≠i \end{matrix}}^{n} \frac{z_{j}}{z_{j} − z_{i}} ) · X(a_{1}, …, 1 + a_{i}, …, a_{n}; b) + (∏_{j=1}^{n}z_{j}) · X(a a a; b).  (3)

In other words, if you replace X by either L or R you get a true statement. Regarding the lefthand 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 curious reader. □

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

Fix a_{1}, …, a_{n}. Notations: a a a = (a_{1}, …, a_{n}),zzz = (z_{1}, …, z_{n}), V_{n}(x x x) = ∏_{1≤i<j≤n}(x_{j} − x_{i}) and x̂ x̂ x̂^{i} = (x_{1}, …, x_{i−1}, x_{i+1} , …, x_{n}). Define the multi-variable polynomial

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

We claim (check! this reduces to the Laplace expansion V_{n}(x x x) = ∑_{i=1}^{n} (−1)^{i}V_{n−1}(x̂ x̂ x̂^{i})x_{i}^{n−1}) that

(n + ∑_{i=1}^{n}a_{i}) F (x x x,zzz)V_{n}(x x x) = ∑_{i=1}^{n}(−1)^{i}V_{n−1}(x̂ x̂ x̂^{i}) · \frac{∂}{∂x_{i}} [∏_{j=1}^{n}(x_{i} − z_{j}) · F (x x x,zzz)].  (4)

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

(n + ∑_{i=1}^{n}a_{i}) ∫_{0}^{z_{n}} … ∫_{0}^{z_{1}} F (x x x,zzz)V_{n}(x x x)dx dx dx

= ∑_{i=1}^{n}(−1)^{i} ∫_{0}^{z_{n}} … ∫_{0}^{z_{1}} V_{n−1}(x̂ x̂ x̂^{i}) · \frac{∂}{∂x_{i}} [∏_{j=1}^{n}(x_{i} − z_{j}) · F (x x x,zzz)]dx dx dx

= ∑_{i=1}^{n}(−1)^{i} ∫_{0}^{z_{n}} … ∫_{0}^{z_{i+1}} ∫_{0}^{z_{i−1}} … ∫_{0}^{z_{1}} V_{n−1}(x̂ x̂ x̂^{i}) dx̂^{i} dx̂^{i} dx̂^{i} × ∫_{0}^{z_{i}} \frac{∂}{∂x_{i}} [∏_{j=1}^{n}(x_{i} − z_{j}) · F (x x x,zzz)]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 x x,zzz)]dx_{i} = ∏_{j=1}^{n}(x_{i} − z_{j}) · F (x x x,zzz)|_{x_{i}=0}^{x_{i}=z_{i}}

= ∏_{j=1}^{n}(x_{i} − z_{j})^{a_{j}+1} ∏_{j=1}^{n} ∏_{\begin{matrix}1≤i′≤n \\ i′≠i \end{matrix}}(x_{i′} − z_{j})^{a_{j}} |_{x_{i}=0}^{x_{i}=z_{i}}

= − ∏_{j=1}^{n}(−z_{j})^{a_{j}+1} ∏_{j=1}^{n} ∏_{\begin{matrix}1≤i′≤n \\ i′≠i \end{matrix}}(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_{n}} … ∫_{0}^{z_{i+1}} ∫_{0}^{z_{i−1}} … ∫_{0}^{z_{1}} V_{n−1}(x̂ x̂ x̂^{i})

× ∏_{j=1}^{n} ∏_{\begin{matrix}1≤i′≤n \\ i′≠i \end{matrix}}(x_{i′} − z_{j})^{a_{j}}dx̂ dx̂ dx̂^{i}

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

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

We now claim that

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

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

In order to prove this, notice that each of the integrands on the left-hand side, and the integrand on the right-hand side, 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(n)}} … ∫_{0}^{z_{π(i+1)}} ∫_{0}^{z_{π(i−1)}} ⋯ ∫_{0}^{z_{π(1)}} f(y_{1}, …, y_{n−1})dy_{1} ⋯ dy_{n−1}

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

Remark: 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

Multi-Integral and determinantal identity  5

6 Appendix: proof of the "exercises"

Proof of Equation (3): Let eee_{i} = (0, …, 0, 1, 0, …, 0) be the i-th unit vector. So, the recurrence (3) reads as

X(a a a; b + 1) = ∑_{i=1}^{n}X(a a a + eee_{i}; b) · ∏_{\begin{matrix}j=1 \\ j≠i \end{matrix}}^{n} \frac{z_{j}}{z_{j} − z_{i}} + X(a a a; b) · ∏_{i=1}^{n}z_{i}.

The integrand on the left-hand side of (1) satisfies this recurrence because it reduces to (check!)

∏_{i=1}^{n} \frac{x_{i}}{z_{i}} = 1 + ∑_{i=1}^{n}(\frac{x_{i} − z_{i}}{z_{i}} ) · ∏_{\begin{matrix}j=1 \\ j≠i \end{matrix}}^{n} \frac{x_{j} − z_{i}}{z_{j} − z_{i}} .  (7)

Induct on n. The case n = 1 is obvious. Let w = x_{1} and treat (7) as a linear equation in w:

\frac{w}{z_{1}} · ∏_{i=2}^{n} \frac{x_{i}}{z_{i}} = 1 + \frac{w − z_{1}}{z_{1}} ∏_{j=2}^{n} \frac{x_{j} − z_{1}}{z_{j} − z_{1}} + ∑_{i=2}^{n} \frac{x_{i} − z_{i}}{z_{i}} · \frac{w − z_{i}}{z_{1} − z_{i}} · ∏_{\begin{matrix}j=2 \\ j≠i \end{matrix}}^{n} \frac{x_{j} − z_{i}}{z_{j} − z_{i}} .

