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II. The Partition Function
20
π 2 + 1 log(1 − e −w )
6w 2

1 1 1 e −kw

+ − + .
k e kw − 1 kw 2
The form of this series is very suggestive. Indeed we recognize any
1 A(kw)
series A(kw) w as a Riemann sum, approximating
k kw
∞ A(t)
the Riemann integral dt for small positive w. It should come
0 t
as no surprise then, that such series are estimated rather accurately.
So let us review the “Riemann sum story”.
Riemann Sums

Suppose that φ(x) is a positive decreasing function on (0, ∞) and
∞
that h> 0. The Riemann sum φ(kh)h is clearly equal to the
k 1
area of the union of rectangles and so is bounded by the area under
∞ ∞
y φ(x). Hence φ(kh)h ≤ φ(x)dx. On the other hand,
k 1 0

the series ∞ φ(kh)h can be construed as the area of this union of
k 0
these rectangles and, as such, exceeds the area under y φ(x).So
∞
this time we obtain ∞ φ(kh)h ≥ φ(x)dx.
k 0 0
Combining these two inequalities tells us that the Riemann sum
lies within h · φ(0) of the Riemann integral. This is all very nice and
ratheraccuratebutitrefersonlytodecreasingfunctions.However,we
may easily remedy this restriction by subtracting two such functions.
Thereby we obtain
∞ ∞

[φ(kh) − ψ(kh)]h − [φ(x) − ψ(x)] h[φ(0) + ψ(0)].
k 1 0
Calling φ(x) − ψ(x) Fàx) and then observing that φ(0) + ψ(0)
is the total variation V of Fàx) we have the rather general result

∞ ∞

F(kh)h − Fàx) h · V(F). (4)
k 1 0
To be sure, we have proven this result only for real functions but
in fact it follows for complex ones, by merely applying it to the real
and imaginary parts.