MULTIPLE DIRICHLET SERIES AND AUTOMORPHIC FORMS 19
For an interchange in the order of summation to work nicely one would like to
have the following hold:
/o 9fi\ V^ L(s,7r x Xd0)P{s,dodl) _ yy L(w, Xm0)Q(w, mprnp
Here the Q(w,rriQm\) should be Dirichlet polynomials with Euler products similar
to P. In fact, for the functional equations to work out properly we should have
(3.27) Q(w, ra0m?) =
m\2wQ{l
~ w, ra0m?)
with
Q(w,m0m2)
=
n ^  
m i
(HmoP2p, 0) + b(m0p2^ l)p~w + b(m0p2^ 2)p~2™ + • • • +
6(m0p2/3,4/3)p4^)
and the recursion relation
(3.28)
b(mop20,
k) =
pkpb(m0p2^
2/3  fc),
holding for 0 k 2/3. Notice that we can allow the first term of the Euler product
to equal 1 on one side of the equation, but we do not have that freedom on the
other.
Let us now consider the coefficients of l~s on both sides of (3.26). This is easily
done by letting s — » oo. As the coefficients must be equal, (3.26) implies that
E ( ^ = C ( « ; ) Q K i ) ,
i.e that Q(w,l) = 1. Similarly, letting w — » oo and equating the coefficients of
l~w
we see that
v ; ^ (m0ml)s '
Implying that
6(mom^,0) = c(raoraf)
for all m =
mom2.
We continue now, equating coefficients of p~s on both sides of (3.26). For fixed
squarefree do this yields the relation
^ (dod?)
+
£ , (dod?)
M
'
X p j
^ (dodf) ^
= i
( d o d f )  
As a consequence of ignoring bad primes we are assuming that reciprocity is perfect
(Xd0(p) —
Xp(do))
It
n o w
follows immediately that
a(d0dl,l) = Xd0{p)c(p)
for all p\d\.
Evaluating the coefficient of
p~w
on each side of (3.26) yields, for fixed square
free mo,
T(a^v, \  V^ Xrn0{p)c{m0m\)

^ b{mom\, 1)
(
'
Xp) =
h (««"*)
+
^ l ^ W
As
v^ xP{mom\)c(mQm\)^—
L(s,nxXp) = ^