# An Explanation of Mellin’s 1921 Paper

Wayne M. Lawton1
1Department of the Theory of Functions, Institute of Mathematics and Computer
Science, Siberian Federal University, Krasnoyarsk, Russian Federation.
<EMAIL_ADDRESS>

###### Abstract

In 1921 Mellin published a Comptes Rendu paper computing the principal root of
the polynomial $Z^{n}+x_{1}Z^{n_{1}}+\cdots+x_{p}Z^{n_{p}}-1$ using
hypergeometric functions of its coefficients $x_{1},...,x_{p}.$ He used an
integral transform nowadays bearing his name. Slightly over three pages, the
paper is written in French in a terse style befitting the language. Unable to
find an elementary explanation on the web or in a texbook, we wrote this
expository article to make Mellin’s landmark result accessible to interested
people who are not experts in hypergeometric functions and complex analysis.

2020 Mathematics Subject Classification:32-03;12-08;33C70 111This work is
supported by the Krasnoyarsk Mathematical Center and financed by the Ministry
of Science and Higher Education of the Russian Federation in the framework of
the establishment and development of Regional Centers for Mathematics Research
and Education (Agreement No. 075-02-2020-1534/1).

##  1 Principal Solution $Z$ and Mellin Transform of $Z^{\alpha}$

In the opening paragraph of his paper, Mellin says that it summarizes research
he undertook years ago and was prompted by notes [2] of Richard Birkeland, a
Norwegian mathematician known for his contributions to the theory of algebraic
equations. The 7 numbered equations in Mellin’s paper and this paper coincide.
We prove each of them.

$Z^{n}+x_{1}\,Z^{n_{1}}+\cdots+x_{p}\,Z^{n_{p}}-1=0$ (1)

For integers $p\geq 1,$ $0<n_{p}<\cdots n_{1}<n,$ we define a principal
solution to be an analytic function $Z(x_{1},...,x_{p})$ on $[0,\infty)^{p}$
satisfying (1) and $Z(0,...,0)=1.$

###### Lemma 1

If a principal solution $Z(x_{1},...,x_{p})$ of (1) exists, it is unique.

Proof For $r>0$ define $D_{r}:=\\{z\in\mathbb{C}:|z|<r\\}.$ For sufficiently
small $r$ a principal solution $Z(x_{1},\dots,x_{p})$ of (1) extends to give a
holomorphic solution $\widetilde{Z}(x_{1},\dots,x_{p})$ of (1) for
$(x_{1},...,x_{n})$ in the polydisc $D_{r}^{p}.$ Therefore, for every $n$–th
root of unity $\epsilon,$ the function
$Z_{\epsilon}(x_{1},...,x_{n}):=\epsilon\,\widetilde{Z}(\epsilon^{n_{1}}\,x_{1},...,\epsilon^{n_{p}}x_{p})$
is a holomorphic solution of (1) in $D_{r}^{p}.$ Since a polynomial of degree
$n$ can have at most $n$ distinct roots, these $n$ distinct functions describe
all holomorphic solutions of (1) on $D_{r}^{p}.$ The conclusion follows since
the restrictions of $Z(x,...,x_{p})$ and $Z_{1}(x_{1},...,x_{p})$ to
$[0,r)^{p}$ are equal and analytic. Define
$\Xi:=\\{\,\xi:=(\xi_{1},...,\xi_{p})\,:\,\xi_{1}+\cdots+\xi_{p}\in(-\infty,-1]\,\\},$
$W(\xi):=1+\xi_{1}+\cdots+\xi_{p},$ and the holomorphic function
$\Psi:=(x_{1},...,x_{n}):\mathbb{C}^{p}\backslash\Xi\rightarrow\mathbb{C}^{p}$
by

$x_{i}(\xi):=\xi_{i}\,W(\xi_{1},...,\xi_{p})^{\frac{n_{i}}{n}-1},\,i=1,...,p$
(2)

where $W^{\frac{1}{n}}:\mathbb{C}^{p}\backslash\Xi\rightarrow\mathbb{C}$ is
the branch of $\sqrt[n]{W}$ that satisfies $W^{\frac{1}{n}}(0,...,0)=1.$

###### Lemma 2

If $x_{1},...,x_{p}$ are defined by (2), then $Z:=W^{-\frac{1}{n}}$ satisfies
(1).