If w = z_{1}, equality holds by the induction assumption. Putting w = 0, turns the task into

∏_{i=2}^{n} \frac{x_{i} − z_{1}}{z_{i} − z_{1}} = 1 + ∑_{i=2}^{n} \frac{x_{i} − z_{i}}{z_{i} − z_{1}} · ∏_{\begin{matrix}j=2 \\ j≠i \end{matrix}}^{n} \frac{x_{j} − z_{i}}{z_{j} − z_{i}}

which is again the inductive step with x_{i} → x_{i} − z_{1} and z_{i} → z_{i} − z_{1}. So, the first claim holds. Now, we focus on the right-hand side of (1) and show it, too, fulfils (3). This tantamount

\frac{1 + b}{n + b + 1 + ∑_{i=1}^{n} a_{i}} = 1 + ∑_{i=1}^{n}(−1)^{δ_{i}} \frac{1 + a_{i}}{n + b + 1 + ∑_{i=1}^{n} a_{i}} · (−1)^{i−1}

where δ_{i} = 1 if i is odd; δ_{i} = 0, otherwise. The justification is immediate. □ Proof of Equation (4): After computing the derivatives and cancelling out the term F (x x x,zzz), the claim boils down to

(n + ∑_{i=1}^{n}a_{i}) V_{n}(x x x) = ∑_{i=1}^{n}(−1)^{i} V_{n−1}(x̂ x̂ x̂^{i}) · ∑_{j=1}^{n}(1 + a_{j}) ∏_{\begin{matrix}k=1 \\ k≠j \end{matrix}}^{n}(x_{i} − z_{k})
= ∑_{j=1}^{n}(1 + a_{j}) ∑_{i=1}^{n}(−1)^{i} V_{n−1}(x̂ x̂ x̂^{i}) · ∏_{\begin{matrix}k=1 \\ k≠j \end{matrix}}^{n}(x_{i} − z_{k}).  (8)

If we consider the sum ∑_{i=1}^{n} (−1)^{i} V_{n−1}(x̂ x̂ x̂^{i}) · x_{i}^{m}, it is recognized as the determinant of the Vandermonde matrix M M M where the last row is replaced by the vector (x_{1}^{m}, …, x_{n}^{m}). This, however, is well-known to be V_{n}(x x x)·s_{m−n+1} where s_{μ} is the Schur polynomial. Notice that ∏n ; \begin{matrix}k=1 \\ k≠j \end{matrix} (x_{i} −z_{k}) is a polynomial of degree n − 1, in x_{i}. But, s_{m−n+1} = 1 if m = n − 1; and s_{m−n+1} = 0 for m < n − 1. Thus, equation (8) becomes trivial (n + ∑_{i=1}^{n} a_{i}) V_{n}(x x x) = ∑_{j=1}^{n} (1 + a_{j}) V_{n}(x x x). □

Proof of Equation (5):

Eq. (5) has been elucidated, neatly, using a symmetrizing process as depicted by Eq. (6). Here, we offer yet another verification. To this end, start with the right-hand side of Eq. (5) by replacing each integral with ∫_{z_{1}}^{z_{i}} = ∫_{0}^{z_{i}} − ∫_{0}^{z_{1}} and expand the product (∫_{0}^{z_{n}} − ∫_{0}^{z_{1}}) ⋯ (∫_{0}^{z_{2}} − ∫_{0}^{z_{1}}). Based on the fact that the integrand is an anti-symmetric, any multi-integral involving repeated evaluation ∫_{0}^{z_{1}} ∫_{0}^{z_{1}} vanishes while out-of-order pairs, such as ∫_{0}^{z_{3}} ∫_{0}^{z_{1}} ∫_{0}^{z_{2}}, changes sign where reordered, that is, − ∫_{0}^{z_{3}} ∫_{0}^{z_{2}} ∫_{0}^{z_{1}}. On account of this, we find the (n − 1)-tuple multi-integrals

∫_{z_{1}}^{z_{n}} ⋯ ∫_{z_{1}}^{z_{2}} = (∫_{0}^{z_{n}} − ∫_{0}^{z_{1}}) ⋯ (∫_{0}^{z_{2}} − ∫_{0}^{z_{1}})

= ∫_{0}^{z_{n}} ⋯ ∫_{0}^{z_{1}} − ∑_{i=2}^{n}∫_{0}^{z_{n}} ⋯ ∫_{0}^{z_{i−1}} ∫_{0}^{z_{1}} ∫_{0}^{z_{i+1}} ⋯ ∫_{0}^{z_{2}}

= ∑_{i=1}^{n}(−1)^{i−1} ∫_{0}^{z_{n}} ⋯ ∫_{0}^{z_{i−1}} ∫_{0}^{z_{i+1}} ⋯ ∫_{0}^{z_{1}} .

The proof follows. □

Acknowledgment: We wish to thank the referee and Peter Larcombe for insightful comments on an earlier draft.

References

[1] A. F. Beardon, T. K. Carne, T. W. Ng, The critical values of a polynomial Constr. Approx., 18 #3 343-354 (2002).

[2] M. Dougherty, J. M. McCammond, Critical points, critical values, and a determinant identity for complex polynomials, Proc. Amer. Math. Soc., 148 5722-5280 (2020).

Author information

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

E-mail: tamdeber@tulane.edu

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

E-mail: DoronZeil@gmail.com