-------.
(a_[mu] - x)[lambda] x^[lambda]

c_{[lambda],[mu]}
The constant term is [Sigma][Sigma]-----------------.
a_[mu]^[lambda]

Let a_[nu] be a value of x which makes the fraction infinite. The residue
of

c_{[lambda],[mu]} [gamma]_[lambda]
[Sigma][Sigma][Sigma]------------------------------ + [Sigma]-----------------------------
(a_[mu] - a_[nu].e^x)^[lambda] a_[nu]^[lambda].e^{[lambda]x}

is equal to the residue of

c_{[lambda],[mu]}
[Sigma][Sigma][Sigma]------------------------------,
(a_[mu] - a_[nu].e^x)^[lambda]

and when [nu] = [mu], the residue vanishes, so that we have to consider

c_{[lambda],[mu]}
[Sigma][Sigma]----------------------------------,
a_[mu]^[lambda].(1 - e^x)^[lambda]

and the residue of this is, by the first lemma,

c_{[lambda],[mu]}
- [Sigma][Sigma]-----------------,
a_[mu]^[lambda]

which proves the lemma.

1 f(x)
Take F(x) = --------------------------------- = ----, since the sought
x^n(1 - x^a)(1 - x^b)...(1 - x^l) x^n

number is its constant term.

Let [rho] be a root of unity which makes f(x) infinite when substituted
for x. The function of which we have to take the residue is

[Sigma][rho]^-n.e^nx.f([rho]e^-x)

[rho]^-n.e^nx
= [Sigma]------------------------------------------------------------.
(1 - [rho]^a.e^-ax)(1 - [rho]^b.e^-bx)...(1 - [rho]^l.e^-lx)

We may divide the calculation up into sections by considering separately
that portion of the summation which involves the primitive qth roots of
unity, q being a divisor of one of the numbers a, b, ... l. Thus the qth
_wave_ is

[rho]_q^-n.e^nx
[Sigma]------------------------------------------------------------------,
(1 - [rho]_q^a.e^-ax)(1 - [rho]_q^b.e^-bx)...(1 - [rho]_q^l.e^-lx)

which, putting 1/[rho]_q for [rho]_q and [nu] = 1/2(a + b + ... + l), may
be written

[rho]_q^[nu].e^[nu]x
[Sigma]---------------------------------------------------------------------