Proof Follows by substitution.

###### Lemma 3

The restriction $\Psi:[0,\infty)^{p}\rightarrow[0,\infty)^{p}$ is a bijection.
Furthermore, the Jacobian of
$\Psi:\mathbb{C}^{p}\backslash\Xi\rightarrow\mathbb{C}^{p}$ satisfies

$\frac{\partial(x_{1},...,x_{p})}{\partial(\xi_{1},...,\xi_{p})}=\left(1+\sum_{k=1}^{p}\xi_{k}\right)^{\frac{n_{1}+\cdots+n_{p}}{n}-p-1}\left(1+\frac{1}{n}\sum_{k=1}^{p}n_{k}\,\xi_{k}\right).$
(3)

The restriction $\Psi:[0,\infty)^{p}\rightarrow[0,\infty)^{p}$ and its inverse
$\Psi^{-1}:[0,\infty)^{p}\rightarrow[0,\infty)^{p}$ are analytic. The
principal solution of (1) is
$Z:=W^{-\frac{1}{n}}\circ\Psi^{-1}:[0,\infty)^{p}\rightarrow\mathbb{[}1,\infty).$

Proof The first assertion is Proposition 1. Define the Kronecker symbol

$\delta_{i,j}=\begin{cases}1,\ \hbox{ if }i=j\\\ 0\ \hbox{ if }i\neq
j.\end{cases}$

Since the matrix

$\frac{\partial\Psi_{i}}{\partial\xi_{j}}=W^{\frac{n_{i}}{n}-1}\left(\delta_{i,j}+\xi_{i}\,(\frac{n_{i}}{n}-1)\,W^{-1}\right),$

is a product of two matrices, it follows that

$\frac{\partial(x_{1},...,x_{p})}{\partial(\xi_{1},...,\xi_{p})}:=\det\frac{\partial\Psi_{i}}{\partial\xi_{j}}=W^{\frac{\sum_{k=1}^{p}n_{k}}{n}-p}\,\det\left(\delta_{i,j}+\xi_{i}\,(\frac{n_{i}}{n}-1)\,W^{-1}\right).$

Proposition 2 implies that

$\det\left(\delta_{i,j}+\xi_{i}\,W^{-1}\right)=1+\sum_{k=1}^{p}\xi_{k}\,\left(\frac{n_{k}}{n}-1\right)\,W^{-1}=W^{-1}\left(1+\frac{1}{n}\sum_{k=1}^{p}n_{k}\,\xi_{k}\right)$

and concludes the proof of the second assertion. Since the Jacobian of $\Psi$
is nozero and holomorphic on $\mathbb{C}^{p}\backslash\Xi$ and
$[0,\infty)^{p}\subset\mathbb{C}^{p}\backslash\Xi,$ assertion three follows
from the inverse function theorem for holomorphic functions ([9], p. 40,
Theorem 2). The fourth assertion then follows from Lemma 2.

Remark 1 Let $n=2,p=1.$ Then
$\Psi:\mathbb{C}\backslash(-\infty,-1]\rightarrow\mathbb{C}\backslash
i((-\infty,-2]\cup[2,\infty))$ is a holomorphic bijection with inverse

$\Psi^{-1}(z)=-1+\left(\,\frac{z}{2}+\sqrt{1+\left(\frac{z}{2}\right)^{2}}\
\right)^{2}.$

