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Riemann Sums
iθ
To modify this result to fit our situation, let us write w he ,
h> 0, −π/2 <θ <π/2, and conclude from (4) that 21
∞ ∞

Fàkhe iθ )h − Fàxe iθ )dx h · V θ (F)
k 1 0
(V θ is the variation along the ray of argument θ), so that

∞ ∞
iθ iθ
Fàkw)w − Fàxe )d(xe ) w · V θ (F).
k 1 0
Furthermore, in our case of an analytic F, this integral is actually
independent of θ. (Simply apply Cauchy’s theorem and observe that
at ∞ F falls off like 1 2 ). We also may use the formula V θ (F)
x
iθ

∞ |F (xe )|dx and finally deduce that

0
∞ ∞ ∞
iθ

Fàkw)w − Fàx)dx w |F (xe )|dx.
k 1 0 0
Later on we show that

1 1 e dx 1
∞ −x
− + log √ , à 5)
x
e − 1 x 2 x
0 2π
and right now we may note that the (complicated) function
2 e −xe iθ e −xe iθ
iθ
F (xe ) − −
3 3iθ
2 2iθ
x e 2x e 2xe iθ
1 e xe iθ
− −
2 2iθ
iθ
x e (e xe iθ − 1) xe (e xe iθ − 1) 2
is uniformly bounded by M 2 in any wedge |θ| <c <π/2(m +
(x+1)
M(c)), so that we obtain

∞ −kw
1 1 1 e 1
− + − log √ Mw (6)
k e kw − 1 kw 2 2π
k 1
throughout | arg w| <c <π/2.