We have mentioned that the series for [[zeta].sub.n](s, [alpha]; a) is

absolutely convergent for [sigma] > 1/2.

From (2.9), we get that the series in (2.7) is

absolutely convergent in D := {z [member of] C : [absolute value of (z)] [less than or equal to] 1} \{0}.

It is easily seen the Dirichlet series is

absolutely convergent for Rs = [sigma] > 1/6.

So [U.sub.[[rho]([absolute value of y'])+[alpha]]] (x) is

absolutely convergent. Now we shall prove the boundary behavior of [U.sub.[[rho]([absolute value of y'])+[alpha]]] (x).

where the integrated term is O(1) because the Dirichlet series is

absolutely convergent, tan s[pi]/2 = O(1) and the integral is O(1).

A double series [[SIGMA].sub.k,l] [a.sub.k,l] is

absolutely convergent if and only if the following conditions hold: (i) There are ([k.sub.0], [l.sub.0]) [member of] [N.sup.2] and [[alpha].sub.0] > 0 such that

is

absolutely convergent for [sigma] < O, since the term in square brackets vanishes for 0 [less than or equal to] x < 1 by (11).

If [C.sub.0] < 1/2, the series is not

absolutely convergent.

We'll prove that if the generic isotropy subgroup [H.sup.1.sub.0] of D is connected semi-simple, then I'([phi]) is

absolutely convergent for every [phi][Epsilon] S(XA), and this convergence is invariant under castling transforms, which means that if two irreducible regular prehomogeneous vector spaces (G,X) and [Mathematical Expressions Omitted] are castling transforms of each other and [Mathematical Expressions Omitted] for (G,X) is

absolutely convergent for every [phi][Epsilon] S([X.sub.A]), so is [Mathematical Expressions Omitted] for [Mathematical Expressions Omitted].

which are

absolutely convergent [16] for [sigma] > 1/2.

In this region we have to take care of the order of the Euler factors participating in the Euler product, since it is not

absolutely convergent. So putting

where H(s) can be written as a Dirichlet series H(s) = [[summation].sup.[infinity].sub.n=1] which is

absolutely convergent for [Real part]s > 1/6.