Letting $\xi_{1}=-1+e^{s+it}$ with $s\in\mathbb{R}$ and $-\pi<t<\pi$ gives
$\Psi(\xi_{1})=2\sinh\frac{s+it}{2}=u+iv$ where $u=2\sinh\frac{s}{2}\cos t/2$
and $v=2\cosh\frac{s}{2}\sin t/2.$ For $t=0$ this gives the curve $v=0.$ For
fixed $t\neq 0$ this gives the branch of the hyperbola described by the
equation $\left(\frac{v}{\sin t/2}\right)^{2}-\left(\frac{u}{\cos
t/2}\right)^{2}=4$ with $tv\geq 0.$

Question 1 For $n\geq 3$ and $p\geq 1,$ is $\Psi$ injective and what is its
image?

Following (3) Mellin says that using the known formula, derived in Proposition
2, in combination with (2) and (3), one can deduce the following result:

###### Lemma 4

The principal solution of (1) satisfies

$\int_{0}^{\infty}\cdots\int_{0}^{\infty}Z(x_{1},...,x_{p})^{\alpha}\,x_{1}^{u_{1}-1}\cdots
x_{p}^{u_{p}-1}\,dx_{1}\cdots
dx_{p}=\frac{\alpha}{n}\,\frac{\Gamma(u)\Gamma(u_{1})\cdots\Gamma(u_{p})}{\Gamma(u+u_{1}+\cdots+u_{p}+1)},$
(4)

where $\alpha>0,$ and the real parts of
$u:=\frac{\alpha}{n}-\frac{n_{1}}{n}u_{1}-\cdots-\frac{n_{p}}{n}u_{p},$ and
$u_{1},...,u_{p}$ are positive.

Proof Equation (2) and Jensen’s inequality implies

$W^{(u_{1}n_{1}+\cdots+u_{p}n_{p})/n}>p^{u_{1}+\cdots+u_{p}}\,x_{1}^{u_{1}}\cdots
x_{p}^{u_{p}}.$

Therefore, since $Z=W^{-\frac{1}{n}}$ and
$\alpha>u_{1}n_{1}+\cdots+u_{p}n_{p},$ the integral $I$ in (4) exists.
Proposition 3 implies that $F:[0,\infty)^{p}\rightarrow[0,\infty)^{p}$ is a
bijective diffeomorphism, so $I$ can be expressed as an integral over
$[0,\infty)^{p}$ with respect to the variables $\xi_{1},...\xi_{p}.$ Hence (2)
and (3) give

$I=\int_{0}^{\infty}\cdots\int_{0}^{\infty}\frac{(1+\frac{n_{1}}{n}\xi_{1}+\cdots+\frac{n_{p}}{n}\xi_{p})\xi_{1}^{u_{1}-1}\cdots\xi_{p}^{u_{p}-1}}{W^{\omega}}\,d\xi_{1}\cdots
d\xi_{p}$

where $\omega:=u+u_{1}+\cdots+u_{p}+1.$ Proposition (2) implies that
$I+I_{0}+I_{1}+\cdots+I_{p}$ where

$I_{0}=\frac{\Gamma(\omega-u_{1}-\cdots-
u_{p})\Gamma(u_{1})\cdots\Gamma(u_{p})}{\Gamma(\omega)}=\frac{\Gamma(u+1)\Gamma(u_{1})\cdots\Gamma(u_{p})}{\Gamma(\omega)},$

and

$I_{i}=\frac{n_{i}}{n}\,\frac{\Gamma(\omega-u_{1}-\cdots-
u_{p}-1)\Gamma(u_{1})\cdots\Gamma(u_{i}+1)\cdots\Gamma(u_{p})}{\Gamma(\omega)}=\frac{n_{i}u_{i}}{nu}\,I_{0}.$

We conclude the proof by observing that

$I=\left(1+\frac{1}{u}\sum_{i=1}^{p}\frac{n_{i}u_{i}}{n}\right)\,I_{0}=\frac{\alpha}{nu}I_{0}=\frac{\alpha}{n}\frac{\Gamma(u+1)}{u}\frac{\Gamma(u_{1})\cdots\Gamma(u_{p})}{\Gamma(\omega)}.$

