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For the second integral,
The Coefficients of q(n) 25

1 Fàz) − φ(z) 1 1
dz · 2exp · 2πØ
2πi B z n+1 2πØ n+1 1 − r

2 1
exp .
r n 1 − r
And this is even smaller than our previous estimate. So combining
the two gives, by (15),

(1 − r) 5/2 π 2 1
p(n) − q(n) M exp . à 17)
r n 6 1 − r

But what is r? Answer: anything we please (as long as 0 <Ø <
1)! We are masters of the choice, and so we attempt to minimize
the right-hand side. The exact minimum is too complicated but the
2
1
approximate one occurs when n(r−1) exp π 1 is minimized and
e 6 1−r
π
this occurs when π 2 1 n(1 − r), i.e., r 1 − √ .Sowe
6 1−r 6n
choose this r and, by so doing, we obtain, from (17), the bound

√
p(n) q(n) + O n −5/4 π n/6 . à 18)
e

The Coefficients of q(n)

The elementary function φ(z) has a rather pleasant definite integral
representation which will then lead to a handy expression for the
q(n).
If we simply begin with the well-known identity

∞ 2 √
e −t dt π
−∞
and make a linear change of variables (a > 0),

√

∞ 2 π
e −(at−bð dt ,
a
−∞