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{\bf TOWARDS A WZ EVALUATION OF THE MEHTA INTEGRAL}

{\it Doron Zeilberger
\footnote{*}{
Department of Mathematics, Temple University, Philadelphia,
PA19122. Supported in part by the NSF.
}}

{\bf Appeared in:} SIAM J. Math. Anal. {\bf 25}(1994), 812-814.

\it
Dedicated with affection and admiration to Dick Askey, on his
$60^{th}$ Birthday.
\rm

Dick Askey [As] proposed the problem of proving
the Mehta (see [M]) integral identity:

$$
 {{1} \over { ( 2 \pi )^{n/2} }} \int_{ - \infty }^{\infty} ... \int_{ - \infty }^{\infty} exp( - x_1^2/2 - ... - x_n^2/ 2 ) \prod_{1<= i < j <= n} ( x_i - x_j )^{2c}  d x_1  ...  d x_{n \,= \,} \prod_{j=1}^{n} {{(cj)!}\over {j!}} \, \,, \eqno(Mehta)
$$     

without using Selberg's integral (see [M]).
This problem was solved by Greg Anderson[An]. Here we
use the method of [WZ1][WZ2] to initiate another Selberg-free proof,
that we believe is of independent interest.
We show that (Mehta) for any given $n$, is equivalent to the following
elegant identity ($d:= n(n-1)/2$) :

$$
\{ {  \sum_{i=1}^{n}  {{\partial^2}\over{\partial x_i^2}} +2c \sum_{ 1<= i < j <= n } {{1}\over{x_i - x_j }} ({{ \partial}\over{\partial x_i }} - {{ \partial}\over{\partial  x_j }} )  }  \}^d  \, \prod_{1<=i < j <= n} ( x_i - x_j )^2  \,= \, 2^d d! n! \prod_{1 <= s < r <= n} (rc +s)
\eqno(Mehta')
$$

$(Mehta')$ is purely routine for any specific $n$, but at the time
of writing we are unable to prove it directly 
for general $n$. Of course, we
do have {\it a} proof, since we are going to show that
(Mehta') and (Mehta) are equivalent, but what we are after is a 
{\it direct} proof. The author is offering a prize of 25 US dollars
for such a proof. 

The present method also extends obviously to
the Macdonald-Mehta integral[M], that was proved by
Beckner and Regev[BR] for the classical root systems (see [M]),
by Garvan[G] for the exceptional root system $F_4$, and by
Opdam[O] for $E_6 , E_7$, and $E_8$. It follows that the
present approach should also yield new proofs for all the exceptional
root systems, at least in principle, but most likely also in
practice. More important,
it seems to have a high chance of producing a uniform, 
intrinsic, classification-independent proof. We leave to the reader,
as an instructive exercise, the task of finding the root-system
analog of (Mehta') that is equivalent to the now-proved
Mehta-Macdonald conjecture, and we offer an additional 25 dollars
for an intrinsic  proof.

Our proposed proof of (Mehta) will be a {\it derivation} rather
than a {\it verification}, and will follow the method of [WZ2].
Let's call the left of (Mehta) $L(c)$, and let's call the integrand
$F(c; x_1 , ... , x_n )$. We know from the general theory of
[WZ2] that for some $r$ and some rational functions
$P_1 , ... , P_n$, in $(c, x_1 , ... , x_n )$,
and some rational functions in $c$, $a_0 (c) , ... , a_r (c)$,

$$
 \sum_{s=0}^{r} a_s (c) F(c+s; x_1 , ... , x_n ) \,= \, \sum_{i=1}^{n}{{ \partial}\over{ \partial x_i }} ( P_i F) \, \,. 
\eqno(WZ)
$$         

Let's be optimistic and try out $r=1$. Without loss of generality, set
$a_1 :=1$. Substituting $F$ in (WZ), performing all
differentiations, and dividing throughout by $F$,
leads to the following equation for the $P_i$
and $a_0$ :

$$
 a_0 +  \prod_{1<=i < j <= n} ( x_i - x_j )^{2 \,= \,} \sum_{i=1}^{n} {{\partial P_i}\over{\partial x_i}} +2c \sum_{i=1}^{n} ( \sum_{j \neq  i} {{1}\over{ x_i - x_j }} ) P_i - \sum_{i=1}^{n} x_i P_i  \, \, \,. 
\eqno(1)
$$     

Let's be even more optimistic, and assume that the $P_i$ are
polynomials, rather than mere rational functions,
in their dependence on $( x_1 , ... , x_n )$,
and are further the components of the gradient of another
polynomial $P$, i.e. $P_i = \partial P / \partial x_i$,
$i=1, ... , n$, for some polynomial $P$. Equation (1) then becomes

$$
 a_0 +  \prod_{1<=i < j <= n} ( x_i - x_j )^{2 \,= \,} \{ \sum_{i=1}^{n} {{\partial^2  }\over{\partial x_i^2}} +2c \sum_{i=1}^{n} ( \sum_{j \neq  i} {{1}\over{ x_i - x_j }} ) {{\partial }\over{\partial x_i }} \}  \,P \,  - \sum_{i=1}^{n} ( x_i{{ \partial}\over{ \partial x_i}} ) P \, \, \,. 
\eqno(2)
$$  

If we can find such a polynomial $P$, and compute the corresponding
$a_0$, then it would follow from (WZ), upon integrating w.r.t
$x_1 , ... , x_n$, that $L(c)$ satisfies the recurrence
$L (c+1 ) = - a_0 (c) L(c)$, which combined with $L(0)=1$, 
would enable one to find $L(c)$. Note that the mere existence
of $a_0$, that we will shortly prove is given by the left side 
of (Mehta'),
implies that $L(c)$ is of {\it closed form}, which from a theoretical
point of view is almost as good as knowing what it is exactly.