##  2 Computing $Z^{\alpha}$ from its Mellin Transform

Mellin continues: “The law of reciprociy relating to the integrals of this
species, demonstrated by us in a previous work [5], allows us to invert as
follows (4):”

$Z(x_{1},...,x_{p})^{\alpha}=$

$\frac{1}{(2\pi
i)^{p}}\int_{a_{1}-i\infty}^{a_{1}+i\infty}\cdots\int_{a_{p}-i\infty}^{a_{p}+i\infty}\frac{\alpha}{n}\,\frac{\Gamma(u)\Gamma(u_{1})\cdots\Gamma(u_{p})}{\Gamma(u+u_{1}+\cdots+u_{p}+1)}\,x_{1}^{-u_{1}}\cdots
x_{p}^{-u_{p}}\,du_{1}\cdots du_{p},$ (5) $\alpha-n_{1}a_{1}-\cdots-
n_{p}a_{p}>0,\ a_{s}>0,\
u:=\frac{\alpha}{n}-\frac{n_{1}}{n}u_{1}-\cdots-\frac{n_{p}}{n}u_{p}$
$u+u_{1}+\cdots+u_{p}=\frac{\alpha}{n}+\frac{n_{1}^{`}}{n}u_{1}+\cdots+\frac{n_{p}^{`}}{n}u_{p},\
\ n_{s}^{`}:=n-n_{s}.$

“This formula constitutes our solution of (1). It supposes that

$-\frac{n_{s}\pi}{2n}<\arg(x_{s})<\frac{n_{s}\pi}{2n}$

but we can extend our domain of validity by suitably deforming the integration
paths.”

The right side of (4) is the Mellin transform of $Z(x_{1},...,x_{p})^{\alpha}$
and (5) represents $Z(x_{1},...,x_{p})^{\alpha}$ by the inverse Mellin
transform. Lacking the luxury of accessing [5] we refer the reader to the
derivation of the inverse Mellin transform via the Fourier transform by
Debnath and Bhatta in section 8.1 of [3]. They address the univariate case,
but the extension to multivariate case is straightforward.

##  3 Generalized Hypergeometric Functions

Let $n$ be a positive integer and
$f_{1},...,f_{p},g_{1},...,g_{s}:\mathbb{C}^{p}\rightarrow\mathbb{C}$ be
entire functions and $F:\mathbb{C}^{p}\rightarrow\mathbb{C}$ satisfy the
following system of functional equations

$F(u_{1},...,u_{s}+n,...,u_{p})=\frac{f_{s}(u_{1},...,u_{p})}{g_{s}(u_{1},...,u_{p})}\,F(u_{1},...,u_{p}),\
\ s=1,...,p.$ (6)

and such that the following integral converges and does not change when we
move the (vertical) integration paths for each $u_{s}$ $n$ units to the right:

$y(x_{1},...,x_{p})=\frac{1}{(2\pi
i)^{p}}\,\int_{u_{1}}\cdots\int_{u_{p}},F(u_{1},...,u_{p})\,x_{1}^{-u_{1}}\cdots
x_{p}^{-u_{p}},\ \ (x_{1},...,x_{p})\in[0,\infty)^{p}.$

Mellon cites a result [5] that $y$ satisfies the following system of partial
differential equations

$f_{s}\left(-x_{1}\frac{\partial}{\partial
x_{1}},...,-x_{p}\frac{\partial}{\partial
x_{p}}\right)\,y=g_{s}\left(-x_{1}\frac{\partial}{\partial
x_{1}},...,-x_{p}\frac{\partial}{\partial x_{p}}\right)\,s_{s}^{n}\,y,\ \
s=1,...,p.$ (7)

Remark 2 (7) is a system of PDE’s of finite order iff $f_{s}$ and $g_{s}$ are
polynomials.

