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II. The Partition Function
28
and using (26) for negative s gives us
2n−2

2 2s s −s
√
s − √
|K n (s)|≤ (1 − s)e n 1 + √ e n
n

2 2s n−1 2
s
s − √
≤ (1 − s)e n e n
√ 2
2
(1 − s)e 2s +1−(1+s/ n)
or
2
|K n (s)|≤ (1 − s)e 2s +1 for s< 0. à 28)
Thus (27) and (28) give the bound for our integral in (21) of
2
2 π
s −2 s− √
2se 2 6 for s ≥ 0,
and
√
2s(s − 1)e 1+π 2/3s for s< 0.
This bound, integrable over (−∞, ∞), gives us the required
dominated convergence, and the passage to the limit is indeed
justified.
Finally we give the following:

Evaluation of our Integral (5). To achieve this let us first note that
as N →∞ our integral is the limit of the integral

−x
∞ 1 1 e dx
(1 − e −Nx ) − +
x
0 e − 1 x 2 x
(by dominated convergence, e.g.). But this integral can be split into

∞ 1 1 dx ∞ e
−x
(1 − e −Nx ) − + (1 − e −Nx ) dx
x
0 e − 1 x x 0 2x
N x −x −(N+1)x
∞ 1 + x − e 1 ∞ e − e
e −kx 2 dx + dx.
k 1 0 x 2 0 x
Next note that
1 + x − e x 1 (1−t)x
− te dt
x 2 0