Let's write $P$ as a sum of its homogeneous parts

$$
  P = \sum_{j=2}^{2d} P^{(j)}, \quad P^{(j)}  \,homog. \,of \,deg. \, j 
$$

where, as above $d$ equals $n(n-1)/2$. Denote
the operator inside the braces of (2) (or (Mehta')) by ${\bf Z}$.
Using Euler's formula, we get
$$
 a_0 +  \prod_{1<=i < j <= n} ( x_i - x_j )^2  \,= \, \sum_{j=2}^{2d}  
{\bf Z} P^{(j)} - \sum_{j=2}^{2d} j P^{(j)}  \,= \, (-2d) P^{(2d)} + \sum_{j=2}^{2d} ( {\bf Z} P^{(j)} - (j-2) P^{(j-2)} ) \,. 
\eqno(3)
$$      

Equating corresponding homogeneous parts, we get

$$
 P^{(2d)} = - (2d)^{-1}  \prod_{1<=i < j <= n} ( x_i - x_j )^2  \,, \, \,  \,P^{(j-2)}  \,= \, {{1}\over {j-2}}  {\bf Z} P^{(j)}  \, \,,  \,j=2d \,, \,2d-2 \,, \, ... , 4 \, \,. 
\eqno(4)
$$        

It is easy to see that  {\bf Z} maps homogeneous symmetric polynomials to
homogeneous symmetric polynomials, and that it reduces the degree by
$2$.  Iterating (4), and comparing the constant part of
(3) finally yields that both $P$ and $a_0$ indeed exist, and that

$$
  a_0 = -  ( 2^d d! )^{-1}     {\bf Z}^d  [ \prod_{1<=i < j <= n} ( x_i - x_j )^2 ]  \, \,. 
$$

Hence Mehta's integral is indeed expressible in closed form, 
and proving that
its value coincides with the value implied by (Mehta) amounts
to proving (Mehta'). QED

The referee noticed empirically the following generalization.
For $a$, any positive integer, we have:

$$
  \{ {  \sum_{i=1}^{n}  {{\partial^2}\over{\partial x_i^2}} +2c \sum_{ 1<= i < j <= n } {{1}\over{x_i - x_j }} ({{ \partial}\over{\partial x_i }} - {{ \partial}\over{\partial  x_j }} )  }  \}^{ad}  \, \prod_{1<=i < j <= n} ( x_i - x_j )^{2a}  \,= \, 
\eqno(Mehta'')
$$ 

$$
  2^{ad} (da)! n!a \prod_{2 <= r <= n}  \, \, \prod_{1 <= s <= ra  \,, \, s \neq  0 \,\,\,mod \,\,\,r} (rc +s)  \, \,. 
$$

It turns out that (Mehta'') follows from (Mehta) exactly the same
way as (Mehta'), just replace $c$ by $ca$, in (Mehta), and
repeat the argument! In particular,
it follows that (Mehta') implies (Mehta'') albeit, indirectly,
via (Mehta). 

{\bf Acknowledgment:} Many thanks are due to Herb Wilf for
countless discussions and inspiration.

{\bf References}

[An] G. Anderson, {\it Letter to R. Askey}.

[As] R.A. Askey, {\it Lost and found mathematics}, Joint MAA-AAAS
invited talk, Columbus, OH, Aug. 1990.

[BR] W. Beckner and A. Regev, \it
unpublished, untitled, and undated
manuscript, c. 1980\rm

[G] F.G. Garvan, {\it Some Macdonald-Mehta integrals by brute force},
in: "q-Series and Partitions", D. Stanton, ed., IMA volumes {\bf 18},
77-98, Springer, New York.

[M] I.G. Macdonald, {\it Some conjectures for root systems},
SIAM J. Math. Anal. {\bf 13}(1982), 988-1007.

[O] E. Opdam, \it
Some applications of hypergeometric shift
operators\rm
, Invent. Math. {\bf 98}(1989), 1-18.

[WZ1] H.S. Wilf, and D. Zeilberger, \it
Rational function certification of
hypergeometric multi-integral/sum/"q" identities\rm
, 
Bulletin of the Amer. Math. Soc. {\bf 24}(1992),143-148.

[WZ2] ", 
\it
An algorithmic proof theory for hypergeometric
(ordinary and "q") multisum/integral identities\rm
, Invent. Math.
{\bf 108}(1992),575-633.

March 1992. Revised April 1993.
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