Mellin calls solutions of (6) hypergeometric type if the factors of
$f_{s},g_{s}$ have the form

$c_{1}u_{1}+\cdots+c_{p}u_{p}+a$

where each $c_{i}$ is a rational real number.

Remark 3 Polynomials are not hypergeometric functions or series as defined in
[4, 10] but they are of hypergeometric type since
$\prod_{k=0}^{m-1}(x\frac{d}{dx}-k)=x^{m}\frac{d^{m}}{dx^{m}}.$

Formally (7) follows from (6) since the functions $x_{s}^{-u_{s}}$ are
eigenfunctions with eigenvalue $-u_{s}$ of the Euler operator
$x_{s}\frac{\partial}{\partial x_{s}},s=1,...,p.$ Clearly the function

$F(u_{1},...,u_{p})=\frac{\alpha}{n}\,\frac{\Gamma(u)\Gamma(u_{1})\cdots\Gamma(u_{p})}{\Gamma(u+u_{1}+\cdots+u_{p}+1)}$

satisfies (6) where $f_{s}$ and $g_{s}$ have the form above, hence
$Z(u_{1},...,u_{p})^{\alpha}$ is of hypergeometric type whenever
$\alpha>n_{1}n.$

Question 2 How large must $\alpha$ be to ensure that $Z^{\alpha}$ is a
solution of (1) of hypergeometric type? The condition $\alpha>nn_{1}$ is
sufficient but not necessary because for $n=1,p=1$ the root
$Z(x_{1}):=-\frac{x_{1}}{2}+\sqrt{1+(x_{1}/2)^{2}}$ is of hypergeometric type.
Semusheva and Tsikh [8] proved this fact directly by deriving the following
Mellin–Barnes integral representation

$Z(x_{1})=\frac{1}{4\pi i}\int_{\Re
z=\frac{1}{2}}\frac{\Gamma(z)\,\Gamma((1+z)/2)}{\Gamma((3+z)/2)}\,x_{1}^{z}\,dz.$

Remark 4 For a comprehensive development of Mellin’s solution of (3) and
systems of differential equations of hypergeometric type see ([1], Chapitre
V).

##  4 Appendix: Crucial Propositions

We prove results required to derive the equations in Mellin’s paper.

Let $e_{1}:=(1,0,...,0),e_{2}:=(0,1,...,0),...,e_{p}=(0,...,0,1)$ be the
standard basis for $\mathbb{R}^{p}.$ For $s>0$ let $H_{s}$ denote the convex
hull of $\\{se_{i}:i=1,...,p\\}$ and let $K_{s}$ denote the convex hull of
$\\{s(1+s)^{\frac{n_{i}}{n}-1}e_{i}:i=1,...,p\\}.$

###### Proposition 1

$\Psi:[0,\infty)^{p}\rightarrow[0,\infty)^{p}$ is a bijection.

Proof For every $s>0$ define the linear map
$L_{s}:\mathbb{R}^{p}\rightarrow\mathbb{R}^{p}$ by

$(L_{s}\,y)_{i}:=y_{i}\,(1+s)^{\frac{n_{i}}{n}-1},\ \ i=1,...,p.$

If $(\xi_{1},...,\xi_{p})\in H_{s},$ then $W(\xi_{1},...,\xi_{p})=1+s$ so
Equation 2 implies that

$\Psi(\xi_{1},...,\xi_{p})=L_{s}(\xi_{1},...,\xi_{p})=\sum_{i=1}^{p}\frac{\xi_{i}}{s}\,s(1+s)^{\frac{n_{i}}{n}-1}e_{i}\in
K_{s},$

hence the restriction $\Psi:H_{s}\rightarrow K_{s}$ is a linear bijection.
Clearly $\Psi(0,...,0)=(0,...,0)$ and $[0,\infty)^{p}$ is a disjoint union of
$\\{(0,...,0)\\}$ and the sets $H_{s},s>0.$ Since for $i=1,...,p,$ the
function $s(1+s)^{\frac{n_{i}}{n}-1}$ is increasing, it follows that
$[0,\infty)^{p}$ is a disjoint union of $\\{(0,...,0)\\}$ and the sets
$K_{s},s>0.$ This concludes the proof.

###### Proposition 2

Let $p\geq 1,$ $y_{1},...,y_{p}$ be indeterminates, and define the matrix

$M(y_{1},...,y_{p}):=\left[\begin{array}[]{ccccc}1+y_{1}&y_{1}&\cdots&y_{1}&y_{1}\\\
y_{2}&1+y_{2}&\cdots&y_{2}&y_{2}\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
y_{p-1}&y_{p-1}&\cdots&1+y_{p+1}&y_{p+1}\\\
y_{p}&y_{p}&\cdots&y_{p}&1+y_{p}\\\ \end{array}\right].$

Then

$\det M(y_{1},...,y_{p})=1+y_{1}+\cdots+y_{p}.$

Proof For $i=1,...,p$ let $\partial_{i}:=\frac{\partial}{\partial y_{i}}.$
Clearly $\det M(y_{1},...,y_{p})$ is a polynomial with constant term $\det
M(0,...,0)=\det I_{p}=1$ and of degree at most $1$ in each variable
$y_{1},...,y_{p}$ so for $i=1,...,p$

$\partial_{i}^{2}\det M(y_{1},...,y_{p})=0.$

Let $M_{i}(y_{1},...,y_{p})$ denote the matrix obtained from
$M(y_{1},...,y_{p})$ by replacing each entry in its $i$–th row with $1$ and
for $j\neq i$ let $M_{i,j}(y_{1},...,y_{p})$ denote the matrix obtained from
$M(y_{1},...,y_{p})$ by replacing each entry in its $i$–row and its $j$–row by
$1.$ Since a determinant of a matrix is a linear function of each of its row
vectors,

$\partial_{i}\det M(y_{1},...,y_{p}):=\det M_{i}(y_{1},...,y_{p})$

and for $j\neq i$

$\partial_{j}\partial_{i}\det M(y_{1},...,y_{p}):=\det
M_{i,j}(y_{1},...,y_{p})=0$

since $M_{i,j}(y_{1},...,y_{p})$ has two identical rows. Therefore Taylor’s
expansion gives

$\det M(y_{1},...,y_{n})=1+\sum_{i=1}^{p}y_{i}\,\partial_{i}\det
M(0,...,0)=1+\sum_{i=1}^{p}y_{i}\,\det M_{i}(0,...,0).$

It suffices to prove $\det M_{i}(0,...,0)=1,\,i=1,...,p.$ This follows since

$M_{i}(0,...0)=\left[\begin{array}[]{ccccccc}1&0&\cdots&0&\cdots&0&0\\\
0&\ddots&\ddots&\ddots&\ddots&\ddots&0\\\ 0&\cdots&1&0&0&\cdots&0\\\
1&\cdots&1&1&1&\cdots&1\\\ 0&\cdots&0&0&1&\cdots&0\\\
0&\ddots&\ddots&\ddots&\ddots&\ddots&0\\\ 0&0&\cdots&0&\cdots&0&1\\\
\end{array}\right].$

Remark 5 Nilsson and Passare ([7], Example 1) used basic calculus to compute
the following integrals for $\Re u_{i}\in(0,1),i=1,2.$

$\int_{0}^{\infty}\int_{0}^{\infty}\frac{\xi_{1}^{u_{1}-1}\,\xi_{2}^{u_{2}-1}}{1+\xi_{1}+\xi_{2}}d\xi_{1}d\xi_{2}=\frac{\Gamma(u_{1})\,\Gamma(u_{2})\,\Gamma(1-u_{1}-u_{2})}{\Gamma(1)}$

The following result uses exterior calculus to extend their computation.

###### Proposition 3

For every positive integer $p,$ complex $u_{1},...,u_{p}$ satisfying
$\Re\,u_{i}>0,$ and $\omega>\max\\{\Re\,u_{i}\\}$ the integral below converges
and satisfies the stated identity.

$\int_{0}^{\infty}\cdots\int_{0}^{\infty}\frac{\xi_{1}^{u_{1}-1}\cdots\xi_{p}^{u_{p}-1}}{\left(1+\xi_{1}+\cdots+\xi_{p}\right)^{\omega}}\,d\xi_{1}\cdots
d\xi_{p}=\frac{\Gamma(u_{1})\cdots\Gamma(u_{p})\Gamma(\omega-u_{1}-\cdots-
u_{p})}{\Gamma(\omega)}.$

Proof Multiplying both sides of the asserted identity in the lemma by

$\Gamma(\omega)=\int_{0}^{\infty}x^{\omega-1}\,e^{-x}\,dx$

gives the equivalent identity
$I_{1}(u_{1},...,u_{p},\omega)=\Gamma(u_{1})\cdots\Gamma(u_{p})\,\Gamma(\omega-
u_{1}-\cdots-u_{p})$ where

$I_{1}(u_{1},...,u_{p},\omega):=\int_{0}^{\infty}\cdots\int_{0}^{\infty}\frac{\xi_{1}^{u_{1}-1}\cdots\xi_{p}^{u_{p}-1}}{\left(1+\xi_{1}+\cdots+\xi_{p}\right)^{\omega}}\,x^{\omega-1}\,e^{-x}\,d\xi_{1}\wedge\cdots\wedge
d\xi_{p}\wedge dx.$

Here we express the volume form using exterior products to implement the
following change of variables: $y:=x/(1+\xi_{1}+\cdots+\xi_{p})$ and
$z_{i}:=y\,\xi_{i},i=1,...,p.$ Then
$I_{1}(\xi_{1},...,\xi_{p},x)=I_{2}(z_{1},...,z_{2},y)$ where

$I_{2}(z_{1},...,z_{2},y):=\int_{0}^{\infty}\cdots\int_{0}^{\infty}z_{1}^{u_{1}-1}e^{-z_{1}}\cdots
z_{p}^{u_{p}-1}e^{-z_{p}}\,y^{\omega-u_{1}-\cdots-u_{p}-1}\,e^{-y}\,\theta$

where

$\theta=\frac{y^{p}}{1+\xi_{1}+\cdots+\xi_{p}}\,d\xi_{1}\wedge\cdots\wedge
d\xi_{p}\wedge dx$

In order to finish the proof it suffices to prove that
$dz_{1}\wedge\cdots\wedge dz_{p}\wedge dy=\theta.$ Since
$dz_{i}=\xi_{i}dy+y\,d\xi_{i}$ and $dy\wedge dy=0$ it follows that

$dz_{1}\wedge\cdots\wedge dz_{p}\wedge dy=y^{p}\,d\xi_{1}\wedge\cdots\wedge
d\xi_{p}\wedge dy.$

Substitute

$dy=\frac{1}{1+\xi_{1}+\cdots+\xi_{p}}\,dx-\frac{x}{(1+\xi_{1}+\cdots+\xi_{p})^{2}}\left(\sum_{k=1}^{p}d\xi_{k}\right)$

and use the fact that $d\xi_{i}\wedge d\xi_{i}=0$ to obtain

$dz_{1}\wedge\cdots\wedge dz_{p}\wedge
dy=\frac{y^{p}}{1+\xi_{1}+\cdots+\xi_{p}}\,d\xi_{1}\wedge\cdots\wedge
d\xi_{p}\wedge dx=\theta$

which concludes the proof.

Acknowledgments We thank Elaine Wong of the Johann Radon Institute for
Computational and Applied Mathematics (RICAM), Linz for her meticulous
proofreading of and corrections to this